Chemistry For Changing Times, 14th Edition John W. Hill Solution Manual

Type:

Solution Manual

Resource:

Chemistry for Changing Times

Edition:

14th Edition

Author(s):

John W. Hill Terry W. McCreary

Chapter 1: Chemistry Overview Chapter 1 introduces the student to the concepts of science, the ability of science to solve problems as well as its limitations. The student is then introduced to the concept of matter and energy and the physical means by which they are measured.

Lecture Outline 1.1

1.2 1.3 1.4 1.5

1.6

Science and Technology: The Roots of Knowledge Many students are confused about the differences between science and technology.

Technology is the direct application of knowledge to solve problems. Science seeks an understanding of underlying principles.

Science: Reproducible, Testable, Tentative, Predictive, and Explanatory The scientific method, from hypothesis to law, theory, or model, is best explained by example.

Science and Technology: Risks and Benefits Solving Society’s Problems: Scientific Research Research can be basic or applied.

Chemistry: A Study of Matter and Its Changes Matter is anything that has mass.

Mass is a measure of the amount of matter present.

Weight is mass times gravitational attraction.

Matter exhibits chemical and physical properties.

Chemical properties tell us how matter will combine to form new and different substances (a chemical change).

Physical properties are directly observable; color, state (solid, liquid, or gas), and texture are examples.

An excellent demonstration is to burn a candle (chemical change) and boil water (physical change).

Classification of Matter A. States of Matter • Solids, liquids, and gases.

B. Substances and Mixtures • Substances have constant composition.

• The composition of a mixture is variable.  Homogeneous mixtures: appear the same throughout (milk, paint, and saltwater are examples).  Heterogeneous mixtures: appear different throughout (pizza, raisin bread, and chocolate chip cookies are examples).

C. Elements and Compounds • Substances are either elements or compounds.  Elements: fundamental building blocks of all matter. (lead, silver, gold, carbon, and oxygen).  Compounds: two or more elements chemically combined in fixed ratios (water, ammonia, and propane).

D. Atoms and Molecules 1 .

Chapter 1: Chemistry

1.7

1.8 1.9

1.10

The Measurement of Matter Most students have been introduced to the metric system during their K12 education; the SI system used in science is based on the metric system. The difference lies in the base units.

• Base units: kilograms (kg) for mass, meter (m) for length, and seconds (s) for time. The four other base units are shown in Table 1.4.

Density Density is the mass-to-volume ratio of matter.

Energy: Heat and Temperature The SI unit of temperature is the kelvin (K); however, the temperature scale used in the chemistry laboratory is normally the Celsius scale. The Celsius scale was developed with the freezing and boiling points of pure water at a pressure of 1 atmosphere as the reference frames. (Students find the history of the development of the Fahrenheit scale much more interesting!)

Critical Thinking

Demonstrations 1. Place samples of various elements (copper, sulfur, zinc, mercury, aluminum, carbon, etc.) in small stoppered bottles or flasks that can be passed around.

2. Compare a yardstick and a meter stick. A meter is slightly bigger than a yard.

A sugar cube is approximately 1 mL. An ordinary paper clip weighs about 1 gram.

A liter is a little bit (6%) larger than a quart.

3. Use Crispix and Raisin Bran to illustrate the difference between a compound and a mixture. Raisins and bran flakes are easily separated, and the ratio of raisins to flakes can vary. Crispix has a 1:1 ratio of corn and rice flakes, and they cannot be easily separated.

4. Ice floats on water because the density of ice is less than that of water. But alcohol has a much lower density. If you use alcohol instead of water, the ice will sink. Place a glass of water and a glass of alcohol side by side and add an ice cube to each one.

5. The “cartesian diver” is a popular demonstration about density. Completely fill a 2-liter plastic bottle with water and put just enough water into a medicine dropper so that it can barely float. Put the dropper into the bottle and cap the bottle. Squeeze and release the bottle to make the “diver” go up and down.

Review Questions 1. Science is testable, reproducible, explanatory, predictive, and tentative.

2 .

Chapter 1: Chemistry

2. A hypothesis is a tentative explanation and must be verified or rejected through experiment.

3. The scientific method can only be used when all the variables in a system can be controlled.

4. Technology is the sum total of the processes by which humans modify the materials of nature; technology need not be based on science.

5. Risk-benefit analysis is an analysis of benefits versus risks, involving an attempt to calculate a desirability quotient (DQ).

6. (a) A benefit is anything that promotes well-being or has a positive effect. (b) A risk is any hazard that leads to loss or injury. A risk-benefit analysis is weighing both the risk and the benefit of a certain technology and determining if the hazard is worth the positive outcome. This tends to be a personal decision and can vary from group to group and person to person.

7. A DQ is benefits divided by risk, also known as a desirability quotient. A large DQ is the result of large benefits and small risks. While scientific investigation can help considerably in determining an accurate DQ, the calculation of benefits is almost entirely a social judgment and the risks and benefits are not always known. Benefits and risks are therefore judgments with a qualitative rather than quantitative aspect.

8. The common units used in the laboratory are (a) the gram (g) for mass and (b) the centimeter (cm) or millimeter (mm) for length.

9. The SI unit of volume is the cubic meter (m3). The volume unit used in the laboratory is the cubic decimeter (dm3) or the cubic centimeter (cm3 or cc).

10.

Prefix Tera Mega Centi Micro Milli Deci Kilo Nano

Symbol T M c μ m d k n

Definition 1012 106 10−2 10−6 10−3 10−1 103 10−9

11. Research projects “a” and “b” are applied research. Research project “c” is basic research.

12. Research project “a” is applied research. Research projects “b” and “c” are basic research.

Problems 13. Penicillin has saved thousands of lives, causing harm to a very few. The use of penicillin for society as a whole has been very beneficial; it has a large DQ. 3 .

Chapter 1: Chemistry

14. The risk of using sodium sulfite is greater in fruit juices than in wines as small children may be sensitive to sulfites and are more likely to consume fruit juice.

15. Hazards are greater than benefits for a person who is exposed daily to the paints than the hobbyist exposed once. The DQ would be increased for the professional painter and the hobbyist if both wore breathing masks capable of filtering out the isocyanate.

16. The risk is much higher for the dentist who stays in the room during every X-ray than for the patient receiving one X-ray per year.

17. The DQ for the use of antibiotics in treating common sore throats is low as there is a small risk to the untreated patient. The DQ for treating more serous influenza is higher as the risk to the untreated patient is greater.

18. There is no benefit for the man or the pregnant woman. There is a significant benefit to the unborn child whose mother has AIDS. 19. 100 g, 2 kg 1" = 23.4" 25.4 mm 20. A2 sized paper is larger. 1” = 25.4 mm thus: 1" 420 mm × = 16.5" 25.4 mm 594 mm ×

21. Both a and b are correct. (c) would be more in the range of 100 kg and (d) would be in the range of 10 kg.

22. No. Weight is mass times the force of gravity! The sample weighed on the moon would have 1/6 th the weight of the sample weighed on Earth.

23. 250 mL 24. 40 cm2 25. Use the following conversions: 1 km = 0.62 mi, (1 km)3 = (0.62 mi)3 = 0.24 mi3 1 km3 3.5 ×108 mi3 × = 1.5 ×109 km3 3 0.24 mi 26. Use the following conversions: 1 km = 0.62 mi, (1 km)2 = (0.62 mi)2 = 0.38 mi2 1 km 2 1.4 × 108 mi 2 × = 3.7 × 108 km 2 2 0.38 mi 27. Use the following conversion: 1” = 25.4 mm, The aluminum tube will fit.

1" 26.3 mm × = 1.03" 25.4 mm 4 .

Chapter 1: Chemistry

28. 18.7 mm 29. (a) Physical

(b) Chemical

(c) Physical

(d) Chemical

30. (a) Physical

(b) Physical

(c) Chemical

(d) Chemical

31. (a) Physical

(b) Chemical

(c) Physical

32. (a) Chemical

(b) Physical

(c) Physical

33. Substances: a, c, d; mixtures: b 34. Substances: a, b; mixtures: c, d, e, f 35. Homogeneous: a, b, c; heterogeneous: d 36. Homogeneous: a, d; heterogeneous: b, c 37. A substance has a definite, or fixed, composition that does not vary from one sample to another.

Glucose samples, regardless of where they are collected, are 8 parts oxygen, 6 parts carbon, and 1 part hydrogen. Therefore, glucose is a substance.

38. Shampoo contains many substances such as water and soap; therefore it is a mixture. “Nothing artificial” reflects the sources of the compounds in the shampoo.

39. Elements: a, b; compounds: c, d 40. Elements: a, c; compounds: b, d 41. (a) Iron

(b) Magnesium

(c) Helium

(d) Nitrogen

42. (a) O

(b) P

(c) K

(d) Ar

43. f 44. d 45. (a) 1.0 dL

(b) 8.5 pg

(c) 1.05 mm

46. (a) 45 mg = 0.000045 kg = 4.5 x 10-5 kg (b) 125 ns = 0.000000125 s = 1.25 x 10-7 s (c) 10.7 µL = 0.0000107 L = 1.07 x 10-5 L (d) 12.5346 kg

5 .

Chapter 1: Chemistry

47. (a) 1000 mL = 1 L, 5.52 × 104 mL ×

1L = 552 L 1000 mL

1g = 0.325 g 1000 mg 1m = 0.27 m (c) 100 cm = 1 m, 27 cm × 100 cm 1 cm = 2.7 cm (d) 10 mm = 1 cm, 27 mm × 10 mm 1 ms = 0.078 ms (e) 1 ms = 1000 µs, 78 µs × 1000 µs (b) 1000 mg = 1 g 325 mg ×

1m = 0.546 m 1000 mm 1 µs = 0.065 µs (b) 1000 ns = 1 µs , 65 ns × 1000 ns (c) 1000 mg = 1 g, 1000 g = 1 k, 1 kg 87.6 mg × = 0.0000876 mg = 8.76 × 10-5 kg 6 1 x 10 mg 1L = 46.3 L (d) 1 dm3 = 1L, 46.3 dm3 × 1 dm3 1 µm = 0.000181 µm = 1.81 × 10-4µm (e) 1 × 106 pm = 1µm, 181 pm × 6 1 x 10 pm

48. (a) 1000 mm = 1 m, 546 mm ×

49. Larger units: (a) cm - centimeters

(b) kg - kilograms

50. 7 x 1027 atoms 51. 31 cm = 310 mm = 0.31 m = 3.1 x 108 nm 52. (a) 352 mL = 0.352 L (b) 26 kL = 26,000 L 53. D =

m 500.0 g = 9.33 g 3 ; D= 3 cm V 53.6 cm

54. D =

m 22.4 g = 0.896 g ; D= mL 25.0 mL V

55. D =

m ; D × V = m ; 1.51 g × 49.1 mL = 74.1 g mL V

56. D =

m ; D × V = m ; 1.43 g 3 × 13.2 cm 3 = 18.9 g cm V

6 .

(c) dL - deciliters

Chapter 1: Chemistry

m m ;V = D V 227 g (a) V = = 343 mL 0.660 g mL 454 g (b) V = = 495 cm 3 0.917 g 3 cm Note: Refer to Table 1.6 for the density data.

57. D =

m m , V= D V 475g (a) V = = 53.1 cm 3 g 8.94 cm 3 253 g (b) V = = 18.7 mL 13.534 g mL

58. D =

59. Three layers will form with mercury on the bottom (greatest density), then water, followed by hexane on the top.

60. The red maple and the balsa wood will float on the hexane, the ice will float on the water, the copper coin will float on the mercury, and the gold coin will sink to the bottom.

61. D =

m , V

1.03 g

D× V = m mL

× 37900 mL = 39037 g ×

1 lb = 86.1 lbs 453.6 g

86.1 lbs + 59.5 lbs = 145.6 lbs The answer is “yes.”

62. D =

m , V

D× V = m

3.785 L 1000 mL = 53.0L x = 53, 000 mL 1 gal 1L 1 kg 0.758 g × 53000 mL = 40200 g × = 40.2 kg mL 1000 g

14.0 gal x

63. Solve the equation for the volume of the crystal ball. The diameter of the crystal ball is 14.8 cm, thus the radius (r) is 7.4 cm.

7 .

Chapter 1: Chemistry

4π r3 4π x (7.4 cm)3 = = 1.70 x 103 cm3 3 3 D×V = m V =

3.18 g

cm3

× 1.70 x 103 cm 3 = 5.41 x 103 g

64. Solve for the total volume required. D =

m , V

V=

m D

500g = 485 mL = 485 cm 3 g 1.031 mL Solve for height (h) of carton.

7.7 cm x 7.7 cm x h cm = 485 cm3 59 cm2 x h cm = 485 cm3 h cm = 485 cm3/ 59 cm2 = 8.2 cm = 82 mm V=

65. ºC = K – 273,

ºC = 77 – 273 = –196 ºC

66. K = ºC + 273,

K = 37 + 273 = 310 K

67. 1 cal = 4.184 J, 1 kcal = 4.184 kJ,

161 kJ ×

68. 1000 cal = 1 kcal, 584 cal = 0.584 kcal

1 kcal = 38.5 kcal 4.184 kJ

0.584 kcal ×

4.184 kJ = 2.44 kJ 1 kcal

69. 1 microcentury = 1 x 10-6 centuries = 1x10-4 years x 365 days/year = 0.0365 days x 24 hours/day = 0.876 hours x 60 minutes/hour = 52.6 minutes.

70. 1 millihelen = 1.0 x 10-3 or 0.001 helens. 1000 ships/helen x 0.001 hellens = 1 ship 71. law or an empirical law 72. A food calorie is 1000 cal or 1 kcal. Thus it takes 79 kcal to melt 1 kg of ice.

73. 200 kilowarhols = 200,000 warhols x 15 min/warhol = 3,000,000 min/60 min/hr = 50,000 hr/24 hrs/day = 2083.3 days/365 days/year = 6 years 74. (1) An experiment

(2) A scientific law

(3) A hypothesis

75. (1) An observation (4) An experiment

(2) A theory (5) An observation

(3) A hypothesis

(4) A theory

76. Subtract the weight of the empty glass container from the weight of the glass container plus the weight of the antifreeze to yield the weight of the antifreeze; then calculate the density of the antifreeze, 8 .

Chapter 1: Chemistry

60.562 g – 48.462 g = 12.100 g antifreeze m 12.100 g D= = = 1.51 g mL V 8.00 mL 77. Calculate the volume of the block, then calculate the mass.

7.6 cm x 7.6 cm x 94 cm = 5400 cm3 m D= , D× V = m V 0.11 g 3 × 5400 cm 3 = 590 g cm 78. yardstick (36 in. = 0.91 m), rattlesnake (1.04 m), 1.21 m chain, 75-in. board (1.9 m) 79. 1.65 kg cabbage, 5 lb bag of potatoes (2.3 kg), 2500 g sugar (2.500 kg) 80. 1.53 m = 5’, 38.5 kg = 104 lb; the woman on the left 81. Find the mass of the chips only by subtracting the paper mass from the total mass.

18.43 g – 1.21 g = 17.22 g m Calculate density using the equation: D = V 17.22 g D= = 5.23 g 3 3 cm 3.29 cm

m , solve for volume.

V m 3180 g V= , V= = 4.0 x 10 2 cm 3 D 7.9 g/cm 3

82. D =

m , solve for volume.

V m 0.00579 g V= , V= = 0.00030 cm3 3 19.3 g/cm D Use length (cm) x width (cm) x thickness (cm) = volume (cm3).

44.6 cm2 x thickness (cm) = 0.0003 cm3 0.0003 cm3 Thickness (cm) = = 6.73 x 10-6 cm 2 44.6 cm 84. Calculate the volume of each stainless steel rod: π x (1.27 cm)2 x 610 cm = 3090 cm3

83. Use D =

9 .

Chapter 1: Chemistry

Calculate the mass of each steel rod: 3 1 kg = 23100 g = 23.1 kg DV = m = 7.48 g 3 x 3090 cm x steel rod steel rod 1000 g steel rod cm 1850 kg = 80 steel rods 23.1 kg steel rod

85. V = 36.1 cm x 36.1 cm x 36.1 cm; V = 47,000 cm3 m = 19.3 g/cm3 x 47,000 cm3; m = 907,000 g 907,000 g x 1 kg/1000 kg = 907 kg 907 kg x 1 metric ton/1000 g = 0.907 metric tons 86. D =

m 425 g = = 8.85 g 3 this is silver 3 cm V 48.0 cm

87. Calculate the volume of each planet.

3 4π (70,000 km ) VJupiter = = 1.44x1015 km 3 3 4π (6400 km ) = 1.10x1012 km 3 3 3 4π (58,200 km ) VSaturn = = 8.26x1014 km 3 3 3

VEarth =

Calculate the density of each planet.

D Jupiter = D Earth =

1.9x10 27 kg g x km 3 = 1.32x1012 kg 3 x1000 15 3 kg km 1.44x10 km 1x1015 cm 3 = 1.3 g

5.98x10 24 kg g x km 3 = 5.44x1012 kg 3 x1000 12 3 kg km 1.10x10 km 1x1015 cm 3 = 5.4 g

D Saturn =

cm 3 cm 3

5.68x10 26 kg g x km 3 = 6.87x1011 kg 3 x1000 14 3 kg km 8.26x10 km 1x1015 cm 3 = 0.69 g

cm 3

Saturn would float on water!

4π (96,600 km ) = 3.78x1015 km 3 3 9.50x10 26 kg g x km 3 D HAT-P-1 = = 2.51x1011 kg 3 x1000 15 3 kg km 3.78x10 km 1x1015 cm 3 = 0.251 g 3

88. VHAT-P-1 =

10 .

cm 3

Chapter 1: Chemistry

89. Green chemistry uses materials and processes that are intended to prevent or reduce pollution at its source and to meet the needs of the present generation without compromising the needs of future generations.

90. The Twelve Principles of Green Chemistry 1) Prevention 2) Atom Economy 3) Less Hazardous Chemical Syntheses 4) Designing Safer Chemicals 5) Safer Solvents and Auxiliaries 6) Design for Energy Efficiency 7) Use of Renewable Feedstocks 8) Reduce Derivatives 9) Catalysis 10) Design for Degradation 11) Realtime Analysis for Pollution Prevention 12) Inherently Safer Chemistry for Accident Prevention 91. a 92. d

11 .

Chapter 2: Atoms Overview Chapter 2 begins with the Greek concept of matter and moves chronologically forward through the development of the atomistic model of matter.

Lecture Outline 2.1

2.2

2.3 2.4

2.5

2.5

Atoms: Ideas from the Ancient Greeks The Greek philosophers had two main ideas regarding matter: the atomistic view (Leucippus and Democritus) the continuous view (Aristotle) Scientific Laws: Conservation of Mass and Definite Proportions Lavoisier was the first great experimentalist. More than anything, he introduced experimental chemistry to Western Civilization. Through his experiments he formulated the law of conservation of mass, which states: Matter is neither created nor destroyed during a chemical reaction.

John Dalton and the Atomic Theory of Matter John Dalton summarized the work of Lavoisier and Proust to form the famous atomic theory.

The Mole and Molar Mass The mole and Avogadro’s number Molar mass, atomic mass Mass and atom ratios Moles Mendeleev and the Periodic Table The periodic table: Mendeleev and Meyer John Dalton established relative atomic masses using hydrogen as a base.

Mendeleev arranged a table of elements according to increasing atomic weights placing elements with similar properties in the same column. This was the beginning of the modern periodic table!

Atoms and Molecules: Real and Relevant

Demonstrations 1. Electrolysis of water.

2. Fill three flasks with zinc (gray powder), sulfur (yellow powder), and zinc sulfide (white powder), and show them to the class to illustrate how the properties of a compound differ from those of the component elements. [Or use mercury (silver liquid), oxygen (colorless gas), and mercuric oxide (red powder).]

3. Use samples of metals and nonmetals to demonstrate the difference in their properties. (Aluminum foil, copper wire, nickel coin, zinc strip; capped bottle of liquid bromine, balloon filled with hydrogen, bottles of carbon, sulfur, etc.)

12 .

Chapter 2: Atoms

4. Using the largest nuts and bolts you can find, start with unequal numbers of nuts and bolts (e.g., 5 and 7). Connect the pairs of nuts and bolts, with some remaining unpaired, to illustrate the law of definite proportions.

5. Heat a sample of HgO in a large test tube with a burner. Mercury droplets form on the inside of the tube. Test for oxygen with a glowing splint, which bursts into flame when inserted into the test tube.

Review Questions 1. (a) The atomic view of matter assumes that matter is made up of small unit particles that cannot be further subdivided and still be the same kind of matter, while the continuous view of matter is just that— continuous throughout with no smaller particle. (b) The Greeks believed that there were four basic elements: earth, air, fire, and water, while today we recognize the existence of elements that are pure chemical substances made of a single atom.

2. Democritus described the tiny ultimate particle as “atomos” (which means “cannot be cut”). Matter appears continuous, and without experimental chemistry there is no way to prove that matter is not continuous. The discoveries are summarized in the laws of definite proportions and multiple proportions.

3. discrete: a, c, e; continuous: b, d 4. Robert Boyle, in The Sceptical Chymist (1661), stated that if a substance could be broken down into simpler substances, it was not an element.

5. A compound contains elements in certain definite proportions and in no other combinations. All samples of water will have 2 atoms of hydrogen and 1 atom of oxygen. The ratio of atoms yields a constant 2:1 volume ratio.

6. Lavoisier is the “father” of modern chemistry. He was the first to carry out a series of experiments in which the masses of reactants and products were carefully recorded. These experiments led Lavoisier to formulate the law of conservation of mass.

7. The law of definite proportions.

8. (a) Dalton’s atomic theory states: “No atoms are created or broken apart in a chemical reaction.”

This is a restatement of the law of conservation of mass. When 3 g of carbon is allowed to react with 8 g of oxygen, 11 g of carbon dioxide is formed. (b) One of the statements of Dalton’s atomic theory is that “compounds are formed when atoms of different elements combine in fixed proportions.” This is a restatement of the law of definite proportions. (c) One statement of Dalton’s atomic theory is that “compounds are formed when atoms of different elements combine in fixed proportions.” This is a restatement of the law of definite proportions but may also explain the law of multiple proportions.

9. The law of multiple proportions.

13 .

Chapter 2: Atoms

10. Elements can combine in more than one set of proportions. (They will still of course combine in small whole-number ratios.) (a) The ratio of the oxygen in ClO2 to oxygen in ClO, the second compound, is 2:1. (b) The ratio of fluorine ClF3 to fluorine in ClF 3 parts to 1 part by mass. (c) The ratio of oxygen in P4O6 to oxygen in P4O10 is 6:10, which reduces to 3:5 by mass.

11. Rectangle C cannot represent the mixture after reaction. Rectangle C contains 15 oxygen atoms; there are 14 oxygen atoms in the initial mixture. The law of conservation of mass has been violated.

12. (a) Avogadro’s number represents the number of particles contained in one mole. (b) 6.02 x 1023 oxygen atoms. (c) 6.02 x 1023 molecules (d) 28.02 g.

13. Dmitri Mendeleev linked the mass of elements to chemical properties and created periodic law and thus the first form of the periodic table.

14. Oersted’s experiment proved that aluminum could not be broken down further; it was an element.

Problems 15. (a) The container would weigh the same regardless of how much time passed. (b) No. The mouse would exchange gases with the surroundings.

16. The iron atoms have not been destroyed; they are in solution.

17. c. Carbon dioxide gas will be generated and released as the anti-acid pill dissolves.

18. (a) The carbon, which forms the diamond, has been converted to carbon dioxide and escaped into the atmosphere. (b) The carbon in the wood is converted to carbon dioxide, but the residual ash is not.

Burning the diamond in an enclosed environment would trap the carbon dioxide for further analysis.

19.

2.02 g H x 78.5 g C = 6.60 g H 24.02 g C

20.

44.01 g nitrous oxide x 48.7 g nitrogen = 76.5 g nitrous oxide 28.01 g nitrogen

21.

36.03 g C x 88.20 g C3 H 8 = 72.06 g C 44.10 g C3 H 8 12.01 g C x 300.06 g CO 2 = 81.88 g C 44.01 g CO2 It takes 81.88 g of carbon to form 300.06 g of carbon dioxide, and you only have 72.06 g of carbon available from 88.20 g of propane. The answer violates the law of conservation of mass.

14 .

Chapter 2: Atoms

22. Mass before the reaction = 2.796 g of zinc + 2.414 g of sulfur = 5.210 g Mass after the reaction = 4.169 g of zinc sulfide + 1.041 g of sulfur = 5.210 g Yes, the experiment obeys the law of conservation of mass.

23. Mass before the reaction = 1.00 g of zinc + 0.80 g of sulfur = 1.80 g Mass after the reaction = 1.50 g zinc sulfide Total mass before 1.80 g – mass after 1.50 = 0.30 g of sulfur (answer b) 24. None of the three methods will destroy the atoms. Incineration is a chemical change.

25. (a)

(b)

26.

2.02 g hydrogen x 775 g water = 86.9 g hydrogen 18.02 g water 16.00 g oxygen x 775 g water = 688 g oxygen 18.02 g water

2.02 g hydrogen x ? kg water = 125 kg hydrogen 18.02 g water 18.02 g hydrogen ? kg water = 125 kg hydrogen x = 1120 kg water 2.02 g water

27.

3.0 parts carbon x 14 kg carbon dioxide = 3.8 kg carbon 3.0 + 8.0 parts carbon dioxide

28.

3.0 + 8.0 parts carbon dioxide x 0.4 g diamond = 2 g carbon dioxide 3.0 parts carbon

29. Statement number two of Dalton’s atomic theory says, “All atoms of a given element are alike, but atoms of one element differ from atoms of any other element.” (a) The different masses of atoms of calcium and vanadium support Dalton’s atomic theory. (b) The same mass of two different elements violates statement two of Dalton’s atomic theory. (Dalton did not predict the existence of isotopes.)

30. Yes, it agrees with statement number two of Dalton’s atomic theory, which says, “All atoms of a given element are alike, but atoms of one element differ from atoms of any other element.”

31. The atoms are neither being created nor destroyed; they are being rearranged.

32. a 33. d;

8 units O 2 atoms H x = 16 1 units H 1 atom O

15 .

Chapter 2: Atoms

34. a;

57 unit F 1 atom N 19 x = by mass 14 units N 3 atoms F 14

35. Assume that there is 1 g of carbon suboxide and that the mass ratio is 0.5296:0.4704 Divide both by 0.5296 to get a mass ratio of 1.0000:0.8882 in carbon suboxide. The ratio of oxygen in all three compounds is 1:2:3.

36. 8.0:85.5 37. Calculate the mass ratio of Sn to O in SnO. 0.742 g Sn / 0.100 g O = 7.4:1.

Calculate the mass ratio of Sn to O in the unknown. 0.555 g Sn / 0.150 g O = 3.7:1 or 7.4:2. Compare the oxygen mass ratios of the two oxides. 1:2; thus the second oxide has 2 oxygen atoms for every oxygen atom in the first oxide. The formula of the second oxide is SnO2 .

38. Divide all oxygen and nitrogen ratios by the smallest oxygen to nitrogen ratio to obtain X 4.00:1.75, Y 2.00:1.75, and Z 1.00:1.75.

39. Calculate the ratio of Fe in hematite to the ratio of Fe in wusitite: 0.558g/0.558 g =1 Calculate the ratio of O in hematite to the ratio of O in wusitite: 0.240g/0.160 g =1.5 Multiply both by 2 to get whole numbers and you obtain Fe2O3 for hematite.

Calculate the ratio of Fe in magnetite to the ratio of Fe in wusitite: 3.35g/0.558 g =6 Calculate the ratio of O in magnetite to the ratio of O in wusitite: 1.28g/0.160 g =8 Reduce to the smallest whole numbers and you obtain Fe3O4 for magnetite. Yes. 40. Divide both carbon and hydrogen percentages by the carbon percent in each compound to obtain C : H ratios V 1.00:0.25 and W 1.00:0.20. The dividing by the smallest hydrogen amount yields hydrogen ratios of 1.00:1.25.

41. 1.000 g C x

1 mol C 6.02x1023 atoms C = 0.08326 mol C x = 5.01x1022 atoms C 12.01 g C 1 mol C

42. 16.00 g/mol O + 2(1.01g/mol H) = 18.02 g/mol H2O 6.02 x 1023 atoms of O and 2(6.02 x 1023 atoms of H) = 1.20 x 1024 atoms of H 43. (5 mol)(12.00 g/mol) = 60.00 g C-12. (5 mol)(12.01 g/mol) = 60.05 g C. (To one significant figure, the correct number of significant figures. To use, both answers are 60 g.)

44. 1.01/12.00 and 1.01/12.01

16 .

Chapter 2: Atoms

0.937 g C 0.0629 g H × 100% = 93.7% C; × 100% = 6.29% H 1.000 g 1.000 g 0.229 g C 0.0153 g H Sample 2 : × 100% = 93.8% C; × 100% = 6.27% H 0.244 g 0.244 g 0.094 g C 0.0063 g H Sample 3 : × 100% = 94% C; × 100% = 6.3% H 0.100 g 0.100 g All ratios are constant to two significant figures. The solid is a pure compound!

45. Sample 1 :

0.862 g C 0.138 g H × 100% = 86.2% C; × 100% = 13.8% H 1.000 g 1.000 g 1.295 g C 0.254 g H Sample 2 : × 100% = 83.6% C; × 100% = 16.4% H 1.549 g 1.549 g 0.826 g C 0.162 g H Sample 3 : × 100% = 83.6% C; × 100% = 16.4% H 0.988 g 0.988 g Sample 1 does not agree with samples 2 and 3. The liquid is a mixture.

46. Sample1 :

47.

1 part by mass F 238 6 F x = 2.09 part by mass U 19 1 U

48. From the first experiment we note that 3.06 g of hydrogen produced 27.35 g of water when allowed to react with oxygen. By applying the law of conservation of mass, 24.29 g of oxygen is required. Next, calculate the ratio by mass of hydrogen to oxygen in water. 3.06 g hydrogen / 24.29 g oxygen = 1 part hydrogen to 8 parts oxygen. Next, compare the ratio (1:8) with the masses of hydrogen and oxygen produced in the electrolysis experiment: 1.45 g hydrogen / 11.51 g oxygen = 1 part hydrogen to 8 parts oxygen. The results are consistent with the law of definite proportions.

49. From the first experiment we note that 0.312 g of sulfur produced 0.623 g of sulfur dioxide when allowed to react with oxygen. By applying the law of conservation of mass, 0.311 g of oxygen is required. Next, calculate the ratio by mass of sulfur to oxygen in sulfur dioxide: 0.312 g sulfur/0.311 g oxygen = 1 part sulfur to 1 part oxygen. The 1.305 g of sulfur in the second experiment requires 1.305 g of oxygen (1:1 by mass); thus, 2.610 g of sulfur dioxide is produced in the second experiment.

50.

92.61 g mercury = 0.9261 g mercury/g mercuric oxide 100.00 g mercuric oxide 0.9261 x (X g mercuric oxide) = 100.00 g mercury 100.00 g mercury X g mercuric oxide = 0.9261 g mercury/g mercuric oxide X g mercuric oxide = 108.0 g mercuric oxide

51. Table 2.1 indicates 1 ratio of 1.000 g oxygen:0.4375 g nitrogen. The nitrogen ratio is 0.4375:0.5836 or 1.000:1.334 or in whole numbers 3:4.

17 .

Chapter 2: Atoms

52. When gasoline is burned, it combines with oxygen from the atmosphere to form carbon dioxide.

53. The law of conservation of mass. The mass of the products is the same as the mass of the reactants.

14.01 g N 4.62 g N 28.00 g N 4.62 g N = ; = 3.03 g H 1.00 g H 6.05 g H 1.00 g H The N and H are in the same ratios; thus 34.05 g of NH3 will be produced.

4.62 g N XgN (4.62 g N)(11.0 g H) = ;X= = 50.8 g N ; 61.8 g NH3 produced b) 1.00 g H 11.0 g H 1.00 g H 4.62 g N X g N (4.62 g N)(2.0 g H) = ;X= = 9.2 g N required c) 1.00 g H 2.0 g H 1.00 g H Remaining N = 11.0 g – 9.2 g = 1.8 g N.

54. a)

55. a) Using Mendeleev’s masses for S and Te: (32+125)/2=78.5 b) Using values from the modern periodic table: (32.06+127.60)/2=79.83 56. a) neither b) rare c) hazardous d) neither e) hazardous 57. Titanium dioxide replacing lead in paints, metal salts replacing lead as stabilizers in plastics, and lead-free adhesives used in electronics.

58. Mercury is toxic and can leach from landfills into groundwater.

18 .

Chapter 3: Atomic Structure Overview Chapter 3 examines the experimental evidence leading to modern atomic theory. The chapter opens with the earliest experiments that indicated the electrical nature of matter and concludes with modern quantum theory.

Lecture Outline 3.1 Electricity and the Atom Humphry Davy’s experiments with the voltaic pile led to the electrolysis of molten salts. Michael Faraday refined Davy’s procedures and became the father of electrochemistry. Cathode ray tube experiments conducted by J. J. Thomson allowed him to calculate the mass-to-charge ratio of the electron. Conducting similar experiments, Eugen Goldstein discovered the proton. Millikan’s oil-drop experiment enabled him to calculate the charge of the electron, and thus its mass.

3.2 Serendipity in Science: X-Rays and Radioactivity Roentgen, in 1895, stumbled upon X-rays while conducting cathode ray experiments.

3.3 Three Types of Radioactivity The alpha particle (α) is a helium nucleus.

The beta particle (β) is an electron.

The gamma ray (γ) is a high-energy photon.

3.4 Rutherford’s Experiment: The Nuclear Model of the Atom The gold foil experiment was strong and convincing evidence that led to the theory of the nuclear atom.

3.5 The Atomic Nucleus The nucleus contains protons (+1 charge), neutrons (neutral), and nearly all the mass of the atom. Electrons are outside the nucleus, have a -1 charge, and only a fraction of the mass of the proton and neutron. It is the number of protons that defines an element! Isotopes are atoms of the same element having different numbers of neutrons, and thus varying in mass.

3.6 Electron Arrangement: The Bohr Model Spectroscopic evidence led Niels Bohr to conclude that only certain energy levels, or orbits, were allowed for electrons.

3.7 Electron Arrangement: The Quantum Model Louis de Broglie showed that the electron had wave properties. Erwin Schrodinger used the de Broglie wave theory and the Bohr quantum theory to derive the quantum-mechanical theory of the atom. The quantum mechanical theory describes the positioning of electrons about the nucleus and electron energies.

19 .

Chapter 3: Atomic Structure

3.8 Electron Configurations and the Periodic Table Figure 3.17 is an excellent tool for teaching students electronic configurations using the shorthand spectroscopic notation based on element locations in the periodic table.

Demonstrations 1. Demonstrate the use of a cathode ray tube, using a magnet to deflect the beam of electrons.

2. Models that can be used to demonstrate the Rutherford gold foil experiment are described in Tested Demonstrations, Vol. II, by George Gilbert et al., on pages L-2,3,4.

3. An overhead projector version of the Rutherford experiment is described by Hau in J. Chem. Educ. 1982, p. 973.

Review Questions 1. (a) William Crookes: developed the cathode ray tube (b) Eugen Goldstein: discovered the proton (c) Michael Faraday: developed electrolysis. (d) J. J. Thomson: discovered the electron; calculated the mass-to-charge ratio of the electron (e) Ernest Rutherford: from the results of the “Gold Foil Experiment” proposed the nuclear theory of the atom (f) James Chadwick: discovered the neutron 2. Radioactivity is spontaneous radiation from an atomic nucleus. Dalton’s atomic theory stated that atoms were indestructible.

3. Both X-rays and gamma rays are forms of high-energy electromagnetic radiation. A gamma ray is more energetic than an X-ray.

4. Goldstein’s cathode ray tube had a perforated cathode that revealed positive particles streaming in the opposite direction as the electrons.

5. Rutherford’s nuclear model proposed a small, dense nucleus containing protons and neutrons. The electrons were proposed to be surrounding the small dense nucleus.

6. (a) Deuterium is a hydrogen atom with a neutron in the nucleus. (b) Tritium is a hydrogen atom with two neutrons in the nucleus. (c) A photon is a quantum of electromagnetic radiation.

7. A & B; A & C; C & D are all (different number of protons) B & C and A & D are yes (different number of neutrons) 8. A and B 20 .

Chapter 3: Atomic Structure

9. Dalton looked on atoms as simple spheres and never attempted to define their inner structure. The nuclear model of the atom is a detailed model that is used to explain many atomic properties.

10. Os, osmium, 190.23 11. Ru, ruthenium, 101.07 12. a discrete unit of energy 13. Given identical atoms, an electron moving from the first shell to the third shell absorbs more energy than an electron moving from the second shell to the third shell.

14. Bohr added the concept of concentric electron energy levels. Schrodinger developed the quantummechanical model of the atom.

15. the quantum-mechanical model 16. (a) 4 (b) 11 (c) 16 (d) 34 (e) 29 (f) 92 17. (a) 4 (b) 11 (c) 16 (d) 34 (e) 29 (f) 92 Problems 18. electrons 19. alpha(mass = 4, charge = 2+), beta(mass = 1/1837, charge = 1–), gamma(mass = 0, charge = 0) 20. a. an ion 21. d. spontanesously give off particles or rays 22. 9 e55 23. (a) 37 17 Cl, chlorine - 37; (b) 25 Mn, manganese - 55

24.

Element Nickel Palladium Nitrogen Iodine

Mass Number 60 108 14 127

Protons 28 46 7 53

Neutrons 32 62 7 74

25. four elements 26. three isotopes of element with Z = 47 21 .

Chapter 3: Atomic Structure

27. 28 28. 9 29. 2+ 30. the d subshell 31. (a) 8 (b) 14 (c) 28 32. (a) 9 (b) 19 (c) 30 33. (a) The electron configuration shown is an excited state; the correct ground-state configuration is 1s2. (b) Incorrect, it is not possible to have 7 electrons on a p orbital. (c) The electron configuration shown is an excited state, the correct ground state configuration is 1s22s2. (d) The electron configuration shown is a ground state.

34. (a) The electron configuration shown is an excited state, the correct ground-state configuration is 1s22s22p2. (b) The electron configuration shown is an excited state, the correct ground state configuration is 1s22s22p3. (c) Incorrect, there is no such orbital as a 2d. (d) Incorrect, the 2s orbital can hold a maximum of 2 electrons.

35. neon 36. (a) neon (b) argon (c) neon 37. Si has 2 3p electrons; Ge has 2 4p electrons; germanium is a larger atom.

38. F and Cl have 5 electrons in the outermost p subshell; F has 5 electrons in the 2p; Cl has 5 electrons in the 3p. Br and I are both expected to have 5 electrons in the outermost p subshell.

39. (a) metal (b) metal (c) metal (d) nonmetal 40. Cs and K 41. Ne and Kr 42. Fe and Mo 43. 4; Ne, P, Kr, N 44. Ra is radioactive; yes 22 .

Chapter 3: Atomic Structure

45. (a) 15 (b) phosphorus (c) 15 (d) 6 (e) 0 46. The atomic number and the number of protons and electrons are the same for a neutral atom. The number of protons and neutrons added yield the mass number.

47. b c

d

a

48. Na emitting at 5890 and 5900 and Ba emitting at 5520 and 5540; Ba 49. Fe 1s22s22p63s23p64s23d6 Sn 1s22s22p63s23p64s23d10 4p65s24d105p2 2 2 6 2 6 2 10 6 2 Pb 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d105p66s24f145d106p2 50. Mg and Al 51. Sr and Y 52. Ca and Sr 53. b 54. c 55. d 56. The experiment indicated that the O2 came from water not CO2, refuting the original hypothesis.

23 .

Chapter 4: Chemical Bonds Overview Chemical bonds determine the properties of matter. Chapter 4 examines ionic and covalent bonds, the symbolism used to represent bonds, and the resulting shape and properties of molecules and compounds that result from different types of bonding.

Lecture Outline 4.1

4.2 4.3

4.4

4.5

4.6

The Art of Deduction: Stable Electron Configurations The most stable electron configurations are those of the noble gases.

Other elements that can alter their electron structures to become like those of noble gases would become less reactive by doing so.

Formation of isoelectronic ions Lewis (Electron-Dot) Symbols Use the periodic table to determine the number of valence electrons.

Lewis dot symbols use “dots” to represent valence electrons.

The Reaction of Sodium and Chlorine Facts: Sodium reacts with chlorine to form sodium chloride (salt) according to the chemical equation: 2 Na + Cl2  2 NaCl (Na+ and Cl ions) Theory: Chlorine takes an electron from sodium, forming the chloride and sodium ions.

The sodium and chloride ions are pulled together by coulombic forces (attraction of positive and negative, the electromagnetic force), forming what is called an ionic bond.

Although the names of atoms and ions are similar, the atom and the ion are quite different chemically, so it is important that we make distinctions and use precise terminology.

Using Lewis Symbols for Ionic Compounds Active metal—active nonmetal compounds form ionic bonds.

Generally speaking, metallic elements in Groups 1A and 2A react with nonmetallic elements in Groups 6A and 7A to form ionic compounds. These are stable crystalline solids.

Atoms gain or lose electrons to have an octet of electrons, similar to the noble gas configurations.

Formulas and Names of Binary Ionic Compounds Learn to write the names and formulas for binary ionic compounds.

Generally, the crossover method can be used to get the correct formula for a binary compound. Write each ion with its charge (positive ion to the left), then transpose the charge numbers (but not the plus and minus signs), and write them as subscripts.

Covalent Bonds: Shared Electron Pairs Generally nonmetal–nonmetal compounds form covalent bonds.

The bond formed by a shared pair of electrons is called a covalent bond.

Formation of double and triple bonds maintains octets.

The systematic names of compounds are discussed.

24 .

Chapter 4: Chemical Bonds

4.7

4.8

4.9 4.10

4.11 4.12

Unequal Sharing: Polar Covalent Bonds Difference of covalent and polar covalent In H—Cl, the chlorine atoms have a greater attraction for a shared pair of electrons than do hydrogen atoms; chlorine is said to be more electronegative than hydrogen.

Electronegativity increases as you move toward fluorine.

Polyatomic Molecules: Water, Ammonia, and Methane In general, many nonmetals often form a number of covalent bonds equal to eight minus the group number.

Hydrogen forms one bond, oxygen forms two bonds (with two lone pairs), nitrogen forms three bonds (with one lone pair), carbon forms four bonds (with no lone pairs).

Polyatomic Ions Table 4.4 lists the common polyatomic ions to be discussed Rules for Writing Lewis Formulas The five steps for writing electron dot formulas appear in this section.

An atom or molecule with unpaired electrons is called a free radical.

Examples are ClO2, NO, NO2.

Molecular Shapes: The VSEPR Theory The common bonding and shapes of molecules appear in this section.

Table 4.6 lists bonding and the shapes of molecules.

Shapes and Properties: Polar and Nonpolar Molecules H—Cl, water, ammonia, and methane are revisited.

Green chemistry and chemical bonds Learning the chemical vocabulary

Demonstrations 1. Use a conductivity apparatus (with light bulb or bell) to test various ionic and covalent compounds.

Answers to Review Questions 1. Sodium metal is a soft, shiny metal that is quite reactive (reacts readily with water); sodium ions are quite stable; NaCl is a white crystalline solid, soluble in water.

2. Potassium metal has 1 valance electron; by losing just one electron it will have a full shell and an octet and become the potassium ion. The potassium ion has a full shell and an octet and is quiet stable.

3. Chlorine atoms are single atoms of chlorine with 7 valence electrons; chlorine molecules are two chlorine atoms held together by a covalent bond; chloride ions have an additional electron, giving them a –1 charge, 8 valence electrons. Chlorine atoms are the most reactive, followed by chlorine molecules, and finally chloride ions, which are quite stable.

4. Duet = 2 and Octet = 8 5. (a) 1+

(b) 2–

(c) 3–

(d) 2+ 25

.

Chapter 4: Chemical Bonds

6. Group 1A: loses one electron to have a complete octet Group 6A: 2 electrons added to complete its octet Group 5A: 3 electrons added to complete its octet Group 2A: loses two electrons to have a complete octet 7. (a) Group 6A

(b) Group 3A

(c) Group 1A

(d) Group 2A

8. (a) Hydrogen forms only one bond. (b) Chlorine generally forms one bond. (c) Sulfur generally forms two bonds. (d) Fluorine generally forms one bond. (e) Nitrogen generally forms three bonds. (f) Phosphorus generally forms three bonds.

9. C, N, O (Oxygen would have a formal charge of.

Answers to Problems 10.

11.

12.

13. (a) (b) (c) (d) 14. (a) (b) (c) (d) 26 .

C

O

Chapter 4: Chemical Bonds

15. (a) True, O2- has 8 electrons in its Lewis formula. (b) False, Mg+2 has zero electrons around it in its Lewis formula. (c) True, it has no electrons around it in its Lewis formula. (d) True, they both have 18 electrons, and no electrons in their Lewis formula.

16. a. Nitride ion is d. N3c. Cuprous ion is f. Cu+ e. Ferrous ion is b. Fe2+ g. Silver ion is h. Ag+ 17. (a) S2(d) F-

(b) Potassium ion (e) Calcium ion

(c) Bromide ion (f) Fe3+

18. (a) Chromium(II) ion

(b) Chromium(III) ion

19. (a) Molybnium(IV) ion

(b) Molybnium(VI) ion

20. (a) V2+

(b) Ti2+

(c) Ti4+

21. (a) Mn2+

(b) Mn3+

(c) Mn7+

(c) Chromium(VI) ion

22. “ic” is used for the more positively charged ion.

23. a. Iron(II) chloride is d. FeCl2 c. Silver fluoride is h. AgF e. Iron(III) bromide is b. FeBr3 g. Sodium oxide is f. Na2O 24. (a) Lithium fluoride (d) AgI

(b) CaCl2 (e) Copper(II) oxide

(c) Magnesium sulfide (f) Cu2S

25. Chromium(III): Cr2O3, chromium(III) oxide Chromium(VI): CrO3, chromium(VI) oxide 26. Sodium and fluorine: NaF, sodium fluoride Tin(II) and fluorine: SnF2, tin(II) fluoride 27. a. Ammonium nitrate is d. NH4NO3 c. Calcium carbonate is h. CaCO3 e. Iron(III) phosphate is b. FePO4 g Calcium acetate is f. Ca(CH3CO2)2 28. (a) KOH (d) FeC2O4 29. (a) Silver nitrite (d) MgHCO3

(b) Magnesium carbonate (e) Copper(II) sulfate (b) Li2CrO4 (e) Cu3PO4

(c) Fe(CN)3 (f) Sodium dichromate

(c) Ammonium sulfite (f) Aluminum permanganate 27

.

Chapter 4: Chemical Bonds

30. Covalent bonds are the sharing of a pair of electrons by two atoms and an ionic bond is when you have oppositely charged ions held together by electrostaic attraction of (+) and (–).

31.

32.

H P H H 33.

Eight electrons around phosphorus H H Si H H 34.

Eight electrons around silicon

35.

Eight electrons around each atom

36.

Eight electrons around each atom

37. A hydrogen atom has one electon that it shares with another atom’s one electron to form a bonding pair. Hydrogen atoms only have one electron.

38. a. Dichlorine monoxide is f. Cl2O c. Chlorine monoxide is b. ClO e. Dichlorine trioxide is h. Cl2O3 g. Dichlorine heptoxide is d. Cl2O7 39. (a) N2O4 (d) NI3

(b) BrCl3 (e) Carbon tetrabromide

40. (a) CS2 (d) Carbon tetraiodide

(b) ClF3 (e) C3O2

(c) Oxygen difluoride (f) Dinitrogen tetrasulfide (or nitrogen sulfide)

(c) Phosphorus pentafluoride (f) Tetraphosphorus trisulfide

28 .

Chapter 4: Chemical Bonds

H

H

H

41. (a)

(b)

C

N

(c) H Each line/bond represents two electrons.

H

H N

(d)

O

H

(e) H

(f)

H C

42. (a)

(b) H

(d)

(e)

O

H HO

C H

2+

S

OH

O

(c)

Cl S Cl

(f) O HO

43. (a)

44. (a)

(b)

C

N

+

P

O

O

(c)

(b)

(c)

45. An electronegativity difference of 0.5–2.0 is typical for a polar bond.

46. An electronegativity difference of less than 0.5 is typical for a nonpolar bond.

47. (a) Polar

(b) Polar

(c) Polar

48. (a) Polar

(b) Polar (difference = 2.0)

(c) Polar

49. (a) H

O

(b)

(c)

50. (a) H

N

(b)

(c) 29 .

Chapter 4: Chemical Bonds

51. a. δ + Si - O δ - (Oxygen is more electronegative.) b. The F - F bond is not polar.

c. δ + N - Fδ − 52. a. δ − O - H δ + b. δ + C - Fδ − c. C=C, it is not polar.

53. (a) K2O is ionic. (b) BrCl is covalent, nonpolar. (c) MgF2 is ionic. (d) I2 is covalent, nonpolar.

Electronegativity differences: < 0.5 = nonpolar, 0.5–1.7 = polar, > 1.7 = ionic 54. (a) Na2O is ionic. (b) CaCl2 is ionic. (c) NBr3 is covalent, nonpolar (due to shape/lone pairs, however the molecule is polar). (d) CS2 is covalent, nonpolar.

55. Difference in brackets: f. P—Cl (0.9) < b. N—F (1.0) < e. Si—Cl (1.2) < a. H—F (1.9) < c. B—F (2.0) < d. Si—F (2.2) 56. a. AlCl3 = d. trigonal planar c. CSe2 = b. linear e. H2Se = f. bent g. SiI4 = h. tetrahederal 57. (a) SiH4: tetrahedral (d) SiF4: tetrahedral

(b) H2Se: bent (2 LPs) (c) PH3: trigonal pyramidal (1 LP) (e) OF2: bent (2 LPs) (f) H2C=O: trigonal planar

58. (a) CHCl3: tetrahedral (d) SF2: bent (2 LPs)

(b) BCl3: trigonal planar (0 LPs) (e) NI3: trigonal pyramidal (1 LP)

59. BeF2 is nonpolar because it is linear, polarity of bonds cancel.

(c) CF4: tetrahedral (f) CCl2F2: tetrahedral F

Be

F .

60. SF2 is polar, the polarity of the bonds do not cancel because the lone pairs on sulfur cause the molecular shape to be bent, not linear.

F

S F

61. Yes (or very close to a nonpolar molecule), bonds would be very close to canceling. polarity

potentially a slight net =

30 .

Chapter 4: Chemical Bonds

62. (a) Bonds polar? No; (b) Bonds polar? No; (c) Bonds polar? No; (d) Bonds polar? Yes; (e) Bonds polar? Yes; (f) Bonds polar? Yes;

bond angle = 109.5º; bond angle = <109.5º; bond angle = <109.5º; bond angle = 109.5º; bond angle = <109.5º; bond angle = ~120º;

molecule polar? No molecule polar? Yes molecule polar? Yes molecule polar? No molecule polar? Yes molecule polar? Yes

63. (a) Bonds polar? Yes; (b) Bonds polar? Yes; (c) Bonds polar? Yes; (d) Bonds polar? Yes; (e) Bonds polar? Yes; (f) Bonds polar? Yes;

bond angle = ~109.5º; molecule polar? Yes bond angle = 120º; molecule polar? No bond angle = 109.5º; molecule polar? No bond angle = <109.5º; molecule polar? Yes bond angle = <109.5º; molecule polar? Yes bond angle = ~109.5º; molecule polar? Yes

64. The lone pair on sulfur results in a bent geometry.

65. Molecular symmetry results in no net molecular polarity.

66. No, CH2Cl2 has a tetrahedral geometry, bond polarities do no cancel.

67. Water is more polar. They both have the same basic shape (bent) but the electronegativity difference from O-H is much greater than S – Cl.

68. a, c: Br and CCl3 are free radicals; they have an unpaired electron.

69. (a) and (b), S and NO2;

31 .

.

Chapter 4: Chemical Bonds

H B

70. (a)

(b)

H

Be

(c)

H

F F

Xe

H

F F

F

F

71. (a)

H

(b)

S

F

F

(c)

(d)

Answers to Additional Problems 72. Neon has an octet of valence electrons (a full outer shell).

δH

73. δ+

S

δ+ H

74. AlP and Mg3P2 75. The electronic shape is tetrahedral (ideal angle of 109.5), but the indicated lone pairs (LP) take up more space and push the Cl bonds together to be less than 109.5.

76.

77. ICl is (slightly) polar covalent.

78. X: 7A, HX,

Na+ X

;

+ Y: 6A, H2Y, 2Na Y

2-

;

Z: 5A, H3Z,

79. No; you would take a potassium salt, such as KCl. See Questions 1 and 2 also.

32 .

Chapter 4: Chemical Bonds

80. Potassium iodide is an ionic crystalline substance and does not come in discrete individual groupings like covalent substances.

81. (a) Ca+2: 1s22s22p63s23p6 (b) Rb+: 1s22s22p63s23p64s23d104p6 (c) S2-: 1s22s22p63s23p6 (d) I-: 1s22s22p63s23p64s23d104p65s2 4d105p6 (e) N3-: 1s22s22p6 (f) Se2-: 1s22s22p63s23p64s23d104p6 82. Na+ is smaller than Na because it has lost its outermost electron (no shell 3 electrons), and there are now 11 protons attracting 10 electrons. Cl- is larger than Cl because it has gained an additional electron, and there are now 17 protons attracting 18 electrons.

83. The halogens only have one unpaired valence electron available for bonding, and they have 7 valence electrons. Forming single bonds by sharing a bond with one other atom gives them an octet of electrons.

84. “Are very light” should be changed to “are gases” or to “are not very dense.”

Methane is a carbon atom connected to 4 hydrogen atoms, not molecules.

85. WO42-and Al3+ gives Al2(WO4)3.

86.

87. Q3(ZX4)2 88. 120° is the furthest (minimal repulsion); three items can be separated around a central point (VSEPR).

89. 109.5° is the furthest (minimal repulsion); four items can be separated around a central point (VSEPR).

90. The shapes (geometric orientation) and intermolecular forces between the molecules 91. Molecular recognition

33 .

Chapter 4: Chemical Bonds

92. (a) Less energy (b) Generate less wastes and nonuseful substances (c) Generate less hazardous materials/waste 93. A better understanding of the biological process will allow us to design new medicines and materials much more efficiently, more safely, at lower cost, using less energy, and developing them more quickly. Chemists can design new drug molecules more effectively to resemble biological molecules and their actions.

34 .

Chapter 5: Chemical Accounting Overview Chapter 5 explains chemical nomenclature, mass relationships, molar masses, and the gas laws.

Nomenclature of metal-nonmetal compounds, nonmetal-nonmetal compounds, polyatomic ions, and binary ionic compounds is discussed. The chapter concludes with balancing chemical equations and mole-mass relationships.

Chapter Outline 5.1 Chemical Sentences: Equations Remember that, matter is neither created nor destroyed in chemical reactions and chemical equations and the balancing of chemical equations make sense.

5.2 Volume Relationships in Chemical Equations 5.3 Avogadro's Number and the Mole: 6.02 x 1023 5.4 Molar Mass: Mole-to-Mass and Mass-to-Mole Conversions 5.5 Solutions Demonstrations 1. Display 1 mole samples of some common elements and compounds. Stoppered flasks make suitable containers for both display and storage. Use 50 mL flasks for 12 g of carbon, 32 g of sulfur, 207 g of lead, 27 g of aluminum, 18 g of water, 58.5 g of salt (sodium chloride), etc. Use a 250 mL flask for 180 g of glucose, and a 500 mL flask for 342 g of sucrose.

2. Put a little water in an empty soda can and heat it to boiling. Lift the can with tongs, and quickly invert it in a beaker of cold water. Air pressure causes the can to collapse.

3. Fill a gallon milk jug (or a 2-liter soda bottle) with very hot water. Then pour out the water and quickly stopper the bottle. The bottle collapses on cooling.

Review Questions 1. (a) the ratios of atoms in one molecule of a substance (b) the average mass of a formula unit of a substance relative to that of a 12C atom. (c) 6.02 x 1023 particles (d) 6.02 x 1023 (e) mass of Avogadro’s number of atoms or molecules (f) volume occupied by Avogadro’s number of atoms or molecules of a gas 2. The atomic mass of chlorine is the mass of a single chlorine atom (Cl). The formula mass of chlorine gas is the mass of the chlorine molecule (Cl2).

35 .

Chapter 5: Chemical Accounting

3. Avogadro’s hypothesis states that equal volumes of gases at the same pressure and temperature contain the same number of molecules. Since molecules react in ratios of small whole numbers, the volumes of reactants and products should be expressed as ratios of small whole numbers.

4. Matter is neither created nor destroyed in a chemical reaction, thus there must be the same number of reactant atoms as product atoms.

5. (a) a homogeneous mixture of solvent and solute(s) (b) greatest fraction of substance present in a solution (c) substance(s) present in smaller fractions in a solution (d) solution in which water is the solvent 6. (a) solution with a large quantity of solute dissolved in a small quantity of solvent (b) solution with a small quantity of solute dissolved in a large quantity of solvent (c) a substance that can be dissolved by the solvent (d) a substance that cannot be dissolved by the solvent Problems 7. (a) 12 (b) 3 (c) 6 8. (a) 6

(b) 3 (c) 7

9. N: 6 P: 3 H: 27 O: 12 10. Fe: 4 C: 36 H: 36 O: 48 11. (a) 2 molecules of hydrogen peroxide (H2O2) decompose to form two molecules of water (H2O) and one molecule of oxygen (O2). (b) 2 moles of hydrogen peroxide (H2O2) decompose to form two moles of water (H2O) and one mole of oxygen (O2). (c) 68.03 g of hydrogen peroxide (H2O2) decompose to form 36.03 g of water (H2O) and 32.00 g of oxygen (O2).

12. (a) 2 moles of magnesium(Mg) react with 1 mole of oxygen (O2) to produce 2 moles of magnesium oxide (MgO) (b) 2 moles of ethane (C2H6) reacts with 5 moles of oxygen (O2) to produce 4 moles of carbon dioxide (CO2) and 6 moles of water (H2O) 13. (a) 2 Mg + O2  2 MgO (b) C3H8 + 5 O2  3 CO2 + 4 H2O (c) 3 H2 + Ta2O3  2 Ta + 3 H2O

36 .

Chapter 5: Chemical Accounting

14. (a) 2 K + O2  K2O2 (b) FeCl2 + Na2SiO3  2 NaCl + FeSiO3 (c) 3 F2 + 2 AlCl3  2 AlF3 + 3 Cl2 15. (a) N2 + O2  2 NO (b) 2 O3  3 O2 (c) UO2 + 4 HF  UF4 + 2 H2O 16. (a) 4 Al + 3 O2  2 Al2O3 (b) CaCO3 + 2 HCl  CaCl2 + CO2 + H2O (c) 2 C6H14 + 19 O2  12 CO2 +14 H2O 17.

+ 18.

+ 20.6 L C7 H16 x7 L CO 2 = 144 L CO 2 1 L C7 H16 29.0 mL O 2 (b) x1 mL C 7 H16 = 2.64 mL C7 H16 11 mL O 2

19. (a)

125 L NH 3 x 2 L N 2 = 62.5 L N 2 4 L NH 3 36 L H 2 O (b) x3 L O 2 = 18 L O 2 6 L H 2O

20. (a)

21.

7 L CO 2 1 L C 7 H16

22.

6 L H 2O 4 L NH 3

37

.

Chapter 5: Chemical Accounting

23. (a) 6.02x1023 molecules S8 (b) 6.02x1023 molecules S8 x 8 atoms S / molecule S8 = 4.82x1024 atoms S 24. (a) 6.02x1023 x 3 = 1.81x1024 calcium ions (b) 6.02x1023 x 2 = 1.20x1024 nitride ions 25. c; 1.20x1024 Br atoms 26. (a) 6.02x1023 barium ions; 1.20x1024 nitrate ions (b) 1.20x1024 nitrogen atoms; 3.61x1024 oxygen atoms 27. (a) AgNO3: 107.9 g/mol + 14.0 g/mol + 3(16.0 g/mol) = 169.9 g/mol (b) Zn(IO)2: 65.4 g/mol + 2(126.9 g/mol) + 2(16.0 g/mol) = 351.2 g/mol (c) Sn(ClO4)2: 118.7g/mol + 2(35.4 g/mol) + 8(16.0 g/mol) = 317.5 g/mol (d) CH3(CH2)3COF: 5(12.0 g/mol) + 9(1.0 g/mol) + 16.0 g/mol + 19.0 g/mol = 104.0 g/mol 28. (a) Bi2O3: 2(209.0 g/mol) + 3(16.0 g/mol) = 466.0 g/mol (b) CuSO4: 63.6 g/mol + 32.1 g/mol + 4(16.0 g/mol) = 159.7 g/mol (c) Ca(CH3COO)2: 40.1 g/mol + 4(12.0 g/mol) + 6(1.0 g/mol) + 4(16.0 g/mol) = 158.1 g/mol (d) (NH4)2C2O4: 2(14.0 g/mol) + 8(1.0 g/mol) + 2(12.0 g/mol) + 4(16.0 g/mol) = 124.0 g/mol 29. (a) 7.57 mol x 233.4 g/mol = 1770 g (b) 4.72 mol x 134.4 g/mol = 634 g (c) 0.250 mol x 342.0 g/mol = 85.5 g 30. (a) 0.460 mol x 137.2 g/mol = 63.1 g (b) 6.15 mol x 152.0 g/mol = 935 g (c) 0.158 mol x 221.9 g/mol = 35.1 g 31. (a) 6.63 g/339.9 g/mol = 0.0195 mol (b) 19.1 g/143.9 g/mol = 0.133 mol (c) 43.4 g/122.0 g/mol = 0.356 mol 32. (a) 16.3 g/146.1 g/mol = 0.112 mol (b) 25.4 g/325.2 g/mol = 0.0781 mol (c) 186 g/165.1 g/mol = 1.13 mol 28.02 g N x 100% = 17.1% N 164.10 g Ca(NO 3 ) 2 14.01 g N (b) x 100% = 26.2% N 53.4 g NH 4 Cl

33. (a)

38 .

Chapter 5: Chemical Accounting

34. C: 40.0% H: 6.7% O: 53.3% 35. (a)

(b)

0.845 mol C 4 H10 x 8 mol CO 2 = 3.38 mol CO 2 2 mol C 4 H10 2.54 mol C 4 H10 x 13 mol O 2 = 16.5 mol O 2 2 mol C 4 H10

2.59 mol C8 H18 x 18 mol H 2 O = 23.3 mol H 2 O 2 mol C8 H18 6.75 mol O 2 (b) x 16 mol CO 2 = 4.32 mol CO 2 25 mol O 2

36. (a)

37. N2 + 3 H2  2 NH3 (a) 2 mol of NH3 are produced for every 3 mol H2 consumed 250 g H2/2.02 g/mol = 124 mol H2 124 mol H2 x 2 mol NH3/3 mol H2 = 82.7 mol NH3 82.7 mol NH3 x 17.0 g/mol = 1.41x103 g NH3 produced (b) 3 mol of H2 are required for every mol N2 consumed 923 g N2/28.0 g/mol = 33.0 mol N2 33.0 mol N2 x 3 mol H2/1 mol N2 = 99.0 mol H2 99.0mol H2 x 2.0 g/mol = 198 g H2 required 38. C7H8 + 3 HNO3  C7H5N3O6 + 3 H2O (a) 3 mol of HNO3 are required for every mol C7H8 consumed 256 g C7H8/92.0 g/mol = 2.78 mol C7H8 2.78 mol C7H8 x 3 mol HNO3/1 mol C7H8 = 8.34 mol HNO3 8.34 mol HNO3 x 63.0 g/mol = 525 g HNO3 required (b) 1 mol of C7H5N3O6 is produced for every mol C7H8 consumed 951 g C7H8/92.0 g/mol = 10.3 mol C7H8 10.3 mol C7H8 x 1 mol C7H5N3O6 /1 mol C7H8 = 10.3 mol C7H5N3O6 10.3 mol C7H5N3O6 x 227.0 g/mol = 2340 g C7H5N3O6 produced 39. M = n/V (a) 23.4 mol/12.0 L = 1.95 M (b) 0.875 mol/1.032 L = 0.848 M 40. M = n/V (a) 2.82 mol/5.75 L = 0.490 M (b) 2.22 mol/0.1933 L = 11.5 M

39 .

Chapter 5: Chemical Accounting

41. M = n/V n = MV (a) 0.500 M x 5.50 L = 2.75 mol x 40.0 g/mol = 110 g NaOH (b) 1.55 M x 0.0650 L = .101 mol x 180.0 g/mol = 18.2 g C6H12O6 42. M = n/V n = MV (a) 0.167 M x 0.125 L = 0.0209 mol x 294.2 g/mol = 6.15 g K2Cr2O7 (b) 0.0200 M x 0.705 L = 0.0141 mol x 158.0 g/mol = 2.23 g KMnO4 43. M = n/V V = n/M (a) 2.00 mol/6.00 M = 0.333 L (b) 8.10 g/180g/mol = 0.0450 mol KH2AsO4 0.0450 mol/0.0700 M = 0.643 L 44. M = n/V V = n/M (a) 1.50 mol/0.250 M = 6.00 L (b) 0.275 g/90g/mol = 0.00306 mol H2C2O4 0.00306 mol/4.25 M = 0.000720 L = 0.720 mL 45. (a) (68.0 mL/650 mL) x 100% = 10.5% (b) (35.5 mL/755 mL) x 100% = 4.70% 46. (a) (35.0 mL/700 mL) x 100% = 5.00% (b) (85.9 mL/1550 mL) x 100% = 5.54% 47. 3750 g solution x 0.082 = 310 g NaCl add 310 g NaCl to 3440 g water 48. 2.54 kg solution x 0.163 = 0.414 kg KOH add 0.414 kg KOH to 2.13 kg water 49. 2.50 L x 0.0200 = 0.0500 L acetic acid add 0.0500 L acetic acid to 2.00 L of water 50. 600.0 mL x 0.300 = 180 mL isopropyl alcohol add 180 mL isopropyl alcohol to 420 mL of water

Additional Problems 51. Al(s) + 2 H+(aq)  Al3+(aq) + H2(g) is not a charge balanced reaction. The correctly balanced reaction is: 2 Al(s) + 6 H+(aq)  2 Al3+(aq) + 3 H2(g) 52. b 40 .

Chapter 5: Chemical Accounting

53. (a) Hg(NO3)2(s)  Hg(l) + 2 NO2(g) + O2(g) (b) Na2CO3(aq) + 2 HCl(aq)  H2O(l) + CO2(g) + 2 NaCl(aq) 54. 2 HgO(s)  2 Hg(l) + O2(g) 18.0 g HgO/216.6 g/mol = 0.0831 mol HgO 0.0831 mol HgO x 1 mol O2/2 mol HgO = 0.0416 mol O2 0.0416 mol O2 x 32.00 g/mol = 1.33 g O2 produced 55. (a) yes, all have 2.00 mol of atoms. (b) C2H2 56. 3 Fe + 2 O2  Fe3O4 50.0 g Fe/55.8 g/mol = 0.896 mol Fe 0.896 mol Fe x 1 mol Fe3O4/3 mol Fe = 0.299 mol Fe3O4 0.299 mol Fe3O4 x 231.4 g/mol = 69.2 g Fe3O4 57. CaCO3  CaO + CO2 2.5 x 105 g CaCO3/100 g/mol = 2.5 x 103 mol CaCO3 2.5 x 103 mol CaCO3 x 1 mol CaO/1 mol CaCO3 = 2.5 x 103 mol CaO 2.5 x 103 mol CaO x 56 g/mol = 1.4 x 105 g CaO 58. NH3 + 2 O2  HNO3 + H2O 1 mol of HNO3 produced for every mol of NH3 consumed 549 g NH3/17.0 g/mol = 32.3 mol NH3 32.3 mol NH3 x 1 mol HNO3/1 mol NH3 = 32.3 mol HNO3 32.3 mol HNO3 x 63.0 g/mol = 2030 g HNO3 59.

16 - fl. oz 1.00 g x 29.6 mL = 473.6 mL x x 0.030 = 14 g H 2 O 2 1 - fl. oz 1 mL 1 mol 14 g H 2 O 2 x = 0.41 mol H 2 O 2 34.0 g

60. (a)

2.59 g NaOH x 100% = 2.52% NaOH 2.59 g NaHO + 100.0 g water

(b) 5.25 mL ethanol x 0.789 g

mL

= 4.14 g

4.14 g ethanol x 100% = 7.65% ethanol 4.14 g ethanol + 50.0 g water 337 mL ethanol x 100% = 94.9% ethanol 355 mL solution 0.00400 L acetone (b) x 100% = 0.258% acetone 1.55 L solution

61. (a)

41 .

Chapter 5: Chemical Accounting

62. (a) NH4NO3  N2O + 2 H2O (b) (c) 1 mol of N2O produced for every mol of NH4NO3 consumed 150.0 g NH4NO3/80.0 g/mol = 1.875 mol NH4NO3 1.875 mol NH4NO3 x 1 mol N2O /1 mol NH4NO3 = 1.875 mol N2O 1.875 mol N2O x 44.0 g/mol = 82.5 g N2O 63. (a) (b)

238.0 g 1x10-24 mol 1 yg x x 1.50 ymol x = 357 yg 1 mol 1 ymol 1x10-24 g 6.02 x10 23 atoms 1x10 -21 mol 602 atoms x = x 1.20 zmol = 722 atoms 1 mol 1 zmol 1 zmol

1 mol = 0.352 mol 90 g n 0.352 mol V= = = 0.410 L M 0.859 M

64. 31.7 g H 2 C 2 O 4 x M=

n V

65. 5.0 L x 0.00165 = 0.00825 L = 8.25 mL alcohol 1 yM = 1.00 yM 1x10 -24 M 6.02x10 23 molecules 1x10 -24 mol 0.602 molecules (b) x = 1 mol 1 ymol 1 ymol

66. (a) 1.00 M x 1 x 10 -24 x

67. 236 mL x

1 mol 1.00 g = 236 g x = 13.1 mol 18 g 1 mL

13.1 mol x

6.02 x10 23 molecules = 7.89x10 24 molecules in 1 cup 1 mol

1x1015 cm 3 1 cup 1.47 x 10 km x = 1.47 x 10 24 cm 3 x = 6.23 x 10 21 cup 3 3 1 km 236 cm 9

68.

3

500 mL x 500 mL x 0.056 mol x 6.02x10 23 molecules = 8430 molecules 24 mL mole 1x10 mL

69. H2SO4 + 2 NaHCO3  Na2SO4 + 2 CO2 + 2 H2O 2 mol of NaHCO3 are consumed for every mol of H2SO4 consumed 50,000,000 g H2SO4 /98.1 g/mol = 510,000 mol H2SO4 510,000 mol H2SO4 x 2 mol NaHCO3 /1 mol H2SO4 = 1,020,000 mol NaHCO3 1,020,000 mol NaHCO3 x 84.0 g/mol = 85,700,000 g NaHCO3

42 .

Chapter 5: Chemical Accounting

70. H2SO4 + Na2CO3  Na2SO4 + CO2 + H2O 1 mol of Na2CO3 are consumed for every mol of H2SO4 consumed 50,000,000 g H2SO4 /98.1 g/mol = 510,000 mol H2SO4 510,000 mol H2SO4 x 1 mol Na2CO3 /1 mol H2SO4 = 510,000 mol Na2CO3 510,000 mol Na2CO3 x 106.0 g/mol = 54,100,000 g Na2CO3 71. 2000 mL x

1.00 g 1 mol = 2000 g x = 111 mol 1 mL 18 g

72. (a) % A.E. = (molar mass of 3 N2/molar mass of 2 NaN3) x100% % A.E. = (84.06/130.04)x100% = 64.6% (b) 5.74 g NaN3 /65.02 g/mol = 0.0883 mol NaN3 0.0883 mol NaN3 x 3 mol N2/2 mol NaN2 = 0.132 mol N2.

0.132 mol N2 x 28.0 g/mol = 3.70 g N2 73. (a) C6H12O6  2 C2H5OH + 2 CO2; C2H4 + H2O  C2H5OH (b) %A.E. = 51.1%; %A.E. = 100% (c) The ethylene reaction is balanced as written and all atoms are converted to product (d) The conversion of glucose is sustainable as it is made from biomass. (e) For industrial purposes ethylene is presently the most economical method.

74. The balanced equation is: C5H12 + 8 O2  5 CO2 + 6 H2O (a) %A.E. = 67.1% (b) 12.0 g C5H12/72.17 g/mol = 0.166 mol C5H12 0.166 mol C5H12 x 5 mol CO2/ mol C5H12 = 0.830 mol CO2 0.830 mol CO2 x 44.01 g/mol = 36.5 g CO2 (c) % Yield = (15.2g/36.5 g) x 100% = 41.6% 75. (a) CuCl2(aq) + Na2CO3(aq)  CuCO3(s) + 2 NaCl(aq); %A.E. = 50.2% (b) 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g); %A.E. = 52.6% (c) 12 KMnO4(aq) + 12 KOH(aq) + 2 KI(aq)  12 K2MnO4(aq) + 2 KIO3(aq) + 6 H2O(l) %A.E. = 82.5%

43 .

Chapter 6: Gases, Liquids, Solids … and Intermolecular Forces Overview Chapter 6 discusses the states of matter and intermolecular forces and introduces the empirical gas laws.

Chapter Outline 6.1 Solids, Liquids, and Gases 6.2 Comparing Ionic and Molecular Substances 6.3 Forces between Molecules 6.4 Forces in Solutions 6.5 Gases: The Kinetic-Molecular Theory 6.6 The Simple Gas Laws Pressure is inversely proportional to volume.

Volume is directly proportional to temperature.

Volume is directly proportional to number of moles.

6.7 The Ideal Gas Law pV = nRT is the ideal gas law.

Demonstrations 1. Display 1 mole samples of some common elements and compounds. Stoppered flasks make suitable containers for both display and storage. Use 50 mL flasks for 12 g of carbon, 32 g of sulfur, 207 g of lead, 27 g of aluminum, 18 g of water, 58.5 g of salt (sodium chloride), and so on. Use a 250 mL flask for 180 g of glucose and a 500 mL flask for 342 g of sucrose.

2. Put a little water in an empty soda can and heat it to boiling. Lift the can with tongs, and quickly invert it in a beaker of cold water. Air pressure causes the can to collapse.

3. Fill a gallon milk jug (or a 2-liter soda bottle) with very hot water. Then pour out the water and quickly stopper the bottle. The bottle collapses on cooling.

Review Questions 1. Solids and liquids consist of particles in close contact with one another. The particles are freer to move about in liquids.

2. (a) a solid changing to a liquid (b) a liquid changing to a gas (c) a gas changing to a liquid (d) a liquid changing to a solid

44 .

Chapter 6: Gases, Liquids, Solids . . . and Intermolecular Forces

3. Top left arrow: melting; Top right arrow: boiling; Bottom right arrow: condensation; Bottom left arrow: freezing 4. (a) melting (b) vaporization 5. Dispersion forces(found in nonpolar substances such as hydrocarbons), dipole–dipole forces(found in polar substances), hydrogen bonds (found in subtances where hydrogen is bonded to oxygen or nitrogen), and ionic bonds (found in salts) 6. (a) ionic forces

(b) dispersion forces

7. Volume is inversely proportional to pressure and directly proportional to absolute temperature.

Problems 8. (a) hydrogen bonding (c) dispersion forces

(b) dispersion forces (d) ionic bonds

9. NaBr 10. b and d 11. N2 12. HBr 13. He, C2H6, NH3, KCl 14. No. KNO3 is an ionic compound and will dissolve in polar solvents.

15. water: a, d

carbon tetrachloride: b, c

16. Use Boyle’s Law: P1V1 = P2 V2 PV 719 mmHg x 1820 mL (a) 1 1 = V2 = = 1740 mL P2 752 mmHg PV 719 mmHg x 1820 mL (b) 1 1 = P2 = = 3790 mmHg V2 345 mL 17. Use Boyle’s Law: P1V1 = P2 V2 P1 V1 4.25 atm x 10.1 m 3 = V2 = = 42.9 m 3 P2 1.00 atm

45 .

Chapter 6: Gases, Liquids, Solids . . . and Intermolecular Forces

18. Use Boyle’s Law: P1V1 = P2 V2 PV 150 atm x 60.0 L (a) 1 1 = V2 = = 9730 L P2 0.925 atm 1 hr 9730 L (b) = 1620 min x = 27.0 hr 60 min 6.0 L min 19. Use Boyle’s Law: P1V1 = P2 V2 P1 V1 1.10 atm x 4.50 L = P2 = = 0.497 atm V2 9.95 L 20. Use Boyle’s Law: P1V1 = P2 V2 P1V1 764 mmHg x 954 mL = P2 = = 1.53x103 mmHg 477 mL V2 21. Use Boyle’s Law: P1V1 = P2 V2 P1V1 0.994 atm x 1.88 L = V2 = = 3.76 L 0.497 atm P2 22. Convert all temperatures to Kelvin and use Charles’s Law:

V1 V2 = T1 T2

V1T2 5.90 L x 351 K = V2 = = 6.93 L 299 K T1 23. Convert all temperatures to Kelvin and use Charles’s Law:

V1 V2 = T1 T2

V1T2 154 mL x 283 K = V2 = = 117 mL T1 373 K 24. T2 =

V2 T1 3V1 x 273 K = = 819 K V1 V1

25. Convert all temperatures to Kelvin and use Charles’s Law:

V1T2 1.00 L x 310 K = V2 = = 1.06 L 291 K T1

46 .

V1 V2 = T1 T2

Chapter 6: Gases, Liquids, Solids . . . and Intermolecular Forces

26. Convert all temperatures to Kelvin and use Charles’s Law:

V1 V2 = T1 T2

V1T2 1.00 L x 310 K = V2 = = 1.18 L 263 K T1 27. Convert all temperatures to Kelvin and use Charles’s Law:

T2 =

V1 V2 = T1 T2

V2 T1 425 mL x 578 K = = 433 K 567 mL V1

28. Convert all temperatures to Kelvin and use Charles’s Law: V1T2 12.7 cm 3 x 838 K = V2 = = 23.6 cm 3 T1 450 K 29. Use the volume of 1 mole = 22.4 L at STP (a) 22.4 L x 1.00 mol = 22.4 L mol (b) 22.4 L x 1.75 mol = 39.2 L mol (c) 22.4 L x 0.225 mol = 5.04 L mol 30. (a) 83.8 g

(b) 2.02 g

31.

222 g/mol = 9.91 g/L 22.4 L/mol

32.

44.0 g/mol = 1.96 g/L 22.4 L/mol

(c) 28.0 g

33. (a) 2.12 g/L x 22.4 L mol = 47.5 g mol (b) 2.97 g/L x 22.4 L mol = 66.5 g mol 34. (a) 1.98 g/L x 22.4 L mol = 44.4 g mol (b) (10.88 g/ 3.33 L) x 22.4 L mol = 73.2 g mol 35. (a) decrease

(b) decrease

(c) increase

36. (a) increase

(b) increase

(c) increase

37. (a) decrease

(b) decrease 47 .

V1 V2 = T1 T2

Chapter 6: Gases, Liquids, Solids . . . and Intermolecular Forces

38. (a) The gas in container A has the greatest density. (b) The gas in container B has the greatest density. (c) The gas in container B has the greatest density.

39. pV = nRT nRT 0.00600 mol x 0.082057 L atm mol K x 304 K (a) V = = = 0.172 L p 0.870 atm nRT 0.0108 mol x 0.082057 L atm mol K x 310 K (b) p = = = 1.04 atm V 0.265 L 40. pV = nRT nRT 1.12 mol x 0.082057 L atm mol K x 335 K (a) V = = = 22.3 L p 1.38 atm nRT 4.64 mol x 0.082057 L atm mol K x 302 K (b) p = = = 29.0 atm V 3.96 L 41. pV = nRT pV 0.918 atm x 0.555 L n= = = 0.0208 mol RT 0.082057 L atm mol K x 298 K 42. pV = nRT 0.933 atm x 0.0745 L pV n= = = 0.00280 mol RT 0.082057 L atm mol K x 303 K m n= fwt m = n x fwt = 0.00280 mol x 28.0 g mol = 0.0784 g CO 43. pV = nRT 0.154 atm x 0.660 L pV n= = = 0.00416 mol RT 0.082057 L atm mol K x 298 K 44. pV = nRT pV 0.555 atm x 20.0 L T= = = 405 K - 273 = 132 o C nT 0.082057 L atm mol K x 0.334 mol 45. pV = nRT nRT 4.55 mol x 0.082057 L atm mol K x 285 K V= = = 14.5 L p 7.32 atm 46. pV = nRT nRT 0.0456 mol x 0.082057 L atm mol K x 302 K p= = = 0.151 atm 7.50 L V 48 .

Chapter 6: Gases, Liquids, Solids . . . and Intermolecular Forces

Additional Problems 47. (a) 1. Particles of a gas are in rapid, constant motion and move in straight lines.

2. The particles of a gas are tiny compared to the distance between them.

3. Because the particles of a gas are so far apart, there is very little interaction between them.

4. Particles of a gas collide with one another. Energy is conserved in these collisions; energy lost by one particle is gained by another.

5. Temperature is a measure of the average kinetic energy of the gas particles. (b) postulate #2 48. postulate #3 49. (a) temperature and amount of gas (c) temperature and pressure

(b) pressure and amount of gas

50. Boyle’s Law 51. 0.75 mol He x 4.0 g mol = 3.0 g 52. PF3 D=

pfwt ; for gases at the same p and T : D α fwt RT

53. Use Charles’s Law:

V1 V2 = T1 T2

V1T2 1.50 L x 77 K = V2 = = 0.394 L T1 293 K 54. (a) All have the same number of atoms (Avogadro’s principle). (b) flask Z (c) flask X (d) All have the same number of moles.

55. pV = nRT mRT pV = fwt mRT 14.4 g x 0.082057 L atm mol K x 298 K fwt = = = 44.0 g mol 1.0 atm x 8.00 L pV 56. Amonton’s Law:

P1 P2 = T1 T2

P1T2 1.32 atm x 283 K = P2 = = 1.25 atm T1 298 K 49 .

Chapter 6: Gases, Liquids, Solids . . . and Intermolecular Forces

57. (a) 4.6 m x 4.9 m x 3.0 m = 68 m3 (b) 68 m3 x 1000 L/m3 = 68,000 L 68,000 L (c) = 3000 mol x 44.0 g mol = 130,000 g CO 2 22.4 L mol pfwt DRT 2.32 g L x 0.082057 L atm mol K x 300 K = 57.9 g mol 58. D = RT ; fwt = p = 0.987 atm C4H10 59. D α p; thus D1 D 2 = p1 p2 D p p2 = 2 1 D1 .50D1 760 mm Hg (a) p 2 = = 380 mm Hg D1 0.050D1 760 mm Hg (b) p 2 = = 38.0 mm Hg D1 0.010D1 760 mm Hg (c) p 2 = = 7.60 mm Hg D1 60. 2 H2O → 2 H2 + O2 1000 g/18 g/mol = 55.6 mol H2O 55.6 mol H2O x 2 mol H2/2 mol H2O = 55.6 mol H2 55.6 mol x 22.4 L/mol = 1250 L H2 61. b, c, and d 62. c 63. 0.601 g/mL x 5.00 mL = 3.00 g/(58.14g mol) = 0.0520 mol x 22.4 L/mol = 1.16 L 64. e 65. scCO2 is inexpensive, convenient to use, nonflammable, and nontoxic.

66. Methylene chloride and perchloroethylene 67. scCO2 may be used again and again.

50 .

Chapter 7: Acids and Bases Overview Acids and bases are important to us in many ways. We consume products that are acids or bases on a daily basis. The balance of acids and bases is crucial for life processes. Chapter 7 defines acids and bases, distinguishes between strong and weak acids and bases, introduces the pH scale, and discusses some acid–base consumer issues.

Chapter Outline 7.1 Acids and Bases: Experimental Definitions 7.2 Acids, Bases, and Salts 7.3 Acidic and Basic Anhydrides 7.4 Strong and Weak Acids and Bases Strong acids and bases undergo 100% dissociation. There are six strong acids: H2SO4, HClO4, HNO3 , HCl, HBr, HI. 7.5 Neutralization 7.6 The pH Scale. pH of 7 is neutral. pH < 7 is acidic, pH > 7 is basic.

7.7 Buffers and Conjugate Acid–Base Pairs 7.8 Acids and Bases in Industry and in Daily Life Demonstrations 1. Place 6 or 8 small beakers on the overhead projector. Put a little (10 mL) dilute acid in half of the beakers and a little dilute base in the other half. Into each acid–base pair add several drops of some selected indicator (e.g., methyl orange, bromocresol green, methyl violet, phenol red, etc.) to show the variety of color changes exhibited by acid–base indicators.

2. Some indicators exhibit more than one color change. In three small beakers on the overhead projector, place small amounts of dilute acid, water, and dilute base. Then add a drop of thymol blue indicator to each sample. The colors will be red, yellow, and blue.

3. On the overhead projector place four beakers containing small amounts of water, 0.01 M HCl, 1 M HCl, and 6 M HCl. To each sample add a few drops of methyl violet indicator solution. The colors will be violet, blue, green, and yellow.

4. On the overhead projector place six small beakers containing 5–10 mL of the following: 6 M NaOH, 1 M NaOH, tap water, distilled water, 1 M HCl, and 6 M HCl. To each sample add a few drops of bromothymol blue indicator. The colors will be violet, blue, green, yellow, orange, and red.

(If the tap water turns blue, add a little more indicator solution.)

5. Place some flower petals in a beaker with about 1/4 cup of rubbing alcohol. (Almost any kind of flower petals can be used.) Use a stirring rod to mash up the petals in the alcohol. (Sometimes it helps to heat the mixture for a few minutes.) Pour off the clear colored extract, filtering it if 51 .

Chapter 7: Acids and Bases

necessary, and pour into several small beakers. Add vinegar to one, ammonia to another, and other acids or bases to the rest.

6. Take several red cabbage leaves and place them in a blender along with about a cup of water. Blend the mixture until the cabbage has been chopped into fine pieces.

Pour off the liquid, using a tea strainer to get rid of the very small bits of solid material. Pour some of the deep-colored liquid into each of five glasses or beakers. Into each container pour one of the following: vinegar, bottled soda water, tap water, baking soda dissolved in water, and clear ammonia.

The colors will be red, pink, violet, blue, and green.

Answers to Review Questions 1. a. hydrogen ion donor; HCl b. hydroxide ion donor; NaOH c. ionic compound; NaCl 2. (a) cause litmus indicator dye to turn red; taste sour; dissolve active metals producing hydrogen gas; react with bases to form water and salts; (b) cause litmus indicator dye to turn blue; taste bitter; feel slippery on the skin; react with acids to form water and salts.

3. no effect on litmus indicator dye; no reaction with iron or zinc 4. According to the Brønsted–Lowry theory, an acid is a proton donor; thus it must have an H to donate. A Brønsted–Lowry base must be a proton acceptor.

5. The proton in acid–base chemistry is solvated by water. The proton in nuclear chemistry is a nucleon and is not solvated.

6. Hydrogen. No, only those that dissociate in water to form H+ ions are acids.

7. Neutralization of an acid occurs when the hydrogen ion is mixed with an equal amount of hydroxide ion. Neutralization of a base occurs when the hydroxide ion is mixed with an equal amount of hydrogen ion. (The concentration of the hydrogen ion is equal to the concentration of the hydroxide ion in a neutral solution.)

8. An anhydride is any substance that forms an acid when mixed with water. Nonmetal oxides are acid anhydrides. Any substance that forms a base when mixed with water. Metal oxides are basic anhydrides.

9. the degree to which they accept protons 10. Magnesium hydroxide is relatively insoluble. The hydroxide ion enters the solution only as it is consumed by a corresponding hydrogen ion.

11. They can cause chemical burns.

52 .

Chapter 7: Acids and Bases

12. Alkalosis is a condition in which the blood gets too alkaline. Ingestion of large quantities of magnesium hydroxide may cause alkalosis.

Problems 13. (a) HClO4 + H2O → H3O+ + ClO4-

(b) . RbOH → Rb+ + OH-

14. (a) HI + H2O → H3O+ + I-

(b) Ba(OH)2 → Ba2+ + 2 OH-

15. H2PO4- + H2O → H3O+ + HPO4216. (a) HBr + H2O → H3O+ + Br-

(b) Sr(OH)2 → Sr2+ + 2 OH-

17. (a) base (b) base (c) base 18. (a) base (b) acid (c) base 19. (a) HCOOH + H2O → H3O+ + HCOO(b) C5H5N + H2O → C5H5NH+ + OH20. (a) HOCl + H2O → H3O+ + Cl(b) (CH3CH2)2NH + H2O → (CH3CH2)2NH2+ + OH21. NH3 + H2O → NH4+ + OH22. HONH2 + H2O → HONH3+ + OH23. (a) hydrochloric acid (b) Sr(OH)2 (c) potassium hydroxide (d) H3BO3 24. (a) RbOH (b) aluminum hydroxide (c) HCN (d) nitric acid 25. (a) phosphoric acid

(b) cesium hydroxide: base

26. (a) magnesium hydroxide: base (b) ammonia:base 27. (a) HNO2

(b) H3PO3

28. (a) H2TeO4

(b) H2TeO3

(c) carbonic acid (c) acetic acid

29. (a) SO3 + H2O → H2SO4; acid (b) MgO + H2O → Mg(OH)2; base 30. (a) 2 K2O + H2O  2 KOH; base (b) CO2 + H2O  H2CO3; acid 31. strong acid; HI (aq) → H+(aq) + I-(aq) 53 .

Chapter 7: Acids and Bases

32. strong base; TlOH(s) → Tl+(aq) + OH-(aq) 33. weak base; CH3NH2 + H2O  CH3NH3+(aq) + OH-(aq) 34. weak acid; HOCN (aq) → H+(aq) + OCN-(aq) 35. (a) strong base (b) strong acid (c) weak acid (d) salt 36. (a) salt (b) salt (c) weak base (d) salt 37. a is highest; b is lowest 38. hydrochloric acid, acetic acid, ammonia, sodium hydroxide 39. (a) KOH + HCl → KCl + H2O (b) LiOH + HNO3 → H2O + LiNO3 40. (a) Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O (b) 2 KOH + H2SO4 → 2 H2O + K2SO4 41. Mg(OH)2 + H2SO3 → 2 H2O + MgSO3 42. H3PO4 + Al(OH)3 → AlPO4 + 3 H2O 43. (a) acidic

(b)neutral

(c) acidic

(d) basic

44. (a) 2 45. pH = –log([H3O+]) pH = –log(1.0 x 10-5) pH = –(–5) pH = 5 46. pH = –log([H3O+]) pH = –log(1.0 x 10-9) pH = –(–9) pH = 9 47. pOH = –log([OH-]) pOH = –log(1.0 x 10-5) pOH = –(–5) pOH = 5 pH + pOH = 14 pH = 14 – pOH pH = 14 – 5 pH = 9 54 .

Chapter 7: Acids and Bases

48. pOH = –log([OH-]) pOH = –log(1.0 x 10-9) pOH = –(–9) pOH = 9 pH + pOH = 14 pH = 14 – pOH pH = 14 – 9 pH = 5 49. pH = –log([H3O+]) -pH = log([H3O+]) + 10-pH = 10log([H3O ]) 10-pH = [H3O+] 10-3 = [H3O+] 1.0 x 10-3 M = [H3O+] 50. pH = –log([H3O+]) -pH = log([H3O+]) + 10-pH = 10log([H3O ]) 10-pH = [H3O+] 10-11 = [H3O+] 1.0 x 10-11 M = [H3O+] 51. 10 and 11 52. 13 and 14 53. (a) acid: HNO3 (b) conjugate base: NO3-

base: NH3 conjugate acid: NH4+

54. (a) acid: CH3CH2COOH (b) conjugate base: CH3CH2COO-

base: H2O conjugate acid: H3O+

55. Al(OH)3 + 3 HCl → 3 H2O + AlCl3 Mg(OH)2 + 2 HCl → 2 H2O + MgCl2 56. (a) HCO3- (b) OH- (c) CO32- (d) CO32Answers to Additional Problems 57. above 7 or basic 58. pink; will turn violet, then blue 55 .

Chapter 7: Acids and Bases

59. no; alcohols contain the OH group and are not bases 60. CaCO3 + 2 HCl → H2CO3 + CaCl2 61. hydroiodic acid (HI)

SrCO3 + 2 HI → H2CO3 + SrI2

62. HS- + H2O → H2S + OH63. (a) pOH = –log([OH-]) pOH = –log(0.001) pOH = –(–3) pOH = 3 (b) pOH = –log(0.01) pOH = –(–2) pOH = 2 (c) pH = –log([H3O+]) pH = –log(0.01) pH = –(–2) pH = 2 pH + pOH = 14 pOH = 14 – pH pOH = 14 – 2 pOH = 12 64. e, b, d, a, c 65. Ba(OH)2 → Ba2+ + 2 OH- a 0.50 M solution of Ba(OH)2 is 1.0 M in OHpOH = -log([OH-]) pOH = -log(1.0) pOH = -(-0) pOH = 0 pH + pOH = 14 pH = 14 – pOH pH = 14 – 0 pH = 14 66. HPO42-(aq) + H2O(l) → H3O+(aq) + PO42-(aq) (acid) HPO42- + H3O+ → H2PO4- + H2O (base) 67. A buffer is a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Sodium hydroxide is a strong base; thus, it will not make a buffer solution. Therefore, ammonia and ammonium acetate will make the best buffer.

68. 2 E-X tums: 2x750 mg = 1500 mg; 1500 mg/500 mg = 3 regular tums 2 ULTRA tume: 2x1000 mg = 2000 mg; 2000 mg/500 mg = 4 regular tums 56 .

Chapter 7: Acids and Bases

69. Al2(CO3)3 + 6 HCl → 3 H2O + 3 CO2 + 2 AlCl3 70. Mg(OH)2 + 2 HCl → 2 H2O + MgCl2 0.400 g Mg(OH)2 = 0.00686 mol Mg(OH)2 58.3 g mol 0.00686 mol Mg(OH)2 x 2 mol HCl = 0.0137 mol HCl 1 mol Mg(OH)2 0.0137 mol HCl x 36.4 g

mol

= 0.499 g HCl

0.499 g HCl = 99.8 g stomach acid 0.005

71. S + O2 → SO2 2 SO2 + O2 → 2 SO3 SO3 + H2O → H2SO4

57 .

Chapter 8: Oxidation and Reduction Overview The process of oxidation and reduction occurs in and around us every second of every day. Chapter 8 begins by defining oxidation-reduction reactions, and then discusses some important types of oxidation-reduction reactions. The chapter concludes with a brief overview of the importance of these reactions to living organisms.

Chapter Outline 8.1 Oxidation and Reduction: Four Views Oxidation is a gain of oxygen atoms. Reduction is a loss of oxygen atoms.

Oxidation is a loss of hydrogen atoms. Reduction is a gain of hydrogen atoms.

Oxidation is a loss of electrons. Reduction is a gain of electrons.

Oxidation increases the oxidation number. Reduction decreases the oxidation number.

8.2 Oxidizing and Reducing Agents The substance oxidized is the reducing agent.

The substance reduced is the oxidizing agent.

8.3 Electrochemistry: Cells and Batteries The passage of the electrons of an oxidation-reduction reaction through a circuit yields energy.

8.4 Corrosion and Explosion Loss of useful metals through the process of oxidation is a costly but unavoidable process. An explosion is a very rapid oxidation-reduction reaction.

8.5 Oxygen: An Abundant and Essential Oxidizing Agent Oxygen is the most importance oxidizing agent. Oxygen exists as O2 and constitutes 21% of the Earth’s atmosphere.

8.6 Some Common Reducing Agents 8.7 Oxidation, Reduction, and Living Things Plants use solar energy and reduce carbon dioxide to form complex carbohydrates. Animals oxidize complex carbohydrates for energy.

Demonstrations 1. Using tongs, hold a strip of magnesium metal in a flame. The oxidation reaction gives off an intense white light.

2. Place a small pile of iodine crystals mixed with zinc dust in the middle of a Pyrex watch glass. The elements do not react. Then add a little water with a medicine dropper. The reaction to produce zinc iodide proceeds with the evolution of heat and purple vapor from the sublimation of iodine.

3. Mix a little shampoo into some fresh 3% hydrogen peroxide and then add a little MnO2. The oxygen gas given off from the oxidation of the hydrogen peroxide produces foam.

58 .

Chapter 8: Oxidation and Reduction

4. To a small pile of fine crystals of potassium permanganate in a Pyrex watch glass add a little glycerol. The reaction starts slowly but produces a lot of white smoke and a violet flame.

Review Questions 1. The oxidation number is increased; the oxidation number is decreased.

2. An electrochemical cell is a two-compartment system, with an oxidation reaction taking place in one compartment (the anode compartment) and a reduction reaction taking place in the other compartment (the cathode compartment). The compartments are connected by a conductor that carries electrons and a salt bridge to carry counter ions. Multiple electrochemical cells are the components of batteries.

3. Oxidation occurs at the anode, and reduction occurs at the cathode. The species at the anode is giving up its electrons, and the species at the cathode is gaining the electrons.

4. In a fuel cell, the oxidizing agent and the reducing agent are supplied contiuously to the cell.

5. The porous partition or salt bridge allows ions to flow from one compartment of an electrochemical cell to the other, thus keeping the solutions neutral.

6. A half-reaction is an oxidation-reduction reaction broken into its two separate reactions, the oxidation reaction and the reduction reaction. The two reactions are added after the electrons of oxidation are made to equal the electrons of reduction.

7. zinc; the zinc case acts as the anode and is oxidized as the battery is discharged.

8. Alkaline cells have a porous zinc case, and the paste is moist manganese dioxide and potassium hydroxide. The dry cell contains a moist paste of graphite powder, manganese dioxide, and ammonium chloride.

9. Lead is converted from the sulfate to metallic lead, and lead oxide and sulfuric acid are produced via the reaction: 2 PbSO4 + 2 H2O  Pb + PbO2 + 2 H2SO4.

10. Lead and lead oxide are converted to the sulfate, and water is produced via the reaction: Pb + PbO2 + 2 H2SO4  2 PbSO4 + 2 H2O. 11. Lithium based-batteries are light weight, may be formed into nearly any shape, provide higher voltage than other battery types, and are rechargeable.

12. Iron is oxidized to iron(III) hydroxide; salt water acts as an electrolyte.

13. Silver tarnishes because it is oxidized by hydrogen sulfide to silver sulfide; use aluminum to reduce the silver sulfide to metallic silver.

14. Bleaches are oxidizing agents and react with dyes and bacteria in clothes.

59 .

Chapter 8: Oxidation and Reduction

15. Photosynthesis is a reduction process; energy for muscle contraction (which produces the heartbeat) is an oxidation.

Problems 16. (a) oxidized, loss of electrons (b) oxidized, loss of hydrogen atoms (c) oxidized, loss of electrons (d) reduced, gain of electrons (e) oxidized, loss of hydrogen atoms 17. (a) oxidation, from 0 to +4 (c) oxidation, from 0 to +4 (e) reduction, from 0 to –4

(b) oxidation, from 0 to +2 (d) oxidation, –2 to –1

18. (a) reduction, gain of hydrogen (c) reduction, gain of electrons (e) oxidation, loss of electrons

(b) reduction, loss of oxygen (d) reduction, gain of hydrogen

19. (a) Ca  Ca2+ + 2 e- (b) Al  Al3+ + 3 e(c) Cu  Cu+ + e- or Cu  Cu2+ + 2 e20. (a) I2 + 2 e-  2 I-

(b) N2 + 6 e-  2 N3-

(c) S8 + 16 e-  8 S2-

21. (a) Cu is oxidized, O is reduced. (b) Mg is oxidized, H is reduced. (c) O is oxidized, Br is reduced. (d) S is oxidized, O is reduced.

22. When photochromic lenses are removed from light, the copper is oxidized from +1 to +2.

23. (a) oxidizing agent: Fe2O3; reducing agent: C (b) oxidizing agent: O2; reducing agent: P4 (c) oxidizing agent: H2O; reducing agent: C (d) oxidizing agent: H2SO4; reducing agent: Zn 24. (a) oxidizing agent: CuS; reducing agent: H2 (b) oxidizing agent: CCl4; reducing agent: K (c) oxidizing agent: H2; reducing agent: C3H4 (d) oxidizing agent: Fe3+; reducing agent: Ce3+ 25. Cu 26. Ag+ 27. Zn is the reducing agent, MnO2 is the oxidizing agent.

28. (a) neither (b) both 60 .

Chapter 8: Oxidation and Reduction

29. The oxidizing agent is at the cathode, the reducing agent is at the anode.

30. Copper 31. oxidizing agent 32. The formation of the oxide ion, O2-, is thermodynamically favorable, thus oxygen undergoes a reduction making it a good oxidizing agent.

2+

-

+

33. (a) Fe(s)  Fe (aq)+ 2 e , oxidation 3+ (b) 2 Al(s)  2 Al (aq)+ 6 e , oxidation 3+

-

34. (a) 2 Al(s)  2 Al (aq)+ 6 e , oxidation 2+ (b) Mg(s)  Mg (aq) + 2 e , oxidation

-

2 H (aq)+ 2 e  H2(g), reduction 3+ 3 Cr (aq) + 6 e  3 Cr(s), reduction +

-

6 H (aq)+ 6 e  3 H2(g), reduction + 2 Cu (aq) + 2 e  2 Cu(s), reduction

35. (a) Fe(s)  Fe2+ + 2 e- oxidation Cu2+ + 2 e-  Cu(s) reduction Fe(s) + Cu2+  Cu(s) + Fe2+ (b) 2 H2O2  2 O2 + 4 H+ + 4 e- oxidation Fe3+ + e-  Fe2+ reduction 2 H2O2 + 4 Fe3+  2 O2 + 4 H+ + 4 Fe2+ (c) WO3 + 6 H+ + 6 e-  W + 3 H2O reduction C2H6O  C2H4O + 2 H+ + 2 e- oxidation 3 C2H6O + WO3  W + 3 H2O + 3 C2H4O 36. (a) 2 I-  I2 + 2 e- oxidation; Cl2 + 2 e- 2 Cl- reduction 2 I- + Cl2  I2 + 2 Cl +

(b) HNO3 + H + e-  NO2 + H2O reduction; +

SO2 + 2 H2O  H2SO4 + 2 H + 2 e- oxidation; 2 HNO3 + SO2  2 NO2 + H2SO4

37. (a) Mn  Mn5+ + 3 e- oxidation; Pb4+ + 2 e- Pb2+ reduction 2+ 2 Mn2+ + 3 Pb4+  2 Mn5+ + 3 Pb 2(b) 2 S2+  2 S2.5+ + e- oxidation; O + 2 e-  O reduction; 4 S2+ + O  4 S2.5+ + O22+

38. (a) SO2 is oxidized; HNO3 is the oxidizing agent. (b) HI is oxidized; CrO3 is the oxidizing agent.

39. (a) Carbon is oxidized and oxygen is reduced. (b) Carbon is oxidized and manganese is reduced.

40. In reaction “a,” nitrogen is oxidized. In reaction “b,” nitrogen is oxidized.

61 .

Chapter 8: Oxidation and Reduction

41. In reaction “a,” manganese is reduced. In reaction “b,” manganese is reduced.

42. reduced, it gains hydrogen.

43. The hydrogen is oxidized, and the unsaturated vegetable oil is reduced. Another way to think of this type of reaction, known as hydrogenation, is that the vegetable oil is gaining H and thus is being reduced.

44.Tantalum (V) oxide is reduced; sodium is the reducing agent.

45. I- is oxidized; Cl2 is reduced.

46. Zr was oxidized; water was the oxidizing agent.

47. reduced, gain of hydrogen atoms 48. The nitrogen in nitrite is reduced to form nitrogen monoxide; ascorbic acid is a reducing agent.

49. oxidized 50. Mn is being reduced, C is being oxidized. MnO2 is the oxidizing agent. Mn goes from +4 to + 2. C goes from 0 to +4.

51. Fe is being reduced, C is being oxidized. Fe(OH)3 is the oxidizing agent. Fe goes from +3 to + 2.

C goes from 0 to +4.

52. (a) SO2 (b) H2O + CO2 or CH3COOH (c) H2O + CO2 53. (a) NO (b) CO2 + S2 (c) CO2 + H2O 54. 0 in all three cases.

Additional Problems 55. (a) +2

(b) +4

(c) -4

(d) +4

(e) 0

56. Na, Ca, Cl, O 57. F, Cl, Br, I 58. Indoxyl is oxidized by exposure to the oxygen in the air. The oxygen in the air undergoes a reduction and is the oxidizing agent.

59. The coating of oil protects the metal tools from oxidation by the oxygen in the air.

2+

3+

60. 3 Cu (aq) + 2 Al(s)  2 Al (aq) + 3 Cu(s) 62 .

Chapter 8: Oxidation and Reduction

61. 2 CO2(g) + 2e-  C2O42-(aq); CO2 is reduced.

62. (a) Pb2+(aq) + H2S(g)  PbS(s) + H2(g) (b) PbS(s) + 4 H2O2(l)  PbSO4(s) + 4 H2O(l) 63. 15 breaths/min x 60 min/hr x 24 hr/day = 21,600 breaths/day 21,600 breaths/day x 0.5 L air/breath = 10,800 L air/day 10,800 L air/day x 0.21 L oxygen/L air = 2,268 L oxygen/day (Note: The final answer should have 1 significant figure, thus be 2000 L oxygen/day) 64. 80 g x 1 g/1000 kg = 0.080 kg 0.080 kg x 2000 L/1.0 kg = 160 L 65. (a) reduction (b) 6 CO2 + 24 H+ + 24 e-  C6H12O6 + 6 H2O 66. 0 to +4 67. +4 to 0 68. (a) oxidation; loss of electrons (b) neither (c) reduction; gain of hydrogen 69. Metallic elements lose electrons, making them candidates for oxidation. Nonmetals act as reducing agents when reacting with metals, but may act as reducing or oxidizing agents when reacting with other nonmetals.

70. TNT decomposes rapidly, producing several gasesous products.

71. H2(g) + Cl2(g)  2 HCl(g); Cl2 is the oxidizing agent.

72. (a) Ca is losing H, oxidation; Na is gaining H, reduction. (b) Ca is gaining electrons, reduction; Na is losing electrons, oxidation.

The loss or gain of hydrogen applies to neutral H. In the problem above, it is H- that is being lost and gained.

73. 2 Al(s) + 6 H2O(l)  3 H2(g) + 2 Al(OH)3; H2O is the oxidizing agent, Al is the reducing agent.

74.

2.0 x 10 4 L H 2 = 893 mol H 2 22.4 L mol 2 mol Al 893 mol H 2 x = 595 mol Al 3 mol H 2 595 mol Al x 27.0 g mol = 16,100 g 63 .

Chapter 8: Oxidation and Reduction

16,100 g = 6.0 x 10 3 cm 3 3 2.70 g cm

75. (a) CH3CH2OH  CH3CHO + 2 H+ + 2 e(b) CH3COCOOH + 2 H+ + 2 e-  CH3CHOHCOOH (c) CH3COCOOH + CH3CH2OH  CH3CHOHCOOH + CH3CHO 76. (a) oxidation; gain of oxygen (b) reduction; gain of hydrogen (c) oxidation; gain of oxygen (d) reduction; loss of oxygen (e) oxidation; loss of hydrogen 77. Industrial chemicals: a, d Specialty chemicals: b, c 78. d 79. The organic part of the catalyst is a target for oxidation.

80. Cr2O72- is the oxidizing agent; C2H6O is oxidized.

64 .

Chapter 9: Organic Chemistry Overview Chapter 9 discusses the chemistry of the carbon atom, the only element that has a major field of chemistry devoted to the study of its compounds. Of the tens of millions of known chemical compounds, more than 95% are compounds of carbon. The types of hydrocarbons are discussed, and various functional groups are introduced.

Lecture Outline 9.1 Aliphatic Hydrocarbons Carbon can form an almost infinite number of molecules of various shapes, sizes, and compositions.

Hydrocarbons: Compounds only containing H and C, often called saturated hydrocarbons.

Alkanes are hydrocarbons with only single carbon-to-carbon bonds.

Formulas and names for the first 10 hydrocarbons and their isomers are discussed.

Properties of the hydrocarbons are discussed.

Cyclic hydrocarbons are ring or loop structures.

Unsaturated Hydrocarbons: Alkenes and Alkynes Alkenes are hydrocarbons with at least one double carbon-to-carbon bond.

Geometric isomers are introduced Alkynes are hydrocarbons with at least one triple carbon-to-carbon bond.

9.2 Aromatic Compounds: Benzene and Relatives The structure of aromatic compounds and their structural features.

Aromatic hydrocarbons have delocalized electrons.

Substituted benzenes are discussed 9.3 Chlorinated Hydrocarbons: Many Uses, Some Hazards Chlorofluorocarbons (CFCs) are a common type of chlorinated hydrocarbon.

9.4 The Functional Group Functional groups determine the chemistry of the organic molecule.

Table 9.4 shows functional groups, and 9.5 shows some common alkyl groups.

9.5 Alcohols, Phenols, and Ethers The general structure and names of alcohols, phenols and ethers Alcohols are R-OH, such as ethanol CH3CH2OH. Phenols (an aromatic hydrocarbon and alcohol) Ethers R-O-R’ a phenol Common alcohols, phenols, and ethers and their uses are discussed.

9.6 Aldehydes and Ketones Some simple aldehydes and ketones, and their important properties are discussed.

OH

O

Aldehyde R–CHO R

C

H

O

Ketone – a carbon chain with a carbon - oxygen double bond R C R 9.7 Carboxylic Acids and Esters Some simple acids and esters, and their important properties are discussed.

General formula of an acid R-COOH Esters: The sweet smell of RCOOR’ 65 .

Chapter 9: Organic Chemistry

The art of organic synthesis: Green chemists find a better way Examples of more efficient and greener synthesis pathways are discussed.

9.8 Nitrogen-Containing Compounds: Amines and Amides Amine-NH2 , common amines are discussed.

Amide-oxygen double bonded to a carbon with a nitrogen next to the carbon.

Heterocyclic compounds: Alkaloids and others Compounds with atoms other than carbon in the ring are heterocyclic.

Demonstrations 1. On the overhead projector, place small beakers containing about 5–10 mL samples of an alkane and an alkene (e.g., cyclohexane and cyclohexene). Add one dropperful of dilute KMnO4 solution to each sample. The purple permanganate will remain unreacted in the alkane, but the alkene will be oxidized by the permanganate, and the purple color will disappear.

2. Place a small beaker containing a little ethyl alcohol on the stage of an overhead projector. Add some acidic dichromate solution and note the darkening of the mixture.

This is the reaction on which the “breathalyzer” test is based.

3. Place a little alcohol in a gallon milk jug, cap the bottle, and shake vigorously. Uncap the bottle, pour out the alcohol, and then replace the cap on the “empty” bottle until ready for demonstration. Hold a flame close to the mouth of the opened bottle and note the violent combustion.

4. Demonstrate a “carbide cannon” in which calcium carbide powder is reacted with water to form acetylene, which is then ignited by a spark.

Review Questions 1. Carbon atoms can bond strongly to one another, carbon atoms can bond strongly to almost all other elements, carbon atoms can bond in a variety of ways, single, double and or triple bonds; rings, chains and branches.

2. (a) Hydrocarbon: A molecule containing only hydrogen and carbon, butane (b) Alkyne: A molecule containing carbon-carbon triple bonds, acetylene (c) Alkane: A molecule containing carbon-carbon single bonds, propane (d) Alkene: A molecule containing carbon-carbon double bonds, ethene 3. Isomers are two or more compounds with the same formula but a different arrangement of atoms and have different properties. Geometric isomers have a different 3D shape but have the same atom connections, like cis-butene and trans-butene. Constitutional isomers are molecules with the same molecular formula but different atom connectivity order, like butane and 2-methylpropane.

4. Unsaturated hydrocarbons are compounds of hydrogen and carbon that contain a ring or double bond or triple bond (such as cyclohexane or ethene, H2C=CH2).

5. The circle indicates the six delocalized or resonance electrons.

66 .

Chapter 9: Organic Chemistry

6. Aromatic compounds (benzene, arenes, and phenols) are those in which electrons in double bonds are not tied down to one location in the molecule; a ring structure with a circle in the center is used to depict aromatic hydrocarbons. They often have alternating double and single bonds in a ring.

7. gases: 1–4 carbons atoms in the alkane liquids: 5–16 carbon atoms in the alkane solids: 18+ carbon atoms in the alkane 8. Liquid alkanes have densities less than 1.0 g/mL; water has a density if 1.0 g/mL. Octane (density of 0.V70 g/mL) and water will not mix, and since octane is less dense it will float as the top layer on the water.

9. (a) Grain: ethanol (b) rubbing: isopropyl alcohol (2-propanol) (c) wood: methanol 10. Excessive consumption can lead to liver damage, brain damage, and high blood pressure.

11. One of the first historical anesthetics; Present: solvent and starter fluid 12. Esters smell sweeter, often fruity. Carboxylic acids have a carboxyl group and are often not very pleasant smelling; esters have a carboxyl group and ether linkage off the alpha carbon.

O R

O

C

OH

R

Acid

13. (a) Alcohol (d) Ether

C

O-R'

R' is a Carbon, not Hydrogen

Ester

(b) Ketone (e) Carboxylic acid

(c) Ester (f) Aldehyde

14. Alkaloids are nitrogen containing heterocyclic bases (ring structures containing nitrogen); some common alkaloids are morphine, caffeine, nicotine, and cocaine.

15. Organic compounds can be identified as those compounds that contain a C-C or C-H bond. (a) organic (b) inorganic (c) inorganic (d) organic 16. (a) organic

(b) organic

(c) inorganic

17. (a) 6

(b) 9

(c) 5

(d) 5

18. (a) 6

(b) 6

(c) 6

(d) 16

(d) inorganic

19. CNH2N+2 (a) 8 carbons: C8H18 (b) 14 carbons: C14H30 20. Subtract 2 hydrogens from the general formula for each ring or pi bond. An alkyne has 2 pi bonds. #H = 8 (general formula for 5C = 12–2–2 = 8)

67 .

Chapter 9: Organic Chemistry

21. (a) butane (b) 1,3-butadiene (c) ethyne (acetylene) 22. (a) cyclobutane

(b) benzene

(c) cyclopentane

23. (a) pentane: C5H12, CH3CH2CH2CH2CH3 (b) decane: C10H22, CH3CH2CH2CH2CH2CH2CH2CH2 CH2CH3 or CH3(CH2)8CH3 24. (a) butane - CH3CH2CH2CH3 25. (a) methyl

(b) isobutene CH3CHCH3

(b) sec-butyl

26. (a) isopropyl 27. (a) ethanol (c) CH3OH

CH3

(b) tert-butyl (b) 1-propanol (d) CH3CH2CH2CH2CH2CH2CH2OH

28. (a) 2-butanol (or butan-2-ol) or sec-butyl alcohol

(c)

(b) 1-pentanol

(d)

OH CH3

1 2

29. (a)

(b)

30. They both have an OH group. Phenols have an OH group attached directly to a benzene ring while alcohols are attached to an alkane carbon. Phenols are significantly more acidic.

31. (a) CH3CH2CH2OCH2CH2CH3 (b) CH3CH2CH2CH2OCH2CH3 32. (a) CH3OCH2CH2CH3 (b) CH3OCH3 ,C2H6O, is the same formula as ethanol CH3CH2OH.

33. (a) (b) CH2O or H2C=O (c) butanal (d) 3-heptanone (or butyl ethyl ketone) 68 .

Chapter 9: Organic Chemistry

34. (a)

O

(b) CH3CH2CHO or CH3CH2CH=O or CH3CH2CH (c) 3-pentanone or diethyl ketone (d) benzaldehyde 35. (a) propanoic acid (b) pentanoic acid (c) CHOOH or

(d) 36. (a) acetic acid (ethanoic acid) (b) hexanoic acid (c) CH3CH2CH2COOH or (d) 37. (a) CH3COOCH3 or (b) CH3CH2CH2COOCH2CH3 or (c) ethyl propanoate

38. (a) CH3CH2CH2COOCH3 or O

(b) CH3CH2CH2COOCH3 or CH3CH2CH2C-OCH2CH3 (c) butyl formate

69 .

Chapter 9: Organic Chemistry

39. (a) (CH3CH2)3N: (b) CH3CH2NHCH3 (c) propylamine (d) ethylmethylamine 40. (a) CH3NH2 CH 3 CH

NH2

(b) CH3 (c) diethylamine (d) aniline 41. (a) same (b) same (c) isomer (they are 2-methylpentane and 3-methylpentane) 42. (a) isomer (they are 1-butanol and 2-butanol) (b) same – each one is a 4-carbon chain with a –NH2 on the second carbon 43. (a) homolog (b) none (different formulas: C5H10 and C5H12) 44. (a) identical – both are 2-methylpentane (b) isomers – both are C5H10 45. (a) unsaturated, alkene (b) unsaturated, alkene 46. (a) unsaturated, alkyne (b) saturated, alkane

47. Two carboxylic acids circled, and an alcohol

48. Carboxylic acid, circled, and an alkene 49. (a) heterocyclic, amine

(b) not heterocyclic, amine (with a cycloalkane)

50. (a) not heterocyclic, ether and cycloalkane

(b) heterocyclic, ether 70

.

Chapter 9: Organic Chemistry

51. (a) C17H26O4 (b) C17H24O3 (loss of 2 hydrogens and one oxygen = loss of water)

(c) Gingerol:

(d) Shogaol: (e) dehydration (loss of water)

52. (a) C6H12 - cyclohexane or HOCH2CHCH2CH2CH2CH2CH3 CH2CH3

(b) C9H20O

NH2 Cl

(c) C6H5Cl2N

Cl

(d) C5H9ClO2 HO O C

(e) C8H9O2N 53. (a) CH3CH2C

N H

CCH3

CH3

2H 2/Ni

(b) CH3CH=CCH2CH2CH2CH3

CH 3CH 2CH2CH2CH3

H2/Ni

CH3CH2CHCH2CH2CH2CH3

CH3

CH3 71 .

Chapter 9: Organic Chemistry

54. (a) (b) OH

(b) CH3CH2CHCH2CH3

55. (a) CH3CH2CH2CH2OH

CH2OH

(c)

(d)

(C6H5CH2OH)

56. homologous series 57. isomerism, all have the same molecular formula, C4H10O 58. 2 CH3OH + 3 O2

2 CO2 + 4 H2O

775 g CH 3OH 2 mol CO 2 x x 44.0 g/mol CO 2 = 1070 g of carbon dioxide 32.0 g/mol CH 3OH 2 mol CH 3OH

59. The equation is balanced already; 1 mol CH 3OH 445 g CH 2 = CH 2 x 46.0 g/mol CH 3CH 2OH =731 g.CH3CH2OH x 28.0 g/mol CH 2 = CH 2 1 mol CH 2 = CH 2

60. (a) pineapple flavor

From butanoic acid (butyric acid):

and ethanol: CH3CH2OH

(b) apricot flavor:

From butanoic acid (butyric acid):

and pentanol: CH3CH2CH2CH2CH2OH 72

.

Chapter 9: Organic Chemistry

61. a. The forces between ether molecules are the weaker dispersion forces (or van der Waals forces). b. the major force of attraction in ethanol is hydrogen bonding, a much stronger force. Since hydrogen bonding is stronger, it is harder (requires more energy) to break the molecules apart and have them boil/enter the gas state. The ether has enough energy to break the intermolecular forces of attraction and to be a gas at room temperature.

O

O

62. diacetyl CH 3 C

C

CH 3 Ketone functional groups (a di-ketone)

O

63. Hydroquinone:

O

para-benzoquinone:

64. (a) 2,3-dimethylbutane will boil lower. (b) Neopentane will boil lower. (c) 2,4-dimethylpentane will boil lower. (d) 2,2,4-trimethylpentane will boil lower Generalization: The more branched isomer will have a lower boiling point.

65. Ideal gas law: PV=nRT Density: D = m/V Moles (n); n = mass (m)/molar mass (M) Substitute moles formula into ideal gas law formula

PV =

m m PM PM RT Rearrange for density = = D= M V RT RT

Methane density = =

(1.00 atm) (16 g/mole) = 0.6651 g/L = 0.0006568 g/mL (0.0821 L atm/n K) (293 K)

Ethane density = =

(1.00 atm) (30 g/mole) = 1.25 g/L = 0.00125 g/mL (0.0821 L atm/n K) (293 K)

Propane density = =

(1.00 atm) (44 g/mole) = 1.83 g/L = 0.00183 g/mL (0.0821 L atm/n K) (293 K)

Butane density = =

(1 .00 atm) (58 g/mole) = 2.41 g/L = 0.00241 g/mL (0.0821 L atm/n K) (293 K)

Note that you can determine which gas is going to be densest. If they behave ideally, all have the same molar volume at STP and the densest gas will be the one with the highest molar mass. The better unit would be g/L. 73 .

Chapter 9: Organic Chemistry

66. 1. Extremely plentiful (but not renewable) 2. Not toxic 3. Not flammable 67. (b) Many organics are not soluble in water (“like dissolves like” and water is not organic).

68. (e) All are true.

69. Four solvents (toluene, tetrahydrofuran, dichloromethane, and hexane) were replaced with ethanol. Ethanol is renewable, plentiful and not as toxic as the other solvents and biodegrades easily in the environment. In addition, overall solvent use was reduced from 60,000 gallons to 6000 gallons per ton of sertraline. Minimizing use and the use of a single solvent over four others reduces waste and cost.

70. 1. Removed a hazardous solvent and now uses no solvent = much less waste and cost; 2. Using glycerol (a waste from biodiesel production) makes what was a waste, useful.

71. Faster reactions and less energy usage. Organics efficiently absorb this high energy form of radiation.

74 .

Chapter 10 Polymers: Giants Among Molecules Overview Chapter 10 introduces the student to the marvelous variety of materials made from polymers. The chapter starts by discussing the history of polymers, then discusses the various types of polymers and their uses. The chapter concludes with a look into the future developments and uses of polymers.

Lecture Outline 10.1 Polymerization: Making Big Ones Out of Little Ones Polymers are huge molecules made from many small molecules, monomers.

Natural Polymers: Wood, wool, and proteins are all natural polymers.

Celluloid: Billiard balls and collars The synthetic material celluloid is derived from natural cellulose Celluloid is used to make billiard balls and historically movie film, and to make collars stiff.

10.2 Polyethylene: From the Battle of Britain to Bread Bags Polyethylene is the simplest and cheapest polymer made from the monomer ethylene (CH2=CH2).

Three main types of polyethylene: LDPE, HDPE, and LLDPE Thermoplastics and thermosetting polymers and their properties 10.3 Addition Polymerization: One + One + One . . . Gives One! Polypropylene, polystyrene, vinyl polymers, and teflon are examples of addition polymers, which are made by adding monomers together to form a macromolecule. Table 10.1 shows some addition polymers.

10.4 Rubber and other Elastomers Natural rubber is isoprene, a naturally occurring polymer.

The importance of crosslinking is discussed, especially vulcanization (the addition of sulfur to isoprene to make the substance harder and more elastic).

Synthetic rubber, neoprene, and SBR are discussed.

10.5 Condensation Polymers Water is eliminated, and the monomers are added together to form polyamides and polyesters, with emphasis on terephthalates.

Nylon, polyester, and bakelite are examples of condensation polymers.

Other polymers: polycarbonates (BPA), composite materials and silicones 10.6 Properties of Polymers The glass transition temperature is an important property of polymers.

In crystalline polymers, the monomers are in line, giving high tensile strength.

In amorphous polymers, the monomers are randomly ordered.

10.7 Plastics and the Environment Problems with plastics; plastics last almost forever. Landfills, incineration, biodegradation, and recycling are ways of disposing of plastics.

Sustainable plastics are discussed, and Greener Polymers Plastics and Fire Hazards: Fumes given off during house fires are very dangerous owing in large part to the polymers in the modern home.

75 .

Chapter 10: Polymers: Giants Among Molecules

Plasticizers and Pollution: Plasticizers make some plastics, such as vinyl polymers, more flexible but resist decomposition and bioaccumulate.

Plastics and the Future: From medical uses to pipes and airplanes Demonstrations 1. Onto a Petri dish about half full of acetone, place a cup made of polystyrene foam. The cup will appear to melt away as the polystyrene dissolves in the acetone.

2. Beads of polystyrene can be mixed with beads of polyethylene and then separated by pouring the mixture into a beaker filled with water. The polyethylene will float, while the polystyrene will sink.

3. A sample of high-density polyethylene (HDPE) can be separated from a sample of low- density polyethylene (LDPE) using a 1:1 mixture of methanol and water. The LDPE will float, while the HDPE will sink. (The HDPE sample can be cut from a gallon milk jug. The LDPE sample can be cut from a waxy squeeze bottle. The samples might be cut in the shapes of an H and an L, respectively.)

4. Insert a copper wire into a cork, and holding the cork, heat the copper wire in a flame. Touch the hot wire to a piece of plastic and return the wire to the flame. If the flame turns green, the plastic contains a halogen. It is likely to be polyvinyl chloride (PVC). (This is the Beilstein test.)

5. If you hold a plastic foam egg carton in a flame, the flame becomes very sooty because the foamed plastic is polystyrene, which has a very high carbon to hydrogen ratio.

6. If you hold a piece of polyethylene in a flame, the odor will be like that of a burning candle.

7. Make “slime” by adding about 5 mL of saturated borax solution to 30 mL of 4% poly(vinyl alcohol) solution in a paper cup. Stir vigorously with a wooden stick until the mixture forms a gel.

Remove the gel from the cup and knead it into a ball. When laid on a flat surface, the ball flattens.

When pulled apart very slowly, it stretches. When pulled apart rapidly, it breaks. The “slime” can usually be stored in an air-tight plastic bag for at least a week. After a while, it will start to mold and should be thrown away.

8. Another polymer gel, similar to “slime,” can be made from a white glue, such as Elmer’s. Mix about 30 mL of the glue with 30 mL of water and mix well. (A few drops of food color can be added, if desired.) Then add about 20 mL of 4%, or saturated, borax solution and stir well until the mixture forms a semisolid mass on the end of the stirrer. Remove the gel and form it into a ball. Try bouncing the ball. The polymer can be treated and stored like “slime.” (Adding a few drops of Lysol to the borax solution will slow down the mold growth.)

9. Polystyrene is a glassy polymer, whereas polyethylene is crystalline. When you step on a transparent polystyrene cup, it shatters. When you step on a cup or bottle made of polyethylene, it bends but does not break.

76 .

Chapter 10: Polymers: Giants Among Molecules

10. Have six or more students stand before the class, an arm’s length apart from each other, with their hands clasped about 6 inches in front of their chests. They represent ethylene molecules. “Initiate the reaction” by unclasping the hands of the student on the far left, placing an object such as an eraser in her right hand. Her left hand then “attacks” the next “molecule,” grabbing his right hand and leaving his left hand free to “attack” the next one, in a “chain reaction.” At the end of the line, the last student has a free left hand, which you can grab as a “chain stopper.”

Review Questions 1. Polyvinyl chloride (PVC) has a chlorine atom in place of one of the hydrogen atoms on every other carbon atom of polyethylene. Polyethylene is prevalent, and some examples are plastic bags, garment bags for dry-cleaned clothing, garbage-can liners, some bottles and pipes, and wire coatings.

2. (a) Macromolecules: Large or giant molecules (b) Elastomer: Materials that when stretched return to their original shape when released (c) Copolymer: A polymer made from at least two different monomers (d) Plasticizer: Material added to a polymer to make it more flexible and soft 3. An addition polymer is one in which all the monomers are added in such a way that the polymeric product contains all the atoms of the starting monomers; the monomers usually contain a double bond.

4. Teflon is a polymer of tetrafluoroethylene (F2C=CF2); Teflon is chemically inert and nonflammable; it is used as a nonstick coating, for electric insulation, and for bearings and gaskets. ~CF2-CF2-CF2-CF2~ 5. Clear drinking cups are made from polystyrene; Foamed plastic coffee cups are styrofoam (polystyrene) and are made from the monomer styrene.

6. (a) Milk jugs: high-density polyethylene (HDPE); (b) 2-L bottles: PETE Polyethylene terephthalate 7. Celluloid is a semisynthetic material made from natural cellulose, like cotton, which is treated with nitric acid, and celluloid (cellulose nitrate) is formed.

8. Thermoplastics are polymers that can be softened by heat and pressure and then remolded; polyethylene is a good example.

9. Most fibers used in the United States today are synthetic; they can have the strength and rigidity of steel but weigh a fraction as much, and they have a wide range of properties.

10. PCBs are polychlorinated biphenyls; they are toxic. They degrade slowly in nature (i.e., they are around a long time), and their solubility in animal fat leads to their concentration in the food chain.

Animals that eat PCB-contaminated food even for short periods of time get liver damage and may die.

77 .

Chapter 10: Polymers: Giants Among Molecules

11. To recycle plastics they must be sorted, chopped, melted, and remolded or spun into fiber.

12. (a) When plastics are discarded into the environment they litter the sides of the roads, parks, and sidewalks and can clog the digestive tract of fish. (b) Plastics last a long, long time and occupy a lot of space in landfills. (c) If the incineration is performed incorrectly, atmospheric pollution can occur.

13. LDPE has a lot of branching carbons chains on the main polymer chain. This prevents the polymers from packing very close together, resulting in a more flexible plastic.

14. LLDPE is a copolymer of ethylene and a small branched alkene. This combination gives a mostly linear polymer (like polyethylene), but with shorter branches than in LDPE. This combination of linear structure with some branches gives it higher strengths than LDPE, while maintaining good flexibility.

15. (a) CH2=CH2

(b)

16. (a) CH3CH2=CH2

(b) H2C=CHCl

17. (a) ~CH2CH(C≡N)CH2CH(C≡N)CH2CH(C≡N)CH2CH(C≡N)~

or

~CH 2CHCH 2CHCH 2CHCH 2CH~ CN

CN

CN

CN

HFH FH FH F ~C-C-C-C-C-C-C-C~ (b) ~CH2CF2CH2CF2CH2CF2CH2CF2~ or H F H F H F H F FFF F FF F F ~C-C-C-C-C-C-C-C~ FFF F FF F F 18. (a) . ~CF2CF2CF2CF2CF2CF2CF2CF2~ or CH3 H ~C

C

C

CH3 C

C

O

OCH3

H

CH 3 H

CH3

C

C

C

C~

C

H O C

H

H

H

(b)

H

C

O

OCH3

OCH 3

O

OCH 3

78 .

Chapter 10: Polymers: Giants Among Molecules

19. (a) ~CH2CH(CH2CH2CH3)CH2CH(CH2CH2CH3)CH2CH(CH2CH2CH3)CH2CH(CH2CH2CH3)~ ~CH2 CHCH2 CHCH2 CHCH2 CH~

Pr

or

CN

H

(b)

Pr

Pr CN

H

C

C

C

H O C

~C

H

Pr CN

H

H

C

C

C C

Where Pr = -CH2CH2CH3

CN C~

H H C O C O

O

OCH3 OCH3 OCH3 OCH3 H

H

H

H

~C H

C

C

O

C H

O

C

O

C

(b)

H

H

H

C

C H

C~

O

C

O

O O C

CH 3

CH3

CH3

20. (a)

H C H

CH3

H

H

H

H

H

H

H

H

~C H

C

C H

C

C H

C~

C

O

C

O

C C H C O

C

OCH3

OCH3

OCH3

OCH3

O

O

21.

22. Vulcanization cross-links the hydrocarbon chains with sulfur atoms, making the natural rubber harder and stronger, and improves its elasticity.

23. The molecules that make up rubber can be coiled, twisted, and intertwined. Elasticity is when stretched rubber is released; the cross-links pull the chains back to their original arrangement.

Rubber bands owe their snap to this sort of molecular structure. Other polymers are not coiled and cannot stretch much.

24. Polybutadiene has only a fair tensile strength and poor resistance to gasoline and oils. Natural rubber is a polymer of isoprene (2-methylbutadiene), while polybutadiene is made from the monomer butadiene (it is lacking the C2-methyl group).

25. Styrene-butadiene rubber (SBR) is a block copolymer made from styrene and butadiene.

79 .

Chapter 10: Polymers: Giants Among Molecules

26. (a) Polybutadiene, it is made from butadiene. (b) Polychloroprene (neoprene), it is made from a CH2=C(Cl)-CH=CH2 a monomer similar to isoprene, but a chlorine replaces a methyl group. (c) Styrene-butadiene rubber (SBR), it is made from styrene and butadiene.

27. Nylon 88: ~CO(CH2)6CO-NH(CH2)8NH-CO(CH2)6CO-NH(CH2)8NH~ 28. ~CH2(C6H10)CH2-O2C(C6H4)CO2-CH2(C6H10)CH2-O2C(C6H4)CO2~

CH2

CH 2

O

O

O

C

C

O

n

or

29. The polymer made from glycolic acid is a condensation polymer.

HO-COCH2-OH + HO-COCH2-OH + HO-COCH2-OH + HO-COCH2-OH  ~O-COCH2-O-COCH2-O-COCH2-O-COCH2~ + nH2O

Or 30. ~NH(C6H4)HN-OC(C6H4)CO-NH(C6H4)HN-OC(C6H4)CO~ NH

NH

O

O

C

C n

or 31. atom and element < compound < polymer

32. (a) Plastics in everyday life generally mean synthetic polymers such as bottles and plastic sacks. (b) In chemistry, plastics mean any long chain polymer, natural, or synthetic.

33. The Glass transition temperature (Tg) is the temperature at which the properties of the polymer change from hard, stiff, and brittle to rubbery and softened; rubbery materials such as automobile tires; squeezable bottles should have a low glass transition temperature; items like computer cases and glass substitutes should have a high Tg. 34. Plasticizers break up the rigid structure of the plastics and lower the glass transition temperature.

35. monomer: a;

repeating unit: b;

polymer: c;

it is an addition polymer

36. condensation polymer, it loses a water molecule (H2O).

37. CH2=CCl2 and CH2=CHCl

80 .

Chapter 10: Polymers: Giants Among Molecules

38. CH2=CH(C≡N) and (C6H6)-CH=CH2 or

39.

CN

CN

CH2=C

CH2=C

CN repeat

CH2C

n

C O

C O

C O

OCH2(CH2) 6CH3

OCH2(CH2) 6CH3

OCH2(CH2) 6CH3 the octyl group

40. (a) Thermoplastics will start to soften upon heating. (b) Thermosetting will start to discolor and decompose upon heating.

41. ~CH2C(CH3)2CH2C(CH3)2CH2C(CH3)2CH2C(CH3)2~ or CH 3

CH 3

CH 3

CH 3

~CH 2C-CH 2-C-CH 2-C-CH 2-C~ CH 3

CH 3

CH 3

CH 3

CH 3

CH3

42. Many answers are possible.

H2 C

H2 C

C

C H

C H

only one iosprene unit shown

CH 3

4 isobutylene units CH 3

CH 3

CH 3

CH 3

~CH 2C-CH 2-C-CH 2-C-CH 2-C=CHCH2 CH2 C~ CH 3

CH 3

CH 3

CH 3

43. HOCH(CH3)CH2COOH same as CH3CH(OH)CH2COOH or

81 .

CH 2

CH 3

Chapter 10: Polymers: Giants Among Molecules

44. (a) Stanyl nylon 46 uses the diacid HOOCCH2CH2CH2CH2COOH. HOOC

COOH

PA4T uses O H N

(b) Nylon 46:

H N

(CH 2)4

(CH 2)4

H N

H N

C

O (CH 2)4

C

O

O

C

C

(c) PA4T

n

n

45. There cannot be a CH3 at the end because it would terminate the sequence (carbon can only have 4 bonds total, this shows 5). This CH3 should not be part of the chain but as a branch to the main chain. The CH is lacking a bond; once again carbon should have 4 bonds, moving the CH3 would fix this issue 46. The monomers of PEN are;

(b) PEN is a polyester 47. An addition polymer from butadiene; elasticity: golf balls > superballs > rubber balls

48.

49. If butanol was substituted, a polymer would not form, but instead the diester below will form:

82 .

Chapter 10: Polymers: Giants Among Molecules

50.

51.

52. They both have a chlorine on carbon 2 and a bromine on carbon 3. Although they are different compounds to start with (they are E/Z stereoisomers of each other) once polymerization takes place, the polymer has all single bonds and the differences no longer exist.

O H2C

O

+

CH2 H2C

S

53. H2C

+

CH2 CH2 O CH2 CH2 O CH2 CH2 O CH2 CH2 O CH2 CH2 O

n

CH2 CH2 S CH2 CH2 S CH2 CH2 S CH2 CH2 S CH2 CH2 S

n

CH2

S

CH2 H2C

CH2

O H 3N

Glycine O

O

C

H N

CH2 N H

54.

O

C

O

C CH2

H N

CH2 N H

O

C

C CH2

n O

55. (b) Polylactic acid is made from corn.

56. No (or much slower), biodegradable polymers usually need specific conditions (temperature and moisture) to degrade. These conditions do not often present themselves in landfills.

57. c. PET and HDPE 58. Thermoplastics can be melted (with heat) and re-formed into useful products again. (Some polymers can be depolymerized (with heat) and remade.)

83 .

Chapter 11: Nuclear Chemistry Overview Chapter 11 discusses the structure of the nucleus, radioactivity, half-life, artificial transmutation, and induced radioactivity. These concepts are then applied to nuclear medicine, radioisotopic dating, the nuclear bomb, and nuclear power.

Lecture Outline 11.1 Natural Radioactivity Background Radiation: Natural sources of radiation The two particles of the nucleus are the proton (+) and the neutron (neutral). Two atoms with the same number of protons but different number of neutrons are isotopes – some of which are radioactive.

11.2 Nuclear Equations The equations are simple if you follow the rule that both mass and charge are conserved. That is, the sums of the nucleons and nuclear charges must be the same before and after a reaction.

Another way to think about the problem is that the sum of the superscripts on the left side of the reaction arrow must equal the sum of the superscripts on the right side of the arrow. The same rule holds for the subscripts.

11.3 Half-Life and Radioisotopic Dating Half-life: The time it takes for half of a radioactive substance to disintegrate.

Use of the half-lives of radioactive isotopes and assumption of constant decay rates over time make it possible to determine the age of items.

11.4 Artificial Transmutation Bombardment of various nuclei with protons, neutrons, alpha particles, or other fundamental particles brings about transmutation, each element being changed into another.

11.5 Uses of Radioisotopes The main use of radioisotopes is as tracers. Because radioisotopes are radioactive, their locations can be traced.

Radioactive isotopes are used for diagnostic tests as well as for therapy. More people have had their lives improved by the use of radioactive materials than have been killed or injured!

Table 11.6 Radioisotopes and Their Medical Applications 11.6 Penetrating Power of Radiation Gamma > Beta > Alpha 11.7 Energy from the Nucleus Einstein and E = mc2 and the Binding Energy Chain reactions, fission, and fusion 11.8 Nuclear Bombs The building of the bomb, how uranium and plutonium bombs are used.

The highly secret Manhattan Project was set up to do build the bomb.

Radioactive fallout and its implications 11.8 Uses of Nuclear Energy From bombs to nuclear power plants Figure 11.16 lists some constructive uses of nuclear energy.

Green remediation of nuclear waste

84 .

Chapter 11: Nuclear Chemistry

Demonstrations 1. Demonstrate the use of a Geiger counter. Using a compound of thorium or uranium, show the large amount of radioactivity in such a compound. (If you can find an old piece of orange dinnerware, it can be compared with a piece of some other color. The old orange pieces were glazed with uranium oxide and are quite radioactive.)

2. Using the alpha, beta, and gamma emitters that came with the Geiger counter (or can be purchased separately), show that a sheet of paper will stop alpha rays, and a sheet of aluminum will stop alpha or beta rays, but neither of these will stop gamma radiation. A thin piece of lead foil will readily block out alpha or beta radiation, but it takes several cm of lead to block out gamma radiation.

3. Show (qualitatively) that as you move the detector away from the source of radiation, the count goes down significantly. (If you move just 3 feet away from a radioactive source, you get only 1/9 as much radiation.)

Review Questions 1. a. rocks and soil, d. radon, and e. cosmic rays 2. b. nuclear medicine and c. nuclear power plants 3. (i) a new compound is formed: (b) chemical (ii) a new element is formed: (a) nuclear (iii) size, shape, appearance, or volume is changed without changing the composition: (c) physical 4. a. Half-life: The time it takes for half of a radioactive sample to disintegrate (t½) b. Positron: Positive electron −10 e c. Background radiation: Radiation present in nature due to natural and artificial sources d. Radioisotope: A radioactive isotope of an element 5. (a) 42 He , The nucleon number decreases by four, and the atomic number decreases by two. (b) 00γ , Neither the nucleon number nor the atomic number changes. (c) 11 p , the nucleon and atomic number each decrease by 1.

6. (a) −01 e , The nucleon number is unchanged; the atomic number increases by 1. (b) 01 n , The nucleon number decreases by one; there is no change in the atomic number. (c) 01 e , The nucleon number does not change; the atomic number decreases by 1.

7. False, 25% would remain.

8. 1.00 to 0.125 = 3 half lives = 3 x 17 = 51 days

85 .

Chapter 11: Nuclear Chemistry

5.00ug 5.00ug = 0.625ug or = 0.625 3 2x2x2 2

9. 33.0 s/11.0 sec = 3 half lives.

10. A naturally occurring transmutation is when an element decays on its own to become a more stable element, such as the alpha and beta decay series of uranium. Artificial transmutation occurs when the nuclear reaction is induced by humans.

11. Technetium-99 produces no alpha or beta particles, and it has a short half-life (both of which result in less patient radiation). It is easily concentrated into tissues of interest.

12. (a) heavy alpha (b) gamma 13. Use shielding; move farther away from the source.

14. The missing mass is converted to energy and is used to hold the protons and neutrons together in the nucleus, it is referred to as the binding energy.

15. 1. Sustaining a nuclear-fission chain reaction 2. Enriching uranium so that it contained about 90% of the fissionable isotope U-235 3. Making plutonium-239 (another fissionable isotope) 4. Constructing a bomb based on nuclear fission 16. Alpha particles do not have a lot of penetrating power, and you can easily protect yourself from alpha emitters. However, once they enter your body, you are no longer protected. They are large particles with lots of energy, so they easily cause damage to unprotected tissue.

17. Nuclear power plants, nuclear weapons, manufacturing radioisotopes for medical diagnostic and therapeutic uses, smoke detectors, and others may apply.

18. (b) comes from many human and natural sources.

19. (b) microwaves.

20. (c) increasing the number of red blood cells.

246 4 21. a. 250 98 Cf → 96 Cm + 2 He 210 0 b. 210 83 Bi → 84 Po + −1 e 117 0 c. 117 53 I → 52Te + 1 e

169 4 22. a. 173 79 Au → 77 Ir + 2 He 138 0 b. 138 53 I → 54 Xe + −1 e 0 104 c. 104 48 Cd + −1 e → 47 Ag

86 .

Chapter 11: Nuclear Chemistry 175 4 23. a. 179 79 Au → 77 Ir + 2 He

b. 126 C + 21 H → 136 C + 11 H 1 1 153 c. 154 62 Sm + 0 n → 2 0 n + 62 Sm 99 m 0 24. A beta particle; 99 42 Mo → 43Tc + −1 e

m 25. 99 43 Tc → 9943Tc + 00γ ; gamma radiation 1 1 24 26. Sodium-24, 24 12 Mg + 0 n → 1 p + 11 Na 4 233 27. 237 93 Np → 2 He + 91 Pa; neptunium - 237 211 4 28. 215 85 At → 83 Bi + 2 He; astatine - 215 1 108 29. 107 47 Ag + 0 n → 47 Ag 108 47

Ag →

0 −1

e+

108 48

Cd

neutron absorption beta emission

210 0 30. 210 85 At → 86 Rn + −1 e 210 86

0 Rn → 210 87 Fr + −1 e B is Francium-210

Half-Life 31. Divide 2.00 μg in half until you get to 0.0625 μg = this series: 2.00, 1.00, 0.50. 0.25, 0.125, 0.0625 = divided in half 5 times.

2.00 = 0.0625 ; solve for n: 2n = 32, nlog(2) = log(8), Or n 2 0.301n = 0.903, n = 5 half lives 5 half-lives x

32.

20,000 y = 100,000 years 1 halflife

12.0 g 6.0 g 3.0 g = 6; = 3; = 1.5 g ; 3 half lives or 3x8.07 days = 24.2 days 2 2 2

33. (b)

87 .

Chapter 11: Nuclear Chemistry

24hrs = 4t 1 / 2 ; 4 half-lives have passed after 24 hours.

6hrs / t 1 / 2 48 = 3 μCi (48 divided in half 4 times) 24 48hrs = 8t 1 / 2 ; 8 half-lives have passed after 48 hours. (b) 6hrst 1 / 2 48 = 0.19 μCi 28

34. (a)

35. From 1000 to 525 counts/s is about half = 80 days.

From 851 to 446 counts/s is about half = 80 days.

From 616 to 323 counts/sec is about half = 80 days.

About 80 days 36. Half the activity is one half life = 5730 years old 25% (one-fourth) of the activity is 2 half-lives, 5730 x 2 = 11460 years old 37. 20,000 counts/min to 10,000 counts/min is 78.1 hours (from table 11.6) 10,000 counts/min to 5000 counts/min = 78.1 hours Total time = 156 hours 38. Determine the number of half-lives by dividing in half until you reach 1.5.

12 12/6/3/1.5 = 3 half-lives. Or n = 1.5 ; solve for n: 2n = 8, n = 3 half lives 2 8.07days × 3 half - lives = 24.2 days. half - life 39. Amount of radioactive potassium-40 = 600 mg x (0.0117 mg/100 mg) = 0.0702 mg 0.070 mg (1.00 g/1000 mg) = 0.0000702 g 1.00 g 0.000070 g ; solve for x. x= 18.5 Bq = 5 x Bq 2.63x10 Bq 40. (b) 16 d. Loss of 75% = 25% remains; = 2 half-lives.

41. Number of half-lives undergone = 10 to 5 to 2.5 to 1.25 = 3 half-lives 75 days/3 half-lives = 25 days/half-life 42. 8 is half of 16, so 1 half-life has passed; 5730 yrs, or 48 293 1 43. 20 Ca + 247 197 Bk → 117Te + 2 0 n Two neutrons released 48 20

294 1 Ca + 247 197 Bk → 117Te + 0 n One neutron released

88 .

16 = 2 2 n = 2 n = 1 half-life 8

Chapter 11: Nuclear Chemistry 262 4 44. 266 109 Mt → 107 Bh + 2 He 262 107

4 Bh → 258 105 Db + 2 He

238 302 45. a. 64 28 Ni + 92 U → 120 X 58 302 b. 26 Fe + 244 94 Pu → 120 X

248 302 c. 54 24 Cr + 96 Cm → 120 X 237 4 46. An alpha particle; 241 95 Am → 93 Np + 2 He

47. (c) 48. Alpha particles do not have a lot of penetrating power and you can easily protect yourself from alpha emitters. Gamma rays easily penetrate into your body.

1 236 236 92 141 1 49. 235 92 U + 0 n → 92 U then 92 U → 36 Kr + 56 Ba + 3 0 n The 3 neutrons released are absorbed by other U-235 atoms, creating a chain reaction.

50. Fission is the splitting of large atoms. Fusion is the joining of small atoms. Both liberate energy because the greatest mass defect occurs at iron-56. Fission of very large nuclei or fusion of very small ones results in greater nuclear stability. See Figure 11.8.

51. a. E = mc2; 1 g = 0.001 kg E = (0.001 kg)(3.00 x 108 m/s)2 = 9 x 1013 kg m2/s2 = 9 x 1013 J 1 cal = 2.15 x 1013 cal 1 cal = 4.184 J; 9 x 1013 J 4.184 J 2.25 x 1013 cal = 2.15 x 1010 kcal b.

2.15 x 1010 kcal = 1.95 x 10 8 bowls = 195 million bowls!

110 kcal/bowl

52. They are separated by their atomic masses and the speed at which their gas phase molecules move. 235 UF6 is slightly lighter than 238 UF6 and the separation is achieved by passing the gas through a series of pinholes. The lighter one moves faster and gradually outdistances the heavier isotope and is isolated (enriched).

53. U-238 was bombarded with neutrons to produce U-239, which underwent two beta decays to first form Np-239 and then Pu-239.

54. The logic behind the design of the bombs was to keep several pieces of fissionable material, below their critical mass, separated from each other in the bomb. Then all the pieces would be forced together using an ordinary explosive, combining them into a single piece above the critical mass, thus triggering a runaway nuclear chain reaction and an explosion.

89 .

Chapter 11: Nuclear Chemistry

55. U-235 absorbs a neutron to form an unstable U-236 nucleus, which then splits (undergoes fission) to produce other elements (such as Sr-90 and Xe-143), and more neutrons and lots of energy are released.

56. Most of the radioactive uranium is still present in the fuel rods. As uranium-235 nuclei undergo fission, the nonfissionable (and very concentrated) uranium-238 nuclei also absorb neutrons and are converted to plutonium-239. Plutonium-239 can be easily separated from the other fission products by chemical means. Plutonium bombs require less sophisticated technology to produce than is needed to make nuclear weapons from uranium. If it falls into the wrong hands, it could be used for weapons instead of energy production.

57. Many apply: Power plants control the rate of the fission reaction, so it is slow, continuous, and controlled. Bombs’ fission reaction rates are almost instantaneous; all the energy is released at once and is not controlled. Power plants use control rods to regulate the amount of neutrons and how much material is above the critical mass. Bombs have no control rods, and it is all brought above the critical mass at once. People can be nearby and live during the operation of a nuclear power plant; you need to be a long ways away during the operation of a nuclear bomb!

58. a. 105 B + 01 n → 42 He + 73 Li 23 0 b. 10 Ne → 23 11 Na + −1 e 1 121 1 c. 121 51 Sb + 1 p → 52Te + 0 n 206 4 59. U-206; 210 94 Pu → 92 U + 2 He

60. 23290 Th + 01 n → 23390Th 233 90 233 91

Th → −01 e + 233 91 Pa Pa →

0 −1

e+

233 92

neutron absorption beta emission

U (U-233 (and 235 are fissionable isotopes.)

219 4 61. 223 88 Ra → 86 Rn + 2 He (alpha emission) 223 88

14 Ra → 209 82 Pb + 6 C (carbon-14 emission)

234 4 62. U-238; 238 92 U → 90Th + 2 He

63. 94 Be + 42 He → 126 C + 01 n ; This was the first experimental detection of the neutron.

64. One half-life = 50%, 2 half-lives = 25%. 13 sec/half-live x 2 half-lives = 26 sec.

65. Divide 100 by 2 until you get below 1: Approximately this series 100,50,25,12,6,3,1.5,0.75 = 7 half-lives. 11.4 days/half-life x 7 half-lives = Approximately 80 days 100 Or n = 1.00 ; solve for n; 2n=100, n = 6.6 or approximately 7 half-lives 2 90 .

Chapter 11: Nuclear Chemistry

66. From 4:00 p.m. Friday to 8:30 a.m. Monday is 64.5 hours.

64.5hrs = 8t 1 / 2 ; 8 half lives have transpired. Divide 500 counts/min by 2, 8 times 8.1hrs / t 1 / 2 500 = approximately 2 counts/min on Monday morning 28 67. Divide 3920 by 2 until you get 980: = 2 half-lives.

5730 years/half-life x 2 half lives = Approximately 11,460 years old 68. 12.5% is 3 half-lives (From the series 100-50-25-12.5) 12.26 years/half-life x 3 half-lives = 36.8 years old Likely bottled around 1977, not authentic!

69. Divide 56 ppm/m3 by 2 until it is below 14 (56-28-14) = 2 half lives It will be at 14 ppm/m3 in 12.7 hrs/half life x 2 half-lives = 25.4 hours or approximately one day.

70. March 15 to April 21 is 37 days.

37 = 2.59 half-lives.

14.28 X = 1.24 x1017 atoms 2.59 2 X = (1.24 x1017 ) x 22.59 atoms Approximately 7.5 x 1017 atoms of P-32 were originally present.

71. a. 3.6 billion years/1.2 billion years = 3 half-lives. 87.5% of patassium-40 has undergone decay.

(12.5% remains) b. Carbon-14 has a half-life of 5730 years; in 3.6 billion years the amount of carbon-14 remaining would be undetectable.

4 124 1 72. (a) 121 51 Sb + 2 He → 53 I + 0 n 0 124 (b) 124 53 I → 1 e + 52Te

73. 0.25 mg = 0.00025 g; (1.26 x 1011Bq/1.0 g) x 0.00025 g = 31.5 x 106 Bq 74. D = m/V rearrange to give V = m/D V = (49,000 g) / (18.75 g/cm3) V = 2,613 cm3 Smaller than a volleyball based on the references below V = (16,000 g) / (18.75 g/cm3) V = 853 cm3 Larger than a baseball 91 .

Chapter 11: Nuclear Chemistry

Volume of a sphere = 4/3 π r3 Volume of Common Spheres Radius (cm) Baseball 3.56 Volleyball 10.54 Basketball 12.17

Volume (cm3) 189 4905 7550

75. d = m/V V = m/d V = (16,300 g) / (19.1 g/cm3) V = 853 cm3 Larger than a baseball (see previous question.)

V = (2,500 g) / (19.1 g/cm3) V = 131 cm3 A little smaller than a baseball

92 .

Chapter 12: Chemistry of the Earth Overview Chapter 12 discusses the components of the Earth and the use that modern civilization makes of the Earth’s constituents. The chapter begins by describing the substances in the Earth’s crust and how they formed. Next, the chapter discusses the uses and abuses which modern civilization makes of the Earth’s resources.

Lecture Outline 12.1 Spaceship Earth: Structure and Composition The Earth is mostly oxygen, silicon, and hydrogen.

Organic: coal, petroleum, and natural gas Inorganic: silicates, carbonates, oxides, and sulfides 12.2 Silicates and the Shapes of Things The silicates are quartz, mica, and asbestos.

Modified Silicates: Ceramics, Glass, and Cement 12.3 Metals and Ores The most important metals are copper, iron, and aluminum.

12.4 Earth’s Dwindling Resources Land Pollution: Solid Wastes The Three R’s of Garbage: Reduce, Reuse, Recycle How Crowded Is Our Spaceship?

Demonstrations 1. Make a “super ball” by adding 6–8 mL of ethyl alcohol to 25 mL of sodium silicate in a paper cup.

Stir with a wooden stick until the mixture forms a solid material. Wearing disposable plastic gloves, press the solid between the palms of the hands and roll it into a ball.

2. Using a plain or a quartered petri dish, place four pieces of rock (e.g., limestone, marble, sandstone, and granite) in the dish. With a dropper, add a little 3 M HCl to each rock. If bubbles form, the rock is probably calcium carbonate. Limestone and marble are two of the many forms of calcium carbonate.

Review Questions 1. enhancement of the effect of one substance by the effect of another; asbestos fibers and cigarette smoke 2. 1:1 layers and 2:1 layers 3. Montmorillonite is a superabsorbent 2:1 clay that swells, causing the cat litter to clump.

4. by adding gold 5. mining operations, particulate matter from crushing operations, and high energy consumption 93 .

Chapter 12: Chemistry of the Earth

6. Copper is easier to obtain from ores than from other metals.

7. Copper and tin are easier to obtain from ores than is iron.

8. Steel is iron mixed or alloyed with carbon. Steel is stronger and more malleable than pure iron.

9. Aluminum is thinly distributed in aluminum-containing clays.

10. Aluminum production creates considerable waste and consumes large amounts of electrical power. Aluminum is highly recyclable.

11. open dumps: infestation of rats and flies; sanitary landfill: land consumption and potential groundwater contamination; incineration: possible air pollution if not properly operated 12. We will not run out of metal. The problem is keeping the metal in a usable form and collected in one place (low entropy).

13. Metals can be recycled because of the large amounts of energy required in the original smelting process as compared with recycling existing metals. The main limiting factor in recycling metals is transporting the metals to a central location where economic recycling can take place.

Problems 14. c; The inner core of the Earth is solid.

15. d 16. lithosphere: the solid portions of the Earth, land hydrosphere: the wet portions of the Earth, oceans atmosphere: the gaseous portions of the Earth, air 17. inorganic (rocks and minerals), organic (human materials), biological (flora and fauna) 18. silicate minerals: asbestos, sand, quartz; carbonate minerals: limestone, marble; oxide minerals: magnetite, corundum; sulfide minerals: galena, cinnabar 19. silicon, hydrogen, sodium, iron, calcium, magnesium, potassium (refer to Table 12.1) 20. silicon, iron, cacium, sodium, potassium, magnesium, hydrogen (refer to Table 12.1) 21. The masses of the various elements are different.

22. living creatures and their remains 23. +3 94 .

Chapter 12: Chemistry of the Earth

24. PbS + 2 HCl  PbCl2 + H2S 25. Fire allowed early humans to modify silicates to form glass, ceramics, and cement.

26. Quartz is pure silicon dioxide (SiO2); Si is surrounded by four O atoms in the shape of a tetrahedron.

27. Silicon is surrounded by 4 oxygen atoms forming a basic unit of SiO4 tetrahedra.

28. The SiO4 are tetrahedra with the Si atom at the center. Diamond consists of carbon atoms bound in a tetrahedron.

29.

30. Muscovite consists of SiO4 tetrahedra arranged in two-dimensional sheetlike arrays.

Wollastonite consists of SiO4 tetrahedra arranged in strands.

31. Mica consists of SiO4 tetrahedra arranged in two-dimensional sheetlike arrays.

32. Asbestos is a double chain of SiO4 bound in a tetrahedron, giving it a fiberlike appearance.

33. a 34. c 35. Oxygen; no 36. Heating breaks the weaker bridging interactions within the crystalline matrix.

37. Weaker bonds break first, causing glass to soften gradually.

38. In the amorphous phase, the bonds connecting the SiO4 tetrahedra differ in length.

39. The addition of extra elements changes the properties of glass.

40. silicon carbide and tungsten carbide 41. Ceramics, like glass, use a heating and cooling process to cause a change in property.

42. limestone (CaCO3) and clay (aluminum silicates)

95 .

Chapter 12: Chemistry of the Earth

43. Concrete is cement mixed with gravel.

44. Aluminum is found as bauxite (Al2O3), copper is found as chalcopyrite (CuFeS2), and iron is found as hematite (Fe2O3).

45. (a) Cu is reduced from +1 to 0. (c) Al is reduced from +3 to 0.

46. (a) 4

(b) 6

(b) Fe is reduced from +3 to 0.

(c) 12

47. chemical reduction; Fe2O3 + 3 CO  2 Fe + 3 CO2 .

48. oxidation; 4 Fe + 3 O2 + 3 H2O  2 Fe2O3 3 H2O 49. (a) The addition of tin hardens the pure copper. (b) bronze (c) classic bronze, containing 10% tin is harder 50. The addition of carbon hardens the iron. Quenching creates a supersaturated carbon-iron alloy.

51. Copper is used in plumbing and electrical wiring. Iron is used in structures such as buildings, bridges, and automobiles.

52. Red-mud is aluminum waste. destroy life in rivers 53. Aluminum is the most energy intensive. The reduction of Al from its ore is achieved by electrolysis.

54. Recycling Al requires 5% of the energy as producing Al from its ore.

55. careful mixing of carbon and steel and differential quenching 56. V2O5 + 5 Ca  2 V + 5 CaO (a) V is reduced. (b) Ca is the reducing agent. (c) Ca is oxidized. (d) V2O5 is the oxidizing agent.

57. 2 ErF3 + 3 Mg  2 Er + 3 MgF2 (a) Er is reduced. (b) Mg is the reducing agent. (c) Mg is oxidized. (d) ErF3 is the oxidizing agent.

58. Al2O3 + 3 C + 3 Cl2  AlCl3 +3 CO (a) Cl2 is reduced. (b) C is the reducing agent. (c) C is oxidized. (d) Cl2 is the oxidizing agent.

59. 2 EuCl3  2 Eu + 3 Cl2 (a) Eu is reduced. (b) Cl is the reducing agent. (c) Cl is oxidized. (d) Eu is the oxidizing agent.

96 .

Chapter 12: Chemistry of the Earth

60. 2 NdF3 + 3 Ca  2 Nd + 3 CaF2 143 g NdF3 / (201.2 g/mol) = 0.711 mol NdF3 0.711 mol NdF3 x 2 mol Nd/2 mol NdF3 = 0.711 mol Nd 0.711 mol Nd x 144.2 g/mol = 102 g Nd 61. 2 InCl3  2 In + 3 Cl2 33.2 g In / (114.8 g/mol) = 0289 mol In 0.289 mol In x 2 mol InCl3/2 mol In = 0.289 mol InCl3 0.289 mol InCl3 x 221.0 g/mol = 63.9 g InCl3 Additional Problems 62. The reaction of interest is 2 Al2O3  4 Al + 3 O2.

7.2 x 1010 g Al/(27.0 g/mol) = 2.7 x 109 mol Al 2.7 x 109 mol Al x 2 mol Al2O3 /4 mol Al = 1.4 x 109 mol Al2O3 1.4 x 109 mol Al2O3 x 102.0 g/mol = 1.4 x 1011 g Al2O3 = 140 million kg Al2O3 140 million kg Al2O3 x 2.1 kg bauxite/1.0 kg Al2O3 = 280 million kg bauxite 63. 17 kWh/1.0 kg Al x 72 million kg Al = 1,200 million kWh 64. 4.5 lb/person day x 14,000 persons x 365 days /y = 23,000,000 lb/y 23,000,000 / (85 lb/ ft3) = 270,000 ft3 (100 ft wide)(100 ft long)(ft deep) = 270,000 ft3 ft deep = 27 ft 65. 4 Au + 8 NaCN + 2 H2O + O2  4 NaAu(CN)2 + 4 NaOH 10 g Au/(197 g/mol) = 0.051 mol Au 0.051 mol Au x 8 mol NaCN/4 mol Au = 0.10 mol NaCN 0.10 mol NaCN x 49 g/mol = 4.9 g NaCN 66. Coal is heated in the absence of air to prevent the complete oxidation of the carbon in the coal.

The carbon in the coal would be completely oxidized: C + O2  CO2.

67. approximately 69% 68. approximately 5% 69. reduce, reuse, recycle 70. 1 lb = 0.4535927 kg (a) 0.4535927 kg

67 bags

x 5.00 x 1011 bags = 3.38 x 10 9 kg

97 .

Chapter 12: Chemistry of the Earth

(b) 67 bags

8.7 bags km

5.00 x 1011 bags

= 7.1 km

8.7 bags km

= 5.7 x 1010 km

71. e 72. Recycling saves energy and reduces the amount of materials sent to landfills.

98 .

Chapter 13: Air Overview Chapter 13 begins by defining the layers and composition of the Earth’s atmosphere. The chapter then moves to cycles. The chapter concludes with detailed discussion about air pollution to discuss the dynamic nature of the atmosphere as seen through nitrogen and oxygen.

Lecture Outline 13.1 Earth’s Atmosphere: Divisions and Composition Composition and layers of the atmosphere. Air is composed of 78% nitrogen.

13.2 Chemistry of the Atmosphere Nitrogen and oxygen cycles Temperature inversions The majority of nitrogen fixation is performed by bacteria.

13.3 Pollution Through the Ages Natural and industrial sources of pollution Methods of pollution control and cleanup Air pollution in China Coal + Fire → Industrial Smog 13.4 Automobile Emissions In urban areas, the automobile’s contribution to smog can exceed 90%.

The main gases in automobile emissions The study of carbon monoxide and catalytic converters 13.5 Photochemical Smog: Making Haze While the Sun Shines Photochemical smog is produced when it is sunny and warm.

The origin of photochemical smog Solutions to photochemical smog 13.6 Acid Rain: Air Pollution → Water Pollution Acid rain is rain having a pH less than 5.6.

Air pollutants that contribute to acid rain The major industrial and consumer sources of acid rain 13.7 The Inside Story: Indoor Air Pollution Indoor air pollution poses a major health concern.

The main indoor air pollutants and their sources Where radon comes from and why it is hazardous 13.8 Stratospheric Ozone: Earth’s Vital Shield Ozone shields humans from UV light.

CFCs and depletion of the ozone layer The consequences of stratospheric ozone depletion 13.9 Carbon Dioxide and Climate Change The greenhouse effect trapping of reradiated solar rays by the atmosphere The mechanism and significance of the greenhouse effect Strategies for reducing the amount of CO2 released into the atmosphere 13.10 Who Pollutes? Who Pays?

Air pollution causes damage to buildings, clothing, and other objects.

Air quality standards and the seven major air pollutants 99 .

Chapter 13: Air

Demonstrations 1. Dip some filter paper (or white paper toweling) into a solution of cobalt(II) chloride, and let it dry.

The paper will be blue when it is dry, but when the atmosphere is humid, it will cause the paper to turn pink. Try holding the paper over a steaming cup of coffee or boiling water.

2. (Air pressure) Pour about 10 mL water into an empty soda can and heat it to boiling. Then with tongs quickly invert the can into a container of cool water. The pressure of the air crushes the can.

Review Questions 1. CFCs were used as dispersing agents, as foaming agents for plastics, and as refrigerants.

2. Hydrofluorocarbons; they are greenhouse gases but are more reactive than CFCs, so they will hopefully break down before reaching the stratosphere. Some problems are that they may break down while in use and they are greenhouse gasses.

3. The smoke and soot from factories or power plants contain solid mineral matter left behind known as bottom ash; fly ash is ash that is carried aloft and settles over surrounding areas; used as plasterboard or shaped into sticks (for blackboard chalk).

4. (a) Bag filters work like giant vacuum cleaner bags, and the filters are cleaned later. (b) A cyclone separator is arranged so that the stack gases spiral upward. The particles hit the outer walls, settle, and are collected at the bottom.

5. A greenhouse gas is a gas that traps heat in the Earth’s atmosphere. Carbon dioxide (CO2) and methane (CH4) are two examples. Another efficient greenhouse gas is water vapor (H2O).

Greenhouse gasses are polar and absorb infrared light (heat).

6. The burning of fossil fuels (coal, gas, oil) and decomposition of organic matter in landfills 7. (a) Ground-level ozone causes respiratory problems and eye irritation. Repeated exposure can make people more susceptible to respiratory infection, cause lung inflammation, aggravate asthma, and decreases lung function. (b) Ozone depletion in the stratosphere (ozone hole) causes increases in UV exposure and cancer.

8. Smog is the combination of smoke and fog (and many other substances such as sulfur dioxide, ash, and soot), resulting in polluted air.

9. Photochemical smog is a secondary pollutant that results generally from the oxidation of primary pollutants. These reactions are started when sunlight causes oxygen atoms to be produced and react with other components in the air (a photochemical reaction).

100 .

Chapter 13: Air

10. The ozone hole and global warming are loosely related CFCs, the main causes of the ozone hole, are greenhouse gases that contribute just a little to global warming. The main greenhouse effect is caused by rapidly increasing concentrations of carbon dioxide (not CFCs) and other greenhouse gases in the atmosphere. The amount of energy reaching the lower atmosphere is not the issue, but whether or not it is being absorbed by greenhouse gasses.

11. Nitrogen is fixed when it forms compounds with other elements such as with hydrogen-forming ammonia or oxygen to form oxides of nitrogen. It is the source of nitrogen for plants. Industrial fixation and application of fixed nitrogen to croplands have led to greater food production but also to water pollution problems.

12. Approximately 78% nitrogen; 21% oxygen; 1% argon; 0.04% carbon dioxide 13. The thermosphere. (67 mi = 107.8 km) 14. The troposphere. (12,000 ft = 3.66 km) 15. (a) 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) (b) 4 Cr(s) + 3 O2(g) → 2 Cr2O3(s) 16. (a.) Ne, He, CH4, H2, N2O, NO2, NH3, and CO are lighter than ozone and will be found above 40 km, the ozone layer. (b.) Kr, Xe, SO2, and I2 are heavier than ozone and will be found below 20 km, the ozone layer. (c.) I2 is the densest of the trace gases (see Chapter 6).

17. Cool, damp, and cloudy.

18. Sulfur and oxygen; industrial smog is made of (smoke, fog, sulfur dioxide, and particulate matter) 19. S + O2 → SO2 20. 2 SO2 + O2 → 2 SO3 21. SO2 + 2 H2S → 3 S + 2 H2O 22. 2 CaSO3(s) + O2 (g) → 2 CaSO4 (s) 23. Limestone and dolomite are pulverized and heated. Heat drives off carbon dioxide to form calcium oxide (lime), a basic oxide that reacts with gas sulfur dioxide to form solid calcium sulfite.

CaCO3(s) + heat → CaO(s) + CO2(g) CaO(s) + SO2(g) → CaSO3(s) 24. The smaller particles (PM2.5) are lighter, and they stay in the air longer and travel farther. Also, since they are smaller, they can get into your body easier when you inhale and can get deeper into the lungs.

25. During extremely hot conditions such as combustion or lightning; N2 (g) + O2(g) → 2 NO(g) 101 .

Chapter 13: Air

26. NO2 initiates photochemical smog with the reaction: NO2(g) → NO(g) + O(g).

27. Peroxyacetyl nitrate (PAN); PAN is formed from hydrocarbons, oxygen, and nitrogen dioxide; it makes breathing difficult and eyes itch.

28. Lower the operating temperature of vehicles and/or run the engine on a richer mixture (with more fuel and less air).

29. (a) and (c) are the significant sources of VOC (oil-based paints and cracked pipeline). The plastics (b) may degrade and give a little VOC. 30. It is easy to give a charge to soot particles and have them attracted to the side. They are also large enough that they will stick together. Individual vapor molecules would not be easily charged or stick together.

31. NO (11 valance electrons, 5 can be paired, and one unpaired,

)

32. CO does not have odor or color; you are smelling other gasses such as unburned hydrocarbons and nitrogen oxides, and you are seeing water vapor condensing.

33. Carbon monoxide binds to hemoglobin, preventing the hemoglobin from carrying O2 causing the heart to work harder.

34. 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g) 35. CO emissions have been decreasing over time. CO is produced from the incomplete combustion of fuels; using newer methods/technology of combustion in automobiles and catalytic converters to finish the oxidation to carbon dioxide has caused reductions.

36. Nitrogen oxide (NO), carbon monoxide (CO), and hydrocarbons are reduced by catalytic converters.

37. Allotrope is the term for the relationship of two chemical forms of the same element.

38. Chloride ions are soluble in water and remain there; they are not an ozone-depleting substance. If somehow they were oxidized to chlorine atoms, they would be very reactive and not survive the trip to the ozone layer before reacting.

39. Alternatives to CFC may not contribute to ozone depletion, but they are greenhouse gasses and may contribute to global warming. They are also more reactive than CFCs, so they can cause equipment to fail sooner, resulting in more production and thus more industrial pollution.

40. Free radicals react with ozone, converting it to atmospheric oxygen. The lowered ozone concentration then allows a greater portion of ultraviolet radiation to reach the Earth.

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41. Tropospheric ozone comes primarily from NO, which is oxidized to NO2. The NO2 then decomposes with sunlight to produce oxygen atoms, which in turn reacts with oxygen to produce ozone.

Stratospheric ozone is depleted primarily by CFCs. CFCs decompose in the upper atmosphere to produce chlorine atoms, which in turn react with (destroy) ozone as well as react with the oxygen atoms nature needs to form more ozone.

42. Increase in skin cancers, such as melanoma 43. Nitric acid (HNO3) and sulfuric acid (H2SO4); the acids of nitrogen oxides and sulfur oxides 44. H2SO4 + CaCO3 → CaSO4 + H2O + CO2 Using CaCO3 as marble 2HNO3 + CaCO3 → Ca(NO3)2 + H2O + CO2 45. 6HNO3(aq) + 2Fe(s) → 2Fe(NO3)3(aq) + 3H2(g) 46. H2O + SO2 → H2SO3 47. Carbon monoxide. Indoor: stoves, furnaces, cigarette smoke. Outdoor: automobiles 48. Increased exposure to high levels of the chemicals in smoke such as tar, carbon monoxide, and nicotine can lead to lung cancer and heart disease. Women who are exposed to secondhand smoke during pregnancy have a higher rate of miscarriages and stillbirths.

49. The concrete better traps the radon inside the house once it enters through cracks in the floor, creating higher levels in the house. The crawl space allows for adequate ventilation, not allowing the radon to enter the house.

50. Better insulation traps pollutants, creating higher levels of pollutants inside than outside.

51. Radon is a gas (a noble gas) and is easily inhaled. Solids and liquids are much easier to see and to sequester.

52. Mold spores, cigarette smoke, and wood smoke.

53. The slow warming of the earth caused by the trapping of infrared energy.

54. Carbon dioxide, methane and water vapor.

55. The CO2 directly exhaled has little impact on atmospheric CO2 concentrations. However, we exhale because we like to eat and live; agricultural production produces a large amount of CO2, and living and going places produce a lot of CO2.

56. Cut energy production from combustion of fuels such as coal or gas by using energy- efficient light bulbs and appliances; keeping your house colder in the winter and hotter in summer; car pool. Encourage/support policy to derive our energy from alternative sources such as solar, nuclear power, biofuels, and wind power.

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57. From about 280 ppm to 402 ppm (rounded 300 to 400) is a percent change of approximately 33%.

Actual is (402–280)/280*100 = 44% 58. Ice can trap air (and CO2) for hundreds of thousands of years.

59. Modification of the environment for our comfort always produces some waste heat. Remember the second law of thermodynamics!

60. (5.2 x 1015 tons)(402 x 10−6) = 2.1 x 1012 tons carbon dioxide 61. 22 m3*(312 µg/m3) = 6864 µg, or 6.8 mg 62. 1. 84,000 L/tree (1 m3/1000L) = 84 m3/tree 2. (from Problem 61) 22 m3 of air per day* 365 days/year * 1000 people= 8,030,000 m3/year needed 3. (8.03 x 106 m3/year) /(84 m3/tree) = 95,600 trees 63. Increased water vapor from pollution is not a concern because water will form clouds/rain and may even lead to cooling by reflecting sunlight back to space. Increased atmospheric temperatures, however, will lead to increased amounts of water vapor in the atmosphere – but probably not warming.

64.

0.5 L 1.3 g 5.2 x 1021 g x = 0.65 g/breath, = 8 x1021 breaths in the atmosphere 1 breath 1L 0.65 g/breath 1 mol 6.02 x 10 23 molecules 0.5 L x = 0.0223 mol; x 0.0223 mol = 1.3 x10 22 molecules in a breath 22.4 L 1 mol There are indeed more molecules in a breath than breaths in the atmosphere.

65. Since there are 1.6 times more molecules in a breath than breaths in the atmosphere, given adequate time for mixing, the probability of one of the molecules being the same is good. See Problem 64.

66. Criteria pollutants are pollutants for which scientific criteria are employed to determine their health effects. CO automobiles; NO2 industry and automobiles both major sourses; O3 industry and automobiles; PM10/2.5 industry; SO2 industry; Pb (lead) automobiles 67. 3 NO2 + H2O → 2 HNO3 + NO 2 NO + O2 → 2 NO2

44.0 Gt CO 2 = 9.9 Gt CO 2 12.0 Gt C 44.0 Gt CO 2 1.6 Gt C x = 5.9 Gt CO 2 12.0 Gt C Total 16 Gt CO2

68. 2.7 Gt C x

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Chapter 13: Air

69. Population estimates china: 1.35 billion, U.S.: 317 billion 9.9 GtCO2/1.35 x 109 people = 0.000000007 GtCO2/person in china 5.9 GtCO2/317 x 106people = 0.00000002 GtCO2/person in U.S. Or per million people: in china 0.007 GtCO2; U.S. 0.02 GtCO2 Per person we produce about 2.5 times more carbon.

70. Thermal inversion traps pollutants near the ground in a lower layer of cold air with a layer of warm air above it. Normally, winds mix the air, and pollutants are diluted with fresh air or blown out of that area.

71. Heat absorbed = mass x specific heat x ΔT (a) 1.00 kJ 1000 J 5.1x1018 kg x x x (15 - 16) K = - 5.1x1021 J added (kg K) (1.00 kJ) (b)

5.1 x 10 21 J added = 10 times more J 5.0 x 10 20 J consumed In comparison, a one degree change in temperature is 10X more energy than the total world annual energy consumption. (c) 5.1 x 10 21 J added = 25 times more J added 2.0 x 10 20 J from hurricane In comparison, a one degree change in temperature is 25X more energy than is released in an average hurricane.

72. (a) 1000 t NF3 x

17000 CO2 = 17,000,000 t CO2 1 NF3

73. (c.) 0.16 x 20,000 = 3200 t NF3

74.

3200 t NF3 x

17000 CO2 = 54,000,000 t CO2 1 NF3

75. R7: May cause fire R8: Contact with combustible material may cause fire R20: Harmful by inhalation R26: Very toxic by inhalation R35: Causes severe burns R36: Irritating to eyes R38: Irritating to skin R41: Risk of serious damage to eyes

105 .

Chapter 14: Water Overview The chapter begins by discussing the special properties of water as compared with other substances.

Next, the water cycle, biological and chemical contamination, and water treatment methods are introduced.

Lecture Outline 14.1 Water: Some Unique Properties Water has a high dipole moment, a high heat of vaporization, a high heat capacity, and a low heat of fusion.

14.2 Water in Nature Saltwater constitutes nearly 98% of the water on the Earth.

The water cycle and natural contaminants, a dynamic equilibrium that is powered by the sun 14.3 Chemical and Biological Contamination Fewer than 10% of the people of the world have access to sufficient clean water.

14.4 Groundwater Contamination: Tainted Tap Water Chemical dumps, garbage dumps, underground storage tanks, and overuse of fertilizers are all potential sources of groundwater contamination.

14.5 Water: Who Uses It and How Much 14.6 Making Water Fit to Drink 14.7 Wastewater Treatment Wastewater treatment plants: Primary, secondary, advanced

Demonstrations 1. (High heat of vaporization of water) Mix ordinary rubbing alcohol with an equal volume of water.

Dip a piece of cloth (such as a handkerchief) into the alcohol mixture, gently squeeze out most of the excess liquid, and then using tongs brush the cloth past a flame so that it catches fire. Wave the burning cloth slowly through the air to show that it really is burning. When the flame goes out hold up the cloth to show that it is still intact. Let a few students feel the cloth to see how cool it is. The heat from the burning alcohol was used up in vaporizing the water.

2. (High surface tension of water) Fill a container several inches deep with water and float a plastic strawberry basket on the surface. The basket is actually heavier than water, but the surface tension of the water keeps it afloat. Adding a few drops of detergent lowers the surface tension of the water, and the basket sinks. (This demonstration can be done with a floating needle on the stage of an overhead projector.)

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Review Questions 1. No, oil will not dissolve in water. Oil, which is less dense than seawater, will float.

2. If food is frozen slowly, the water molecules have time to form large crystals. This expands the volume, ruptures cell walls in the food, and makes it taste awful!

3. Organic matter from human and farm waste and fertilizers from agriculture 4. Fertilizers stimulate the growth of microorganisms which increase the biological demand for oxygen (BOD) in the water. If the BOD is high enough, other organisms cannot survive, leading to a dead zone.

5. Microorganisms that cause disease 6. Cholera, typhoid fever, and dysentery; modern water treatment facilities 7. Fertilizer, industrial runoff, and the like 8. Human or animal waste introduces nitrates and phosphates to water, which may serve as nutrients for the growth of algae. When this extra biomass dies and becomes organic matter itself, the cycle resumes, amplified by the biomass added due to the newly incorporated CO2, further increasing the BOD. This process is called eutrophication.

9. Individuals can conserve water and avoid products that require large quantities of water to produce.

10. Underground storage tanks can rust through and leak such contaminants as gasoline and oils into groundwater.

11. Chlorinated hydrocarbons are unreactive and do not break down readily.

Problems 12. Due to hydrogen bonding between water molecules, when water freezes it expands forming large crystals, while in the liquid state water forms hydrogen bonds with four other water molecules. This “hyperdense” hydrogen bonding makes liquid water denser than ice.

13. Water expands when it freezes, lowering the density of ice with respect to liquid water. The lower density of ice allows it to float and makes life in northern lakes possible in the liquid water below the ice layer.

14. Specific heat is the amount of heat required to raise the temperature of 1 g of a substance 1 oC. The high specific heat capacity of water moderates the Earth’s temperature fluctuations.

15. q = mcΔT = 750 g x 1.00 cal g o C x (45.0 o C - 12.0 o C ) = 2.48 x 10 4 cal 16. q = mcΔT = 750 g x 0.107 cal g o C x (45.0 o C - 12.0 o C ) = 2.65 x 10 3 cal 107 .

Chapter 14: Water

17. Liquid water requires roughly twice the amount of energy to heat than it takes to heat ice.

18. Heat of vaporization is the energy required to vaporize a substance. The heat of vaporization of water is high, as compared to other substances, due to hydrogen bonding.

19. Vaporization of water, in the form of sweat, cools our bodies.

20. The lake exchanges heat more slowly than the surrounding land. Thus, during the day it is cooler by the lake, and at night it is warmer by the lake.

21. Vaporization of water from your skin is a cooling effect.

22. Water is polar in nature and dissolves most ionic substances, such as salt. No. 23. Rainwater picks up dust and dissolves atmospheric gases, such as carbon dioxide and oxygen, as it passes through the atmosphere. During thunderstorms, the water reacts with fixed nitrogen to form nitric acid.

24. Anaerobic decay produces CH4, NH3, H2S, amines, and sulfur compounds such as CH3SH. 25. Aerobic bacteria are chemical oxidizers and therefore require oxygen. Anaerobic bacteria are chemical reducers and do not require oxygen.

26. b 27. 1 gal = 3.785 L 3.0 gr x 64.8 mg = 51 mg L 3.785 L 1 gr 28. No, the upper limit in 18 gal would be 3.5 g. (see below).

3.0 grains x 18 gal = 54 grains x 64.8 mg = 3500 mg = 3.5 g 1.0 gal 1 gr 29. Bodies of water gain acidity from acid rains; abandoned mines and fertilizer runoff also contribute.

30. b 31. Limestone, used to neutralize acidic waters, introduces Ca2+ ions into the water.

32. Acids mobilize the aluminum ion, Al3+, which kills young fish.

33. about 1.0 ppm by mass 34. Excessive fluoride leads to mottling of the tooth enamel, as well as many internal problems.

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35. yes; by irradiation with UV light 36. Methemoglobinemia, known as the blue baby syndrome, is caused by ingestion of excessive nitrates.

37. about 2 L 38. Cattle consume large quantities of water both directly and indirectly through consumption of vegetation.

39. Bottled water tastes better than most tap water, but it costs from 240 to 10,000 times more than tap. Production of the plastic bottle containing the water consumes both petroleum and water.

40. The EPA regulates drinking water from municipal water supplies, and the FDA regulates bottled water.

41. Primary sewage treatment involves the collection of sewage into a pond where the solids settle.

The water is then discharged into a stream or lake. Many suspended and dissolved particles still remain.

42. Secondary sewage treatment passes the effluent from primary sewage treatment through filters and aerates the effluent so that aerobic bacteria can convert the organic waste to water, carbon dioxide, and inorganic substances.

43. Organic molecules, inorganic ions such as nitrates, and phosphates remain.

44. The sewage is placed in tanks and then aerated with large blowers. Flocs form and filter contaminants, while aerobic bacteria convert organic matter into sludge.

45. (a) tertiary

(b) primary

(c) secondary

46. Spreading sewage sludge on farmland solves the disposal problem and provides compost to the farmland; there is a possibility of crop contamination by pathogenic organisms in the sludge.

47. 2 HNO3 + CaCO3 → Ca2+ + 2 NO3- + CO2 + H2O 48. Al2(SO4)3 + 3 Ca(OH)2 → 2 Al(OH)3 + 3 CaSO4 49. Ca5(PO4)3OH(s) + F-(aq) → Ca5(PO4)3F(s) + OH-(aq) 50. CO2(g) + H2O(l) → H2CO3(aq) 51. (a)

11x10-6 g benzene 1x10 6 11 g benzene x = = 11 ppb 3 6 1x10 g solution 1x10 1x109 g solution

(b)

0.014 g BaCO3 1x10 4 140 g BaCO3 x = = 140 ppm 4 100 g solution 1x10 1x10 6 g solution 109 .

Chapter 14: Water

52. (a)

25x10-6 g trichloroethylene 2.6x10-9 g = g solution 9.5x103 g solution 2.6x10-9 g 1x109 2.6 g x = = 2.6 ppb 9 9 g solution 1x10 1x10 g solution

(b)

3,800 g Cl 2 38 g Cl 2 1x10 2 = = 3,800 ppm x 4 2 1x10 6 g solution 1.0x10 g solution 1x10

53. No. the measured amounts are: Cu 9 μg/L or 9 x 10-3 mg/L and NO3- is 8 mg/L 54. No. The measured amount is: 8 μg/L or 8 x 10-3 mg/L. 55. Heptachlor 4gF 1 mol 1000 g solution x x = 2x10 -4 mol/L 6 1.0x10 g solution 19.00 g 1L -4 3 x 10 mol/L is more concentrated than 4 ppm.

56. 4 ppm =

Additional Problems 57. d 58. a 59. b 60. d 61. d 62. (a) 3.1 ppm =

3.1 g F 1 mol 1000 g solution x x = 1.6x10 -4 mol/L 6 1.0x10 g solution 19.00 g 1L -

27 g NO3 1 mol 1000 g solution x x = 4.3x10-4 mol/L 6 1.0x10 g solution 61.01 g 1L (c) Use of ppm and ppb avoids the use of very small numbers. (b) 27 ppm =

63. Chlorine is reduced. Chlorine is the oxidizing agent, and sulfur dioxide is the reducing agent.

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64. (a) benzene

(b) toluene

(d) methylene chloride (CH2Cl2)

(c) chloroform (CHCl3)

(e) trichloroethylene

“Methylene” is a term used to describe a “CH2.”

65. dispersion forces 66. d 67. (a) 50 gal x 3.8 L/gal = 190 L x 1 kg/L = 190 kg x 1 kcal/kg oC x (60 oC–10 oC) = 9500 kcal (b) 9500 kcal x 4.184 kJ/cal = 40,000 kJ x 1mol/890 kJ = 45 mol 45 mol x 16.0 g/mol = 720 g (c) 22.4 L/mol x 45 mol = 1000 L x 1ft3/28.3 L = 35 ft3 (d) 3.0 gal/min x 10 min = 30 gal x 3.8 L/gal = 110 L x 1 kg/L x 1 kcal/kg x (60 oC–10 oC) = 5500 kcal x 4.184 kJ/kcal = 23,000 kJ x 1 mol/890 kJ = 26 mol 22.4 L/mol x 26 mol = 580 L x 1ft3/28.3 L = 20 ft3 20 ft3 x $1.22/100 ft3 = $0.24 68. 140 ppm Ca2+ needs to be added; 1 ppm = 1mg/L 140 mg/L x 1000 L = 140,000 mg x 1 g/1000 mg = 140 g 140 g Ca/40.1 g/mol = 3.5 mol Ca2+ The formula of calcium nitrate is Ca(NO3)2; thus 7.0 mol NO3- is added.

7.0 mol x 62 g/mol = 434 g x 1000 mg/g = 434,000 mg/1000 L = 434 ppm 69. c 70. Silicon, aluminum, and oxygen 71. a 72. (a) Remove 45 ppb: 45 ppb x 2000 g = 0.000090 g = 0.09 mg 0.09 mg/(500 mg/g) = 1.8 x 10-4 g (b) (15,000 ppb) x (3.2 x 1012) g = 4.8 x1010 mg 4.8 x1010 mg/(500 mg/g) = 9.6 x 107 g 73. Microporous: pores less than 2 nm diameter Mesoporous: pores between 2 nm and 50 nm diameter Macroporous: pores larger than 50 nm diameter

111 .

Chapter 15: Energy Overview Chapter 15 begins by discussing the sources of energy and the laws of thermodynamics. Next, the chapter discusses the energy requirements and costs of a technologically advanced culture.

Lecture Outline 15.1 Our Sun, A Giant Nuclear Power Plant The sun is the source of energy on planet Earth.

15.2 Energy and Chemical Reactions Chemical reactions depend on several factors and are either endothermic or exothermic.

15.3 Reaction Rates 15.4 The Laws of Thermodynamics The First Law: Energy Is Conserved The Second Law: You Can’t Break Even 15.5 Power: People, Horses, and Fossils 15.6 Coal: The Carbon Rock of Ages 15.7 Natural Gas and Petroleum 15.8 Convenient Energy 15.9 Nuclear Energy 15.10 Renewable Energy Sources Demonstrations 1. Place several tablespoonfuls of calcium oxide (quicklime) in a plastic zip-lock bag. Add water (and some red food color, if desired). The bag will get hot and can be passed around.

2. Place some ammonium nitrate or ammonium acetate in a zip-lock bag. Add water (and some blue food color, if desired). The bag will get cold and can be passed around.

Answers to Review Questions 1. Nuclear fusion.

2. Wood.

3. Coal; petroleum.

4. Advantages: Water vapor is the only exhaust, cuts down air pollution and greenhouse gases, and reduces reliance on foreign oil and political and economic costs.

Disadvantages: Producing hydrogen gas economically, re-forming emits carbon dioxide, storage, transportation, and dispensing of fuel because hydrogen is lighter than air.

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Problems 5. (a) C3H8 + 5 O2 → 3 CO2 + 4 H2O (b) 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 6. (a) 2 C3H8 + 7 O2 → 6 CO + 8 H2O (b) 2 C4H10 + 9 O2 → 8 CO + 10 H2O 7. C13H28 + 20 O2 → 13 CO2 + 14 H2O 8. C + O2 → CO2 9. Hydrocarbons react with oxygen to produce carbon dioxide and water.

10. 2 C + O2 → 2 CO 11. CH4 + 2 O2 → CO2 + 2 H2O 12. 2 CH4 + 3 O2 → 2 CO + 4 H2O 13. A reaction goes faster at a higher temperature.

14. The burning of wood is an oxidation–reduction reaction, with oxygen serving as the oxidizing agent.

15. (a) (b) 413 kJ/mol x 4 mol = 1650 kJ (c) 498 kJ/mol x 2 mol = 996 kJ (d) -798 kJ/mol x 2 mol = -1600 kJ (e) -467 kJ/mol x 4 mol = -1870 kJ (f) -824 kJ; energy is released to the surroundings

16. (a) (b) 946 kJ/mol x 1 mol = 946 kJ (c) 432 kJ/mol x 3 mol = 1300 kJ (d) -391 kJ/mol x 6 mol = -2350 kJ (Note: There are 3 moles of N-H bond in each ammonia molecule.) (e) -104 kJ; energy is released to the surroundings.

17. 803 kJ/mol x 24.5 mol = 19,700 kJ 18. 572 kJ/2.00 mol x 30.0 mol = 8580 kJ 113 .

Chapter 15: Energy

19. 1060 cal x 4.184 J/cal = 4,435 J x 1 kJ/1000 J = 4.435 kJ 20. 15.8 g H2/2.02 g/mol = 7.82 mol H2 483.6 kJ/2 mol H2 x 7.82 mol H2 = 1890 kJ 21. The reaction is being driven in reverse; 803 kJ. 22. The reaction is being driven in reverse; 483.6 kJ/2 mol H2O x 1 mol H2O = 241.8 kJ. 23. Energy is neither created nor destroyed; it is changed from one form to another.

24. All spontaneous processes lead to an increase in entropy of the universe.

25. A measure of randomness of the distribution of energy; increased.

26. Nitrogen in the atmosphere.

27. Potassium in KCl. 28. When the form of energy is changed, some of it is randomly distributed or lost to entropy.

29. Energy flows from the higher temperature of the surroundings to the ice cream, and it melts.

Liquid ice cream has higher entropy than the solid.

30. There is a lot of coal, and the oxidation of coal is a very exothermic process. Coal, as a solid fuel, is hard to move from one location to another and the oxidation leads to significant air pollution.

31. Petroleum and natural gas are more useful in two ways: They are easier to transport than coal, and they are much cleaner fuels.

32. 1526 billion barrels/32.6 billion barrels/year = 46.8 years 33. (a) 33.4 billion barrels/6.7 billion barrels/year = 5.0 years (b) 10.4 billion barrels/6.7 billion barrels/year = 1.6 years 34. 6854 trillion ft3/120 trillion ft3/year = 57.1 years 35. 323 trillion ft3/24.5 trillion ft3/year = 13.2 years.

36. Petroleum is thought to be the remains of ancient marine organisms. Coal is thought to be the remains of ancient plants and trees.

37. Because it is a liquid, petroleum is easy to move from location to location and it has a large chemical potential. Petroleum is valuable for many other uses besides as a fuel; current petroleum consumption may lead to shortages in the near future and to increased air pollution.

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38. Fractional distillation separates the various components of petroleum by their boiling points, which is proportional to the size of the petroleum molecule.

39. Octane rating is an arbitrary performance standard of gasoline, with isooctane given an octane rating of 100.

40. Tetraethyl lead boosts octane, thus saving on the refining process, but it is an environmental hazard.

41. Asphalt is the residue of the petroleum refining process.

42. 2,3,3-trimethylpentane [CH3CH(CH3)C(CH3)2CH2CH3] due to branching.

43. about 20% 44. about 74% 45. Nuclear power plants use only about 3% of uranium-235, whereas nuclear bombs require about 90% uranium-235.

1 239 46. 238 92 U + 0 n → 92 U

239 92

U → −10e + 239 93 Np

239 93

Pa → −10 e + 239 94 Pu

47. No. Reactor-grade uranium is about 3% of uranium-235, whereas nuclear bombs require about 90% uranium-235.

48. Most nuclear waste is from the productions of bombs.

49. A breeder reactor converts nonfissionable fuel into fissionable fuel by bombardment with neutrons during the fission process. No, energy is not being created.

50. Highly reactive sodium must be used as a coolant; the plutonium could be used to make bombs.

51. Plentiful fuel, cleaner than fission. Plasma is a mixture of atomic nuclei and separated electrons at temperatures such as those on the sun.

52. Extremely high temperatures are required; no physical container is able to withstand the reaction conditions.

53. 23290 Th + 01n → 23390Th 233 90

Th → −10 e + 233 91 Pa

233 91

Pa → −10e + 233 92 U

54. 21 D + 31T → 01n + 42 He 115 .

Chapter 15: Energy

55. CH4 + 2 O2 → CO2 + 2 H2O + 803 kJ 1.8 x 1010 kJ/803 kJ/mol = 2.2 x 107 mol x 16.0 g/mol = 3.6 x 108 g 3.6 x 108 g x 1 kg/1000 g x 1 t/1000 kg = 360 metric tons 56. 1.8 x 1010 kJ/393 kJ/mol = 4.6 x 107 mol x 12.0 g/mol = 5.5 x 108 g 5.5 x 108 g x 1 kg/1000 g x 1 t/1000 kg = 550 metric tons 57. A cell that converts light directly to electricity.

58. Solar energy is not concentrated; it is diffuse, making it difficult to transform into useful forms.

59. 1.3 kW x 1000 W/1 kW = 1300 W 1300 W/(100 W/m2) = 13 m2 The area required and steady sunlight.

60. 1.5 MW x 1,000 W/1 kW = 1,500 kW/1.3 kW/house = 1150 houses The limits are the area for wind turbine construction and a constant wind source.

61. Plant material used as fuel. Inefficient and expensive, but renewable 62. Fuel is fed into a fuel cell continuously, and the electrodes serve only to conduct electricity, taking no part in the reaction.

63. (a) Clean, free, and abundant; the wind doesn’t always blow. (b) Useful where geological formations bring steam or hot water near the surface; few sites; wastewater is salty. (c) Renewable; limited availability and expensive (d) Clean, efficient; best dam sites taken. (e) Renewable, expensive up-front costs, and may need to be augmented with other heat sources.

64. (a) C + 2 H2 → CH4 (b) C + H2O → CO + H2 65. (a) CO + 2 H2 → CH4OH (b) 2 H2 + O2 → 2 H2O Additional Problems 66. Endothermic; the process absorbs heat from the surroundings.

67. Exothermic; the process releases heat to the surroundings.

68. (a) endothermic

(b) exothermic

(c) endothermic

69. 2100 kcal x 4184 J/kcal = 8.8x106 J/86400 s/day = 102 W 102 W/745 W/1 hp = 0.14 hp 116 .

(d) endothermic

Chapter 15: Energy

70. 0.745 kW/1 hp x 740 hp = 551 kW 71. (a) 170 kcal x 4184 J/kcal = 7.1x105 J/7200 s = 99 W (b) 170 kcal x 4184 J/kcal = 7.1x105 J/0.0012 s = 5.9x108 W Both contain the same potential energy; power is energy per unit time.

72. (a) Hydrogen: 286.6 kJ/mol x 1 mol/2.0 g x 1000 g/1 kg = 143,300 kJ/kg Isobutane: 2868 kJ/mol x 1mol/58.0 g x 1000 g/1 kg = 49,450 kJ/kg (b) Hydrogen: 143,300 kJ/kg x 8.988x10-5 kg/L = 12.9 kJ/L Isobutane: 49,450 kJ/kg x 2.51x10-3 kg/L = 124 kJ/L 73. The reaction is: CH4 + 2 O2 → CO2 + 2 H2O 19.0 t x 1x106 g/t = 1.90x107 g CO2/(44 g/mol) = 4.32x105 mol CO2 4.32x105 mol CO2 x 1mol CH4/1mol CO2 = 4.32x105 mol CH4 4.32x105 mol CH4 x 16 g/mol = 6.9x106 g CH4 = 6.9 metric tons CH4 74. Gasesous fuels are more convenient for distribution through pipelines. Liquid fules are more convenient for use in mechanized transportation.

75. (a) S + O2 → SO2 (b) 2600 tons coal x 0.0065 tons S/ton coal = 17 tons S 17 tons S x 2 tons SO2/ton S = 34 tons of SO2.

34 tons SO2 x 9.1x105 g/ton = 3.1x107 g SO2 45 km x 60 km x 0.40 km = 1080 km3 x (1000 m)3/1 km3 = 1.1x1012 m3 3.1x107 g SO2/1.1x1012 m3 = 2.8x10-5 g SO2/m3 No, 2.8x10-5 g SO2/m3 x 1x106 μg/g = 28 μg SO2/m3.

76. The reactions are: C + O2 → CO2 + 393.5 kJ CH4 + 2 O2 → CO2 + 2 H2O + 803 kJ 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O + 5754 kJ (a) CH4 is the lowest at 2.75 g (b) CH4 is the lowest at 0.055 g/kJ 77. (a) 9.92x106 TWh/1.32x105 TWh/yr = 75.2 yr (b) Demand would rise by approximately 4 times: 9.92 x 106 TWh/(5.28 x 105 TWh/yr) = 18.8 yrs (c) We had better find alternative renewable energy sources fast!

78. payback = .05 x 30 yr = 1.5 yr 79. natural gas is contained in the tar sands 80. reuse materials, use real time analysis, to prevent pollution

117 .

Chapter 16: Biochemistry: A Molecular View of Life Overview Chapter 16 discusses the chemistry of life, ranging from the substances required for energy to the molecules that define the structure of life.

Lecture Outline 16.1 Energy and the Living Cell The major parts of the cell, structure, and energy production are discussed along with photosynthesis and the sources of energy for animals and metabolism.

16.2 Carbohydrates: A Storehouse of Energy Carbon and water (“carbo hydrate”) are the basis for the structures of the simple sugars (monosaccharides) to the complex carbohydrates, starches, glycogen, and cellulose.

16.3 Fats and Other Lipids Fats are esters of the fatty acids. Fat structures may be saturated, with no carbon-to-carbon double or triple bonds, or unsaturated, with one or more carbon-to-carbon double or triple bond(s).

16.4 Proteins: Polymers of Amino Acids Muscle, skin, and hair are all mostly proteins. Here the structure and properties of amino acids and proteins are discussed.

16.5 Structure and Function of Proteins The four levels of proteins are described (primary, secondary, tertiary, and quaternary.)

How enzymes work as catalysts 16.6 Nucleic Acids: Parts, Structure, and Function Four simple base pairs define everything about you!

The two types and functions of nucleic acids are described Replication is the copying or duplication of DNA molecules.

16.7 RNA: Protein Synthesis and the Genetic Code In transcription, the DNA is transcribed to mRNA. The mRNA travels to the ribosome and is then translated into a protein.

The important characteristics of the genetic code are described.

16.8 The Human Genome Various topics of recombinant DNA technology, genetic testing, and research are discussed, ranging from DNA profiling and DNA fingerprinting to gene therapy and genetically modified organisms.

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Chapter 16: Biochemistry: A Molecular View of Life

Demonstrations 1. (Test for starch) To several mL of a dilute solution of soluble starch add a drop of iodine in KI solution. The starch-iodine complex is dark blue. (This can be shown on the overhead projector.)

2. (Biuret test for proteins) To 5 mL of 1% albumin or gelatin in a small beaker add 5 mL of 2 M NaOH, and then add several drops of 0.5% copper sulfate. (Do this on the stage of an overhead projector.) A lavender color confirms the presence of peptide bonds.

3. (Enzyme action) Dissolve a little urea in about 10 mL of water in a small beaker. Add a little neutralized indicator solution (e.g., a yellow-green solution of bromthymol blue). When a small amount of urease is added, the solution turns blue because of the formation of ammonia.

CO(NH2)2 + H2O  CO2 + 2 NH3 4. (Steroid test) Into several mL of CH2Cl2 (in a small beaker on the overhead projector) dissolve a few crystals of cholesterol. Then add 1 mL of acetic anhydride and about 5 drops of concentrated sulfuric acid. The solution will turn pink, then violet; then it will slowly turn blue and finally green.

Review Questions 1. Animals cannot directly use the energy of sunlight. They must get their energy by eating plants or by eating other animals that eat plants. Photosynthesis is the process by which plant cells capture the energy of the sun and convert it to chemical energy, which is then stored as carbohydrate molecules in plants. Plant cells can also convert these carbohydrate molecules to fat molecules and, with the proper inorganic nutrients, to protein molecules. Animals obtain energy from three major types of foods: carbohydrates, fats, and proteins.

2. (a) Cell membrane: encloses the cell and allows for the diffusion of nutrients and wastes into and out of the cell (b) Cell nucleus: contains the genetic material of the organism (c) Chloroplasts: occurring only in plants; responsible for the conversion of solar energy to chemical energy (d) Mitochondria: energy-producing structure of the cell (e) Ribosomes: site of protein synthesis 3. Proteins are found in every cell; muscles, skin, nails, and hair.

4. Carbohydrates and fats all have C, H, and O; proteins also have N and possibly S. 5. Proteins contain carbon, hydrogen, oxygen, nitrogen, and sulfur. About 20 different amino acids are incorporated by the genetic code. They are polyamides.

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6. A peptide bond, indicated below, is an amide functional group that joins two amino acid units and is also known as an amide.

O

N

C

7. The difference between a polypeptide and a protein is size/mass. If the molar mass of a polypeptide exceeds 10,000 g, it is called a protein.

8. Plant cells contain chloroplasts in which energy from the sun is converted to chemical energy, which is stored in the plant in the form of carbohydrates.

9. Hydrogen bonds 10. Both replication and transcription rely on DNA as a template strand. In each case, doublestranded DNA is converted to single-stranded DNA via the breaking of hydrogen bonds in a localized region. This is accomplished by DNA helicase in the case of replication and by the RNA polymerase complex in the case of transcription. Once single-stranded regions of DNA are available, the DNA is utilized as a template for the formation of either a new strand of complementary DNA (replication) or RNA (transcription). Replication will be conducted through the entire strand of template DNA, while transcription only takes place in defined sections called “genes.” The end result of replication is semiconservative replication of chromosomes where each chromosome consists of a new strand and old strand of DNA. The end result of transcription is that the section of DNA utilized re-forms the original hydrogen bonds, while an independent RNA molecule is formed that no longer associates with the original DNA template.

11. DNA is deoxyribonucleic acid (missing oxygen from the 2’ carbon), while RNA is ribonucleic acid (contains an OH on the 2’ carbon). DNA is generally double stranded, whereas RNA is single stranded. Finally, RNA nucleotides associate with bases that include guanine, thymine, adenine, and uracil. In DNA uracil is replaced by cytosine.

12. Nucleic acid is a series of deoxyribose associated with purines (adenine and guanine) or pyrimidine (cytosine and thymine) called nucleotides. Sections of these nucleotides can be transcribed into RNA, which is further translated into proteins. These sections are called genes.

Chromosomes are composed of both gene and nongene sections of DNA. 13. PCR is an in-vitro procedure that produces millions of copies of a specific DNA sequence. It relies on DNA polymerase enzymes, primers (small oligonucleotides) and template DNA in a reaction that amplifies DNA in an exponential manner. The primers dictate the region of DNA to be amplified. A thermal cycler heats and cools the DNA in a stepwise manner that allows enzymes to copy template DNA into many copies rapidly.

14. Redundancy refers to the fact that the same amino acid can result from several three-base codons.

For example, both CCC and CCA codons result in a proline. Does redundance exist for every amino acid? No. Both methionine (AUG) and tryptophan (UGG) rely on a single codon.

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15. The main steps in recombinant DNA technology are: Step 1: The gene is isolated and amplified.

Step 2: The gene is spliced and inserted into a plasmid.

Step 3: The plasmid is then inserted into a host organism.

Step 4. The plasmid replicates making copies of itself.

16. A functional or “good” gene is placed in a virus. The virus is allowed to infect a cell, and as it does, the functional gene is inserted into the host cell. In most cases this gene will not replace the “bad” or nonfunctional gene. However, the host cell now has a functional gene that can be used to produce a functional protein.

17. a. c and d are carbohydrates. They have the formula C3H6O3. The ratio of hydrogen to oxygen is 2:1, H2O. 18. Disaccharides are carbohydrates that can be hydrolyzed into two monosaccharides. Examples of disaccharides are sucrose, lactose, and maltose.

19. Glycogen is animal starch with branched chains of glucose units.. Amylose is a plant starch with glucose units joined in a continuous chain. Amylopectin is a plant starch with highly branched chains of glucose units.

20. Both amylose (a type of starch) and cellulose are polymers of glucose. The acetal linkages between glucose units are alpha in starch and beta in cellulose.

21. monosaccharides: c, d (mannose and glucose) 22. monosaccharides: b, c (galactose and fructose) 23. In Problem 21 none are disaccharides, and in Problem 22 sucrose and lactose are disaccharides.

24. In Problem 21 amylose and cellulose are polysaccharides, and in Problem 22 none are polysaccharides.

25. (a) The hydrolysis product of amylose is glucose. (b) The hydrolysis product of glycogen is glucose.

26. (a) The hydrolysis product of amylopectin is glucose. (b) The hydrolysis product of cellulose is glucose.

27. Lactose is a disaccharide of galactose and glucose.

121 .

Chapter 16: Biochemistry: A Molecular View of Life H

O C

Galactose

H

OH

HO

H

HO

contains an aldehyde and alcohol.

H

4

H

OH CH 2OH

O HC

Glucose

H

OH

HO

H

H

OH

H

OH

contains an aldehyde and alcohol.

CH2OH

28. Arabinose is a monosaccharide; heating it in water would hydrolyze it to form other substances if it was a disaccharide or polysaccharide.

29. It is a disaccharide of glucose; hydrolysis yields twice as many moles of glucose.

CHO

30. Manose:

HO

C

HO

C

H

H

C

OH

H

C

OH

2

H

Aldehyde and alcohols

CH 2OH

31. Structurally, oils are very similar to fats except that they incorporate a higher proportion of unsaturated (more C-to-C double bonds) fatty acid units. Properties: At room temperature, most animal fats are solids. Oils are liquid fats, obtained principally from vegetable sources.

32. (b) Glycerol 33. saturated: a. palmitic and d, stearic acid

unsaturated: b. linoleic, and c. oleic

34. monosaturated: oleic

polyunsaturated: linoleic

35. (a) linoleic: 18

(b) palmitic: 16

(c) stearic: 18

36: (a) lauric: 12

(b) oleic: 18

(c) butyric: 4 122

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Chapter 16: Biochemistry: A Molecular View of Life

37. Corn oil (in the bottle); oils have more C=C double bonds (unsaturated) than do fats.

38. Solids would be lower (lard), they usually are saturated. Liquid margarine would be higher; it has more double bonds.

39. (d) Proteins 40. A dipeptide is a molecule with two amino acid units. A tripeptide is a molecule with three amino acid units. A polypeptide is a molecule with more than 10 amino acid units.

41. Lysine and glycine are essential.

42. (a)

(b)

Glycine, no R group, just another H

O CH3CH CH

C

OH

OH NH2

43. (a) threonine

(b) phenylalanine

44. (a) glutamic acid and (b) cysteine

45. (a)

(b)

46. serylleucine and glyclcysteine 47. aspartylphenylalanine (from aspartic acid and phenylalanine.)

48. The other dipeptide would be phenylalanine and aspartic acid.

O H2 N

CH CH2

C

O NH

CH

C

OH

CH2 C

O

OH

49. Hydrogen bonds, ionic bonds, disulfide linkages, and dispersion forces

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50. (a) When the substrate induces the proper alignment of the active site that enables the enzyme to perform its catalytic function. The substrate must fit in the active site of the enzyme somewhat like a key must fit in a tumbler lock in order to open it. (b) An inhibitor attaches to a part of the enzyme, which changes the shape of the enzyme, so that the substrate is no longer able to bind to it.

51. (a) DNA (b) RNA

(c) RNA

52. (a) RNA

(c) DNA

(b) DNA

53. ribose; uracil 54. ribose; cytosine 55. (a) guanine

(b) thymine

(c) cytosine

(d) adenine

56. (a) uracil

(b) cytosine

(c) adenine

(d) guanine

57. Each parent strand will remain hydrogen bonded to a new daughter strand.

58. The protein coding regions of DNA are transcribed into mRNA. Once processed, the mRNA will exit a nuclear pore and assemble with a complex of proteins, including ribosomes and other RNA molecules (tRNA) where the nucleic acid sequence governs the order of amino acids in a growing amino-acid chain. The amino-acid chains will fold into functional proteins.

59. Transcription utilizes DNA to produce mRNA (and snRNA, tRNA, and rRNA). Translation does not utilize DNA; tRNA and mRNA are involved.

60. mRNA is a protein coding molecule. It contains the genetic information that dictates the order of amino acids in a forming protein. tRNA transfers those amino acids from the cytoplasm to the translational proteins.

61. (c) transcription 62. (b) CAA to TAA 63. (d) a ribosome 64. (a) amino acids 65. (c) thymine 66. (d) a ribosome 67. (a) amino acids 68. lysine 124 .

Chapter 16: Biochemistry: A Molecular View of Life

69. (a) primary

(b) secondary

70. (a) polyglycine +NH3 CH2

C

NH

O

CH2

C

NH

CH2

O

C

NH

O

CH2

C

O-

O

(b) polycysteine:

71. (a) pyrimidine

(b) pyrimidine

72. (a) pyrimidine

(b) DNA

73. (a) purine

(b) RNA

74. CTC Glutamine; CTT to CTA Lysine; to CAC Stop codon 75. (a) Threonine-Serine-Methionine-Alanine. (b) Threonine-Serine-Alanine-Alanine 76. Plants can be a source of lysine, so if they escaped, say during a really bad storm, they could eat plants from outside sources.

77. CGU. (Also, taking the wobble hypothesis into account, CGC, CGA and CGG would also work.)

78. a. Ala-Ser-Cys-Phe-Gly-Gly b. ASCFGG: Alanine (Ala or A), Serine (Ser or S), Cysteine (Cys or C), Phenylalanine (Phe or F), Glycine (Gly or G), Glycine (Gly or G).

79. water (H2O) 80. (b) carbohydrates and lipids 81. (d) all of the above 82. (b) Separation of products is difficult.

125 .

Chapter 17: Food: Molecular Gastronomy Overview In this chapter we discuss food, what it’s made of, and the substances added to it. The three main classes of foods are carbohydrates, fats, and proteins. For proper nutrition, our diet should include balanced proportions of these three foodstuffs, plus water, vitamins, minerals, and fiber.

Lecture Outline 17.1 Carbohydrate in the Diet Dietary carbohydrates, the sugars and starches, and metabolism are discussed.

17.2 Fats and Cholesterol The fats contain abundant calories, but they are harmful in excess.

Saturated and unsaturated fats are described.

Lipids and their function are described.

17.3 Proteins: Muscle and Much More Proteins are built from the amino acids. Here the essential amino acids and the diseases associated with their deficiency are described.

17.4 Minerals, Vitamins, and Other Essentials The minerals are inorganic substances required for life. The roles of vitamins, minerals, water, and fiber are studied.

17.5 Starvation, Fasting, and Malnutrition The effects of starvation, fasting, and malnutrition are discussed.

17.6 Flavorings: Spicy and Sweet The common food additives and their purpose are studied in detail. Some substances are added to food to make it taste better. Artificial flavors, flavor enhancers, and sweeteners are studied.

17.7 Other Food Additives: Beneficial or Dangerous?

The benefits of food additives, such as preventing spoilage and adding minerals and vitamins to the diet, are studied.

17.8 Problems with Our Food The main problems with harmful substances in our food such as carcinogens and contaminants, as well as how to avoid them, are discussed.

Demonstrations 1. Label reading: Sobe Lifewater yumberry pomegranate drink bottle has many of the things discussed in this chapter: a zero calorie sugar substitute (erythritol), carrot juice for color, and vitamins listed by name such as cyanocobalamin (B12).

2. Fill a small beaker (100–150 mL) about 1/3 full of sugar (sucrose). Carefully add 5–10 mL of concentrated sulfuric acid and step back. A black cylinder of foamed carbon will rise up out of the beaker, and there will be smoke and fumes. The demonstration should be done in a hood or in a well-ventilated area.

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3. (Cornstarch putty) To 1 cup of cornstarch slowly add about 1/3 cup water and mix. The gooey mixture has some unusual properties. It can be cut with a knife or with scissors, but it flows like a viscous liquid. It looks wet, but if you rub it between your fingers, it gets powdery. It can be rolled into a ball, but then the ball will flatten. It is a non-Newtonian fluid.

4. Put several spoonfuls of baking soda into a glass soda bottle. Add about a cupful of vinegar, and then quickly spread the opening of a balloon over the mouth of the bottle. The balloon will be blown up by the carbon dioxide generated. The balloon can be removed from the bottle at any time.

Review Questions 1. The function of carbohydrates in the body is primarily as a source of energy through a complex set of reactions to produce carbon dioxide, water, and energy. Excess carbohydrates can be converted to fats and stored. Cellulose helps in digestion, and our taste buds use simple carbohydrates for sweet flavor!

2. The function of fats is to provide an energy source and thermal insulation and to protect organs.

3. Vitamins are organic, whereas minerals are inorganic.

4. The protein is broken down into amino acids that in turn are used to synthesize new proteins for growth and repair, like making muscles, hair, and enzymes.

5. An excess of a fat-soluble vitamin is more dangerous because a fat-soluble vitamin will be stored in adipose tissue instead of being excreted, unlike an excess of a water-soluble vitamins.

6. The human body is totally deprived of food.

7. A food additive is a substance other than basic foodstuffs that are present in food as a result of some aspect of production, processing, packaging, or storage.

8. Intentional additives (aspartame, bread enriched with nutrients such as vitamins and iron added); incidental additives (insect parts, antibiotics).

9. Monosodium glutamate, a sodium salt used as a flavor enhancer.

10. Botulism is poisoning from a toxin that is caused by Clostridium botulinum, the bacterium that produces botulism poisoning and is sometimes present in improperly preserved foods; aflatoxins are compounds produced by molds growing on stored grains and peanuts.

11. Vitamin E is a fat-soluble antioxidant. Vitamin C is a water-soluble antioxidant 12. Saccharin, cyclamates, aspartame, acesulfame K; Saccharin, aspartame, and acesulfame K are currently approved.

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13. Aspartame is the methyl ester of the dipeptide aspartylphenylalanine.

O H2N

CH

C

CH 2

O NH

CH

C

OCH3

CH 2

COOH

14. Some incidental additives that have been found in foods are pesticide residues, insect parts, and antibiotics.

15. Some functions of food additives are to provide nutritional value, enhance flavor and color, retard spoilage, provide texture, sanitize, bleach, ripen, control moisture, and prevent or enhance foaming.

16. FDA; Since 1958, all companies have had to prove that additives are safe for the intended use.

17. a. C-to-O-to-C linkage 18. a. glucose

b. sucrose

c. fructose

19. Hydrolysis and combustion are the two types of reactions that ultimately form CO2 and H2O. 20. Glucose is formed when starch is digested.

21. The triglycerides (fats) are broken down in a hydrolysis reaction to form fatty acids and glycerol, with mono- and diglycerides formed along the way.

22. c. 4 kcal and 9 kcal, respectively 23. Most animal fats are solids; most vegetable oils are liquids; animal fats generally have fewer C-toC double bonds than vegetable oils (oils have a high percentage of monounsaturated fatty acids).

24. A trans fat is an unsaturated fat in which a double bond has the trans-conformation (the two hydrogens on the double bond are on the opposite side). They are made during the production of partially hydrogenated oils.

25. An adequate protein supplies all the essential amino acids in the quantities needed for the growth and repair of body tissues. If the body does not consume adequate protein in nutrition, then muscle will also waste away as more vital cellular processes (e.g., respiration enzymes, blood cells) recycle muscle protein for their own requirements. Lean meat, milk, fish, eggs, and cheese supply adequate protein.

26. A limiting reactant is the reactant that is present in the least amount. In protein synthesis, the essential amino acids are considered limiting reagents because if they are absent, protein synthesis cannot be accomplished, and so our body cannot produce essential amino acids as needed.

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27. a. Corn is lacking lysine and tryptophan; b. beans are lacking methionine.

28. Kwashiorkor is a protein deficiency; children.

29. It will provide adequate protein because wheat floor lacks lysine, and the pea flour lacks methionine; thus, each has what the other flour is missing.

30. Carbohydrates and fats are primarily C, H, and O, while proteins also contain nitrogen and sulfur.

31. a. Iodine is necessary for the thyroid gland. b. Iron is necessary for the proper functioning of the oxygen-transporting compound hemoglobin. c. Calcium is necessary for the proper development of bones and teeth, coagulation of the blood, and maintenance of the heartbeat rhythm. d. Sodium ions are important to the exchange of fluids between cells and plasma.

32. Ca, Cl, Na, P; they are structural or macromolecules of the cell.

33. Water-soluble (polar and generally have –OH groups): pantothenic acid, biotin; fat-soluble: phylloquinone (nonpolar and may contain long carbon chains).

34. a. yes (but debatable; it is considered to be “essential”) b. fat soluble 35.

Compound ascorbic acid calciferol cyanocobalamine retinol tocopherol

Correct Designation vitamin C vitamin D vitamin B12 vitamin A vitamin E

36. a, c (niacin and pyridoxine) 37. a. scurvy: vitamin C c. beriberi: thiamine (B1)

b. pellagra: niacin (B3) d. pernicious anemia: cyanocobalamine (B12)

38. a. fat soluble c. water soluble

b. water soluble d. water soluble

39. a. fat-soluble D vitamin c. water-soluble B vitamin

b. water-soluble B vitamin d. fat-soluble E vitamin

40. b. caffeine 41. d. scavenging free radicals 42. b. cyclamate 129 .

Chapter 17: Food: Molecular Gastronomy

43. a. preservative of fats (antioxidant) 44. a. sweetener c. spoilage inhibitor for meat

b. adds color

c. sweetener

b. improves nutrition d. inhibit molding of bread and cheese

45. (d) none of the above 46. No, it has been modified or synthesized from natural products.

47. Antioxidants are reducing agents, usually free-radical scavengers added to foods to prevent fats and oils from becoming rancid; vitamin E; BHT and BHA. 48. A chain reaction occurs when intermediates are formed during a reaction that keep it going. An example is when a fat molecule reacts with oxygen to form a free radical, then the radical reacts with another fat molecule, and the reaction repeats indefinitely. Antioxidants stop the chain reaction by reacting with the free radical.

49. [7 g protein x (4 kcal/g)]/110 = 25.5% from protein [14 g carbs x (4 kcal/g)]/110 = 50.9 from carbohydrates [2.5 g fat x (9 kcal/g)]/110 = 20.5% from fat (Approximation errors on g/cal; does not add up to exactly 100%) 50. [8 g protein x (4 kcal/g)]/130 = 24.6% from protein [23 g carbs x (4 kcal/g)]/130 = 70.8% from carbohydrates [1.5 g fat x (9 kcal/g)]/130 = 10.4% from fat 51. 3 g protein x (4 kcal/g) = 12 kcal from protein 6 g carbs x (4 kcal/g)] = 24 kcal from carbohydrates 5 g fat x (9 kcal/g)] = 45 kcal from fat Total kcal = 81 45 kcal/81 = 55.6 % from fat 52. 23.5 g protein x (4 kcal/g) = 94 kcal from protein 6 g fat x (9 kcal/g) = 54 kcal from fat Total = 94+54 = 148 kcal 170-148 = 22 kcal would be from carbohydrates X g carbs x (4 kcal/g) = 22 kcal from carbohydrates; solve for X X= 5.5 g of carbohydrates 53. [X g fat x (9 kcal/g)]/1200 = 0.25 from fat; solve for X X g fat x (9 kcal/g) = 300 X = 33.3 total grams of fat allowed 30 % off 33.3 total grams = 10 g saturated fat allowed 54. [X g fat x (9 kcal/g)]/3000 = 0.30 from fat; solve for X X g fat x (9 kcal/g) = 900 X = 100 g maximum of fat allowed 130 .

Chapter 17: Food: Molecular Gastronomy

55. Omega-6 fatty acids are c. linoleic acid, d. linolenic, and e. arachidonic acid 56. Part 1. (b) is saturated.; (a) is polyunsaturated Part 2. (a) is linolenic acid and (b) is stearic acid 57. (70 calories from fat/160 total calories)*100 = 43.75 % from fat 58. From the label (12%/2000 cal) x 2200 cal diet = 13.2 % 59. From the label (25%/2000 cal) x 2200 cal diet= 27.5% 60. 9 g/cup x 3 cups = 27 g of protein. 27/60*100 = 45 % obtained from the milk 61. a. calories from sugars = 12 g x 4cal/g = 48 cal. 48/90 x 100 = 53.3% b. calories from protein = 9 g x 4 cal/g = 36 cal. 36/90 x 100 = 40% c. lactose 62. a. They would all likely be water soluble (all have –OH groups and II is charged/ionic). b. Based on vitamins being specific organic compounds that are required in the diet (in addition to the usual proteins, fats, carbohydrates, and minerals) to prevent specific diseases: none would be a vitamin, (II) and (III) could be argued, as they are essential nutrients, but by definition they are not vitamins.

63. Riboflavin and cyanocobolamine are water soluble and may need to be replaced as they are excreted.

64. Sodium saccharin is an ionic substance and is soluble in water like most sodium salts due to strong ion–dipole interactions with water, while sucralose has strong hydrogen-bonding interactions (dipole–dipole) with water.

65. Liquid fats/oils have the double bonds that allow the molecules to undergo oxidation/radical formation much more readily and the fats will go rancid faster.

66. Cattle convert only a small proportion of the plants consumed into animal protein. Much of the food ingested is used for metabolic processes to keep the cattle alive.

67. β-Carotene is converted to vitamin A, so it is a vitamin additive as well as color.

68. They are very similar, but sucrolose has three chlorines substituted on the sucrose.

69. a. 4;

b. 3;

c. 2;

d. 1

70. a. Agricultural waste is 3. Straw b. Processing waste is 1. Orange peel c. Postconsumer waste is 2. Canteen leftovers

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71. Many apply: Limonene used as food flavoring and as a cleaner Lycopene from grapefruit, used as food coloring Vitamin C, used as a nutritional supplement in foods Pectin as a thickening agent in foods Ethanol from digestion of the waste to be used as fuel 72. Food Supply Chain Waste (FSCW) a. Reduced dependence on fossil fuels: Food waste can be used directly as fuel but can be more efficiently converted to methane and used. b. Minimized energy use and derivative use: Using waste materials as fuel would minimize other fuels costs. FSCWs provide alternate supplies for chemical derivatives that currently only have fossil fuels as viable supplies. c. Saved money: Waste minimization and reduced fuel costs save money. d. Improved business opportunities. New products and methodologies are being derived from food wastes and potentially at a reduced cost.

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Chapter 18: Drugs: Chemical Cures, Comforts, and Cautions Overview Chapter 18 examines the world of pharmaceuticals, ranging from aspirin to opiates. The type and action of the common classes of drugs are discussed A drug is a chemical substance that affects the functioning of living things and is used to relieve pain, to treat illness, or to improve one’s state of health.

Lecture Outline 18.1 Scientific Drug Design.

Natural and synthetic drugs Chemotherapy (chemical therapy) can be used to treat disease.

18.2 Pain Relievers: From Aspirin to Oxycodone The wonder drugs! The common OTC analgesics, NSAIDS, antipyretics, and antiinflammatory drugs are identified and how each works is discussed.

The potential for addiction with narcotics Chemistry, allergies, and the common cold 18.3 Drugs and Infectious Diseases Antibacterial drugs are reviewed, effective against bacteria, not viruses!

Viruses and antiviral drugs are discussed along with AIDS. Structure function relationship and combinatorial chemistry are discussed.

18.4 Chemicals against Cancer The aim of cancer research is to kill the cancerous tissue without killing normal cells.

The type and action of the common anticancer drugs are described.

18.5 Hormones: The Regulators Regulate growth, metabolism, reproduction Hormone, prostaglandin, and steroid are discussed along with how each works.

The types of sex hormones, how each acts, and how birth control drugs work The chemistry and social revolution related to “The Pill” are discussed.

18.6 Drugs for the Heart The types of drugs for treating heart disease are discussed, along with how they work High blood pressure is the most common cardiovascular disease 18.7 Drugs and the Mind How the brain amines norepinephrine and serotonin affect the mind Three classes of drugs: stimulant, depressant, and psychotropic Chemistry of the nervous system and neurotransmitters Brain amines and their effects are discussed.

The importance of chirality PEA could govern emotions of romantic relationships.

Anesthetics and their effects, from general anesthetics to ketamine and PCP Depressant drugs Alcohol is the most used and abused depressant drug.

Barbiturates and antianxiety drugs are discussed

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Chapter 18: Drugs: Chemical Cures, Comforts, and Cautions

Stimulant drugs: Cocaine, caffeine, and nicotine are discussed. Amphetamine was once used for weight loss; it is now used in treating mild depression and narcolepsy. Methamphetamine is inexpensive and widely abused.

Hallucinogenic drugs: These drugs change the way people see things; from LSD to marijuana to Ecstasy.

18.8 Drugs and Society Drug abuse and drug misuse are discussed.

Placebo and nocebo effects Demonstrations [None] Review Questions 1. Natural drugs are extracted directly from plant or animal sources by physical processes, as opposed to materials formed by chemical reactions. Semisynthetic drugs rugs are made by modifying molecules derived from natural sources to improve the substances’ properties. Synthetic drugs have no relationship to naturally occurring substances; the basic parts have some natural components, but they are predominantly human-made.

2. Antihistamines relieve the symptoms of allergies—sneezing, itchy eyes, and runny nose—by inhibiting the release of histamine. An antitussive is a cough suppressant, and expectorants help bring up mucus from the bronchial passages.

3. An antibiotic is a drug that kills or slows the growth of bacteria; originally limited to formulations derived from living organisms.

4. A hormone is a chemical messenger, sending signals for physiological changes to other parts of the body.

5. Cortisol: a hormone released in the body during stressed or agitated states; prednisone: a synthetic hormone similar to cortisone.

6. Birth control pills work by estrogen preventing the ovaries from releasing an egg. If no egg is released, there is nothing to be fertilized by sperm and the woman cannot get pregnant.

7. Prostaglandins act as mediators of hormone action; they act near the site where they were produced, they are rapidly metabolized, and they have different effects on different tissues; hormones act throughout the body.

8. Barbiturates: Mild sedation (small doses) and sleep (larger doses). Long-term use presents some hazards: They are strongly addictive; and they lead to a tolerance, as a result of which an everlarger dosage is required to produce the same intoxication.

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9. Many compounds may be used for this answer; the primary dissociatives are similar in action to phencyclidine (PCP), and include ketamine and dextromethorphan (DXM). Also included are nitrous oxide (laughing gas), salvia divinorum, and muscimol.

10. A synergistic reaction takes place when one drug enhances the action of another drug, such as using a depressant (alcohol) with a sedative such as a barbiturate.

11. A narcotic is a drug that produces stupor and relief of pain; usually refers to opioids given to relieve pain.

12. Paclitaxel: Natural Heroin: Semisynthetic (made from morphine) Morphine: Natural Penicillin: Natural (many semisynthetic derivatives exist) Cisplatin: Synthetic 13. Brain amines are made from dietary amino acids; for example, high carbohydrate diets produce high serotonin levels in the brain.

14. Psychotropic drugs are those that affect the human mind.

15. Tranquilizers are not used to cure schizophrenia; they just relieve or control some symptoms.

16. A hallucinogenic drug alters the way people perceive things.

17. Many apply. Natural: Morphine; Semisynthetic: Codeine; Synthetic: Cisplatin.

18. Codeine is semisynthetic; fluoroquinoline is semisynthetic; quinine is natural; vinblastine is natural; taxol is natural; 6-mercaptopurine is semisynthetic.

19. Chemotherapy is the use of chemicals that are more toxic to disease organisms than to human cells to control or cure infectious diseases.

20. Many apply: Cure disease; relieve physical pain; improve mental health; improve athletic performance, enjoyment/relaxation, addictions, sleep disorders, and sedation for medical procedures.

21. Aspirin is acetylsalicylic acid.

22. a. Both are analgesics and antipyretics: They relieve minor aches and reduce fever. b. Aspirin inhibits the synthesis of prostaglandin derivatives and reduces the inflammatory process; it is an NSAID. 23. Aspirin is an anticoagulant and can cause/increase bleeding.

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24. A COX-2 inhibitor is an enzyme called cycloxygenase (COX). Blocking this enzyme impedes the production of the chemical messengers (prostaglandins) that cause the pain and swelling of arthritis inflammation. COX-2 inhibitors are a class of NSAIDs. Newer NSAIDS, like Celebrex, are said to selectively block the COX-2 enzyme and not the COX-1 enzyme, which is involved in most of the side effects.

25. Acetaminophen contains a phenol and an amide.

26. Aspirin contains a carboxylic acid and an ester.

27. (c) heroin 28. (b) caffeine; it is similar to bases like adenine and guanine.

29. (a) acetaminophen 30. They have similar effects (analgesics and addictive) but differ in intensity. Heroin is less polar, so it can enter the brain more rapidly and produce stronger feelings of euphoria and pain relief.

31. Codeine is a less potent analgesic and is less addictive; both are analgesics and addictive.

32. a. The needles may stimulate the nerves and release peptides/endorphins that block pain signals. b. The body secretes its own painkillers/endorphins.

O

O

SNH 2

COH

O

33.

H 2N

H 2N

sulfanilamide

para-aminobenzoic acid

34. Penicillins are β-lactam antibiotics obtained from Penicillium molds. Penicillin works by inhibiting enzymes that the bacteria use to make their cell walls. The bacterial cell walls are made up of mucoproteins, polymers in which amino sugars are combined with protein molecules.

Penicillin prevents cross-linking between these large molecules. This leaves holes in their cell walls, and the bacteria swell and rupture.

35. They are effective against a wide variety (broad spectrum) of bacteria.

36. Fluoroquinolones are broad-spectrum antibiotics; they work by inhibiting bacterial DNA replication.

37. (b) broad spectrum 38. No, they do not cure; they can be prevented by vaccination.

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39. The three types of antiviral drugs are: 1. Nucleoside analogs: The substitution of an analog for a nucleoside in the viral DNA cripples it, slowing down its replication.

2. Reverse transcriptase inhibitors (HNNRTIs): These drugs prevent reverse transcriptase from working properly.

3. Protease inhibitors: These block enzymes that would allow the virus to make new copies.

40. Antibiotics target bacteria, not viruses.

41. (b) Cyclophosphamide is an anticancer drug.

42. The two classes of anticancer drugs are antimetabolites (like 5-FU) and alkylating (like Cisplatin) agents.

43. Yes, for example, nitrogen mustard causes lung, mammary, and liver tumors in mice but can reduce some human tumors.

44. Cisplatin, an alkylating agent, binds to DNA and blocks its replication. (It inhibits cell division.)

45. 6-Mercaptopurine. Chemotherapy: It is an antimetabolite (inhibiting cell division and the multiplication of cancer cells).

46. Cyclophosphamide (Cytoxan). Chemotherapy: It is an alkylating agent.

47. Many apply: Prostaglandins differ from hormones in that they (1) act near the site where they are produced, where hormones can react throughout the body, (2) can have different effects in different tissues, and (3) are rapidly metabolized. Another difference is that prostaglandins are synthesized in the body from arachidonic acid, whereas most hormones are proteins, peptides, or amino acids.

48.

49. Progesterone contains two ketones and an alkene.

50. a. Estrogen: female sex hormone b. Androgen: male sex hormone 51. RU-486 works by blocking the pregnancy from being established by inhibiting the action of progesterone.

52. The addition of the ethynyl group (-C≡C-H), and resulting hydroxyl group formation –(OH), make it more likely to be absorbed orally (more water soluble).

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53. Emergency contraception refers to contraceptive measures that, if taken after sex, may prevent pregnancy. They work by blocking a pregnancy from being established, or maintained by preventing ovulation or fertilization by halting development of the uterine lining.

54. (b) encephalin.

55. (a) causing the kidneys to excrete more water and thus lower the blood volume.

56. (c) acting as vasodilators, causing the muscles around blood vessels to relax.

57. (c) Nitric oxide, NO 58. (a) angina pectoris 59. Psychotropic drugs are those that affect the mind.

60. Hallucinogenic drugs are drugs that alter qualitatively the way we perceive things.

61. (a) antimetabolite 62. (d) decreasing the stimulant action of epinephrine and norepinephrine 63. (b) lidocaine 64. (a) anesthetic 65. (c) coniine (a poisonous alkaloid found in poison hemlock) 66. (a) Diethyl ether is highly flammable.

67. a. Nitrous oxide: Given too much, you may not get enough oxygen, resulting in brain damage. b. Halothane: higher rate of miscarriages c. Diethyl ether: nausea, slow onset, and slow recovery 68. a. An agonist mimics the action of the drug. b. An antagonist inhibits the action of a drug by blocking the receptors. For example, a morphine antagonist inhibits the action of morphine by blocking the receptors.

69. (b) low cost 70. (c) Illegal drug use affects only the users.

71. (b) Some test subjects receive the drug, and others receive a placebo.

72. Drug misuse is using a drug for the wrong purpose or effect. Drug abuse is to use drugs for their intoxication effects.

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73.

N F

COOH O

74. Because it is not anti-inflammatory.

75. (a) 140 lb (1 kg/2.205 lb) = 63.5 kg 40 mg/kg x 63.5 kg = 2540 mg/day; 2540/3= 847 mg/dose (0.847 g) if taken every 8 hrs (b) (0.847 g/dose)(1000 mg/g)(1 capsule/500 mg) = 1.7 capsule/dose 1.7 pills (about 2) for a pill containing 500 mg of active ingredient 76. Diuretics are used as drugs to lower blood pressure; they may also be the most effective drugs for treating hypertension. They work by causing the kidneys to excrete more water and thus lower the blood volume.

77. All steroids have the fused four-ring system shown here:

78. Water soluble; The compound is relatively polar; it is ionic / a salt.

79. Fat soluble; The compound is relatively nonpolar; 17 C atoms with only 1 O atom and 1 N atom.

80. Barbiturates all have the pyrimidine-like ring drawn below; different drugs are made by substituting various groups on the ring or by replacing one of the carbonyl O atoms by an S atom.

O

N

N

O

O

NH 2

81.

82. PCP and ketamine. When a drug is soluble in fat and has no appreciable water solubility, it is stored in fatty tissue and released when the fat is metabolized, resulting in a flashback.

83. Pentobarbital (Nembutal) and phenobarbital (Luminal). Most barbiturates would apply.

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84. Naproxen, because it takes a smaller amount (lower LD50) to kill 50% of the population of test animals.

85. The timber rattlesnake, because it takes a smaller amount (lower LD50) to kill 50% of the population of test animals.

86. 1500 mg/kg x 10 kg = 15,000 mg; 15,000 mg/325 mg/tablet = 46 tablets 87. Nicotine is more toxic.

Procaine: 50 mg/kg x 50 kg = 2500 mg for a lethal dose Nicotine: 1 mg/kg x 50 kg = 50 mg for a lethal dose 88. 5 mg/kg x 70 kg = 350 mg ; no, they are not always directly related to humans.

89. Codeine: CH 3 N

H3 C O

O

OH

90. Heroin

91. Oxycodone: CH 3 N

H3C O

O

O

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92. Nalorphine:

93. GBL is a cyclic ester, or lactone, shown below. It is a hygroscopic, colorless, oily liquid with a weak characteristic odor and is soluble in water. GBL is a common solvent and reagent in chemistry and is used as an aroma compound, as a stain remover, as a superglue remover, and as a paint stripper. Hydrolysis of GBL will result in GHB, as shown below.

OH

H2O O

O

(hydrolysis)

HO O

94. 1,4-Butanediol is used industrially as a solvent and in the manufacture of some types of plastics and fibers. It can be converted to GHB by an oxidation reaction. [ox]

OH

OH HO

HO

O

95. (b) < (a) < (c) 96. a, b, c, cholesterol, prednisone, and progesterone.

97. NSAIDs inhibit the production of prostaglandins by blocking COX enzymes from producing them. Inflammation is caused by an overproduction of prostaglandin derivatives, and inhibiting their synthesis reduces the inflammatory effect.

98. Removal of the hydroxyl group from amoxicillin will result in it being less polar and less able to hydrogen bond, thus (a) decreasing its water solubility and (b) increasing its fat solubility.

99. It represents Ketoprofen. It is related to diphenylketone (diphenylmethanone). It contains a ketone and a carboxylic acid (-COOH).

100. (a) Changing to the palmitate ester would increase its fat solubility, it is becoming much less polar and much more “fat like.” (b) Changing it to a sodium salt would increase its water solubility. Sodium salts are soluble in water.

101. It is a prostaglandin (prostaglandin E2 or PGE2); it contains carboxylic acid(-COOH), ketone(C=O), and two alcohol (-OH) functional groups.

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102. It is a steroid (c), see Problem 77 for the common steroid structure.

103. TDF E-Factor = 70*657 tons/year = 46,000 tons waste per year TAF E-Factor = 85*21.9 tones/year = 1860 tons of waste per year 44,100 tons less waste generated using TAF 104. Principle 5: Safer solvents—using water Principle 2: Atom economy Principle 9: Using enzymatic/catalytic processes

142 .

Chapter 19: Fitness and Health Overview Chapter 19 connects the world of fitness, how we feel, with the type and quantity of chemical compounds we ingest. The chapter begins by discussing the food categories and how much of each category should be in our diet. Next, vitamins, minerals, and electrolytes are introduced. The chapter concludes with the chemistry of sports materials.

Lecture Outline 19.1 Calories: Quantity and Quality Most calories should come from carbohydrates. No more than 25% of calories should come from fats.

19.2 Vitamins, Minerals, Fluids, and Electrolytes Table 19.1 is a table of recommended daily adult intakes of vitamins.

Sodium, potassium, and chloride ions are the major essential electrolytes.

Water is the best drink when you are thirsty!

19.3 Weight Loss: Diets and Exercise To lose a pound, you must burn off 3500 kcal.

The healthy way to lose weight is to exercise. Not only will you lose weight, but you will also look and feel better.

19.4 Measuring Fitness Men should have more than 3% body fat; women need 10–12%.

19.5 Some Muscular Chemistry Muscles consume ATP (adenosine triphosphate) for energy.

Aerobic muscle contraction occurs in the presence of oxygen.

Anaerobic muscle contraction occurs when sufficient oxygen is not available.

19.6 Drugs, Athletic Performance, and the Brain Restorative drugs are painkillers and anti-inflammatory drugs.

Stimulant drugs are used in an attempt to improve performance.

Anabolic steroids are used to build muscle mass.

Some runners get a euphoric high during and after a long, hard run.

Smoking is one of the worst things you can do to your physical well-being.

Review Questions 1. Animal products; they are high in saturated fats.

2. The individual’s age, sex, activity level, and general health condition must be considered.

3. The loss of 5 pounds in one week is mostly loss of water; no.

4. Recent research indicates that obesity is a combination of genetics and environment.

5. The development of a better understanding of muscle physiology, pharmacological studies, and the development of improved sporting equipment 143 .

Chapter 19: Fitness and Health

6. Athletes need more calories; by eating more carbohydrates.

7. a) skin-fold calipers: inaccurate and measure water retention as well as fat b) displacement of water: cannot get all air out of the lungs 8. The heart must supply blood to 200 miles of capillaries for every pound of fat; thus it works harder.

Other health risks include diabetes.

9. Saturated fats and cholesterol are implicated in the clogging of arteries.

10. Balding, development of extra body hair, deep voices, and menstrual irregularities may all be caused by anabolic steroid use by females.

Problems 11. Athletes need more calories than the average sedentary individual. The extra calories should come from carbohydrates.

12. Protein metabolism produces more toxic wastes that tax the liver and kidneys.

13. Vitamin B12 helps in healthy regulation of the nervous system, reducing depression, stress, and brain shrinkage; helps maintain a healthy digestive system; protects against heart disease by curbing and improving unhealthy cholesterol levels; protects against stroke and high blood pressure. It is essential for healthy skin, hair, and nails; and helps protect against cancers, including breast, colon, lung, and prostate cancer. Deficiency leads to pernicious anemia.

14. Scurvy is caused by a lack of vitamin C and has symptoms of lethargy, followed by formation of spots on the skin, spongy gums, and bleeding from the mucous membranes. Vitamin C prevents scurvy.

15. Vitamin B6 provides relief from arthritis and lowers blood cholesterol.

16. with an intake of β-carotene 17. Vitamin D promotes absorption of calcium and phosphorus; the upper limit is 2000 IU. 18. They prevent oxidation from occurring in other tissues and blood vessels.

19. They promote water loss; keep water in the body.

20. If urine is cloudy, you are dehydrated.

21. No, thirst is a delayed response to water loss.

22. No, many of the sports drinks are so concentrated in electrolytes that consumption may lead to diarrhea. The best way to rehydrate is to drink water.

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23. Loss of glycogen: It will be replaced as soon as carbohydrates are eaten.

Loss of water: It will be regained as soon as water is drunk.

24. The brain requires glucose, and if not provided in the diet it is obtained from protein, leading to muscle loss.

25.

420 kcal = 2.0 h 210 kcal h

26.

40 kcal km x 6 km day x 365 day = 11 kg 7700 kcal/kg

27. 0.8 g/kg x 220 lb x 0.4536 kg/lb = 80 g of protein 28. 0.8 g/kg x 110 lb x 0.4536 kg/lb = 40 g of protein 29.

2000 kcal = 20 km 100 kcal/km

30.

100000 kcal = 1000 km 100 kcal/km

31. a) Leptin controls appetite and metabolic rate. b) Ghrelin is linked to short-term eating behaviors.

32. a) Melanocortins act on the brain to control food intake. b) PYY acts as an appetite suppressant.

33.

705 x 186 lbs = 26.8 (5ft x 12 in/ft + 10 in) 2

34.

705 x 300 lbs = 38.6 (6 ft x 12 in/ft + 2 in) 2

35. Exercise makes muscles larger, more flexible, and more efficient in their use of oxygen. No, you do not create new muscle.

36. actin and myosin 37. No, exercise results in larger muscles.

38. Aerobic exercise: muscle contractions that occur in the presence of oxygen.

Anaerobic exercise: muscle contractions that occur with insufficient oxygen.

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39. a) anaerobic

b) aerobic

40. Actomyosin forms the structural complex in muscles and acts as an enzyme for the removal of a phosphate group from ATP. 41. provides faster oxygen transport and increased respiratory capacity 42. Type I-aerobic; Type IIB-anaerobic 43. It allows ATP to be generated rapidly and also to be hydrolyzed rapidly during intense muscle activity.

44. Slow-twitch Type I muscles have high amounts of myoglobin, faster oxygen transport, and increased respiratory capacity. Slow-twitch Type IIB muscles have low respiratory capacity and low myoglobin levels, but have a high capacity for glycogen use.

45. No, the best way to build muscles is by weightlifting.

46. Weight training develops fast twitch muscle fibers. These fibers increase in size and strength with repeated anaerobic exercise.

47. Cortisone derivatives reduce swelling in damaged joints and tissues. Side effects of cortisone use include fluid retention, hypertension, ulcers, and disturbance of sex hormone balance.

48. steroid hormones used to increase muscle mass; males: testicular atrophy and loss of function, impotence, acne, liver damage, edema, elevated cholesterol levels, and growth of breasts; females: lowering of voice, extra body hair, increased masculinity, balding, and menstrual irregularities 49. In the “runner’s high” endorphins are produced in the body, which block pain.

50. Runners feel down when they cannot run, and they have to run increasingly farther to get the same high.

51. Exercise increases the brain’s blood supply and produces neurotophins, substances that enhance the growth of brain cells.

52. screening blood and urine samples for illegal drugs 53. heart disease, stroke, lung cancer, emphysema, pneumonia, cancer of the pancreas, bladder, breast, kidney and cervix 54. Smokers have more stillborn babies, premature babies, and babies who die before the first month.

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Additional Problems 55. (15 kcal/lb) x 160 lb = 2400 kcal 56. (0.2 L/min.kg) x 50 kg = 10 L/min oxygen required (10 L/min) x (45 s) / (60 min/s) = 7.5 L oxygen required (4 L/min) x (45 s) / (60 min/s) = 3.0 L available Oxygen Debt = 4.5 L 57. 90 kg/110 L = 0.82 kg/L x 1000 g/1 kg x 1 L/1000 mL = 0.82 g/mL; fat 58. 14 lb x 3500 kcal/lb x 1 yr/365 days = 134 kcal/day 59. Generate a graph of the data or estimate values directly from the data table.

40.00

35.00

% Body Fat

30.00

25.00

20.00

15.00

10.00

5.00 1.010

1.020

1.030

1.040

1.050

1.060

1.070

1.080

1.090

Body Density (g/cm3)

a) 26.4 % b) Dperson = (Mperson)/(Massperson-Massunderwater) Convert measurements to relevant units and take into account 1.57 L of air retained in the lungs, which adds 1.5 lb to the mass of the water.

165 lb x 0.4536 kg/lb = 74.8 kg = Mperson 14 lb x 0.4536 kg/lb = 6.4 kg + 1.6 kg = 8.0 kg = Munderwater Dperson = (74.8 kg)/(74.8 kg – 8.0 kg) = 1.1 kg/L = 1.1 g/cm3 3.6% Body fat c) person B d) Dperson = (89.34 kg)/(89.34 kg – 5.30 kg) = 1.06 g/cm3 Air in the lungs allows the person to float.

60. the continuation of a high-calorie diet and a lack of continued exercise

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61. a)

b)

705 x 199 kg x 2.2 lb kg = 27 (8 ft x 12 in/ft + 11 in) 2 705 x 486 kg x 2.2 lb kg = 144 (184 cm x 0.3937 in/cm) 2

62. Pheasants have Type IIB; great blue herons have Type I. 63. a) 15 g Hb/100 mL x 100 mL x 1.34 mL O2 = 20 mL O2 b) 15 g Hb/100 mL x 6000 mL x 1.34 mL O2 = 1200 mL O2 64. energy efficiency and less pollution 65. 1.4 mi x 2 x 80 x

45 k cal 1 lb = 10080 k cal x = 2.88 lb mi 3500 k cal

66. d 67. c

148 .

Chapter 20: Chemistry Down on the Farm . . .

Overview This chapter discusses modern food production and associated technologies. Fertilizers, pest control, herbicides and defoliants, and sustainable agriculture are all discussed from a conventional, modern farming techniques perspective to modern organic/alternative techniques.

Lecture Outline 20.1 20.2 20.3 20.4 20.5 20.6

Farming with Chemicals: Fertilizers Discuss how we use and produce the three primary plant nutrients: nitrogen, phosphorus, and potassium.

The War Against Pests Discuss how we control substances that kill organisms that we consider pests as well as how to handle pest resistance.

Pesticides and biological insect controls Herbicides and Defoliants Discuss how to control weeds as well as how to handle weed resistance.

Sustainable Agriculture Discuss what is needed to produce food and fiber indefinitely.

Sustainable agriculture and organic farming Some Malthusian Mathematics Rule of 72 and population growth and its implications Can We Feed a Hungry World?

Demonstrations [none] Review Questions 1. Carbon dioxide (CO2) and water (H2O) 2. Farming must include crop diversification, integrated pest management, disease prevention by careful crop management, and genetic improvement of crops. These will result in a sustainable agriculture without causing irreparable damage to the ecosystem and accomplish the main goals of (1) providing a healthy environment, (2) achieving farm profitability, and (3) attaining social and economic equity.

3. A preemergent herbicide kills weeds before crop seedlings emerge, and stops germination.

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4.

Energy Labor Profitability Yields

Organic 2.3 times less 12% more Similar* 10 % lower

Conventional 2.3 times more 12% less Similar* 10% higher

* Organic farms generate more of their profits from operating efficiency and price markup, while conventional farms generally derive more of their profits from volume of sales.

5. Thomas Robert Malthus predicted that populations increase faster than the food supply. Unless the birth rate is controlled, poverty and war would have to serve as restrictions on the increase.

Malthus’s predictions were based on simple mathematics: Population grows geometrically, while the food supply increases arithmetically.

6. The primary challenges are opposition to genetic modification of crops, which can increase yields, resistance, and nutritional value along with transportation issues, and getting the food to the people who need it due to cost of transportation and political issues involved.

7. Carbon, hydrogen, and oxygen 8. Nitrogen, phosphorus and potassium 9. Nitrogen is fixed synthetically mainly by the Haber process, which catalyzes the reaction: 3 H2 + N2 → 2 NH3. Other high-energy processes also can fix nitrogen (combustion). Natural sources of nitrogen fixation include lightning and bacteria.

10. The Haber process catalyzes the reaction: 3 H2 + N2 → 2 NH3 11. Anhydrous ammonia, without water, is NH3(g) and is a gas; it is used as fertilizer. When the ammonia is dissolved in water, it is ammonium hydroxide, NH4OH(aq), a solution.

12. Phosphate rock is treated with phosphoric acid to make calcium dihydrogen phosphate.

Phosphates play an important role in the energy-transfer processes of photosynthesis.

13. Ion exchange occurs; the potassium ion is exchanged for the hydronium ion (H+).

14. Eutrophication occurs when phosphate runoff enters lakes, ponds, and rivers, causing rapid growth of blue-green algae and other aquatic plants in waterways.

15. Advantages: DDT is very effective against insects, Disadvantages: it remains toxic long after initial use—environmental persistence.

16. DDT interferes with calcium metabolism and the formation of eggshells.

17. Chlorinated hydrocarbons are soluble in fats and bioaccumulate.

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Chapter 20: Chemistry Down on the Farm . . .

18. a. A narrow spectrum insecticide is a pesticide effective against only a few insects, such as carbaryl, carbofuran, and aldicarb. b. A broad-spectrum pesticide is one that kills many kinds of insects, such as malathion or DDT. 19. Pheromones are used to attract a mate, so they are used to attract the males to monitor infestations to determine when best to use a pesticide. The attractant can also be used in quantities sufficient to confuse and disorient males, who detect a female in every direction but can’t find one to mate with.

20. Juvenile hormones control the rate of development of the young; they can be synthesized and used to keep insects in the juvenile stage.

21. SIT sterile insect technique. Males are sterilized by radiation, chemicals, or crossbreeding, and then released. The process is very expensive.

22. Atrazine and glyphosate. Atrazine is a nonselective herbicide, used mostly on weeds in cornfields, which are resistant to it. Glyphosate is a nonselective herbicide that kills all plants, unless they have been genetically modified to resist it.

23. 70 kg x (0.50 g/kg) = 35 g DDT 24. 12 kg x (2.0 mg/kg) = 24 mg Carofuran (0.024 g) 25. 56 kg x (600 mg/kg) = 33,600 mg (33.6 g) diquat dibromide 56 kg x (150 mg/kg) = 8400 mg (8.4 g) paraquat 26. $1000/($125/Month) = 8 months. Arithmetic growth: A constant amount is added for each growth period.

27. Thirty-two gerbils as shown in the table below; geometrically. Note: You’d better like gerbils!

Weeks 6 12 18 24

# Gerbils 4 8 16 32

28. Apply the rule of 72; 72/1.35 = 53 years; changes in birth rate or death rate (politics, war, famine, birth control . . . ).

29. Apply the rule of 72; 72/0.493 = 146 years to double. Changes in birth rate or death rate (politics, war, famine, birth control.)

30. 38.6% of land, roughly 37 billion acres of land on the Earth.

(0.386 x 37 x109 acres)/106 = 14,000 million acres 31. 14.9% used for urban land, roughly 37 billion acres of land on the Earth.

(0.149 x 37 x109 acres)/106 = 5500 million acres 151 .

Chapter 20: Chemistry Down on the Farm . . .

32.

(40 - 7) x 100 = 470% increase. Increase in oil prices and profitability of ethanol production 7 make this an enticing business, however, since we are burning food, it removes significant amounts of food that could be used for eating.

33. 6 CO2 + 6 H2O → C6H12O6 + 6 O2 34. 3 H2 + N2 → 2 NH3 35.

Boric acid Diatomaceous earth Lead arsenate Malathion Rotenone

Organic Chemical

Organic Farming

Both Chemical and Farming

No No

Yes Yes

No No

Organic in No Way at All? False False

No Yes Yes

No No Yes

No No Yes

True False False

Organic Chemical

Organic Farming

Both Chemical and Farming

Yes Yes Yes Yes Yes

Yes No No Yes No

Yes No No Yes No

Organic in No Way at All False False False False False

36.

Acetic acid Atrazine 2,4-D Gluten meal Glyphosate

37. KNO3; potassium and nitrates.

38. (a) (NH4)2HPO4; (b) nitrogen and phosphorus; (c) from ammonia and phosphoric acid 2 NH3 + H3PO4 (NH4)2HPO4 39. (a) codling moth: alkenes and an alcohol (b) juvenile hormone molecule: epoxide (a cyclic ether), alkene and an ester. (c) methoprene: ether, alkene and an ester 40. Pyridium ion, chloride ion, 41. Lime is calcium oxide, CaO. It is used to neutralize acidic soils.

42. Eugenol contains an alkene, ether, arene, alcohol (arene + alcohol = phenol) 152 .

Chapter 20: Chemistry Down on the Farm . . .

43. 2 KOH + CO2 → K2CO3 + H2O (Formation of potassium carbonate) K2CO3 + CO2 + H2O → 2 KHCO3 (Formation of potassium bicarbonate) It is not organic in a chemical sense (no C-C or C-H bonds) and not organic in the natural sense, it is a synthetic compound as its synthesis is described above. But it does occur naturally, and as stated in the problem, it is an organic pesticide.

44. Dicamba: Cl

O

OH

O

CH3

Cl

45. (a) (37.5-34.5)/34.5*100= 8.70% increase (b) Apply the rule of 72; 72/8.70 = 8.3 years to double. (c) Changes in birth rate or death rate (politics, war, famine, birth control…)

46. The soap only acts while wet and in direct contact with the insect. Once it dries out you mush “wash” the plants with water two to three hours after applying. Repetition may be needed four to seven days later, not practical large scale.

47. (a) increased yield of food production (b) reduced crop failure 48. Pesticides are often toxic to other organism than the target organism (broad range). If they are stable, they can also bioaccumulate.

49. d. All of the above.

153 .

Chapter 21: Household Chemicals Overview The modern home is stocked with a variety of chemical products. Chapter 21 introduces the student to the chemistry of soaps, detergents, bleaches, paints, cosmetics, perfumes, and other chemical compounds used in and around the home.

Lecture Outline 21.1 A History of Cleaning Fat + Lye  Soap. Soap is a substance formed by mixing a base and a fatty acid.

21.2 Synthetic Detergents Modern detergents are made in the laboratory.

ABS detergents: nonbiodegradable LAS detergents: biodegradable Laundry detergent formulations Dishwashing detergents 21.3 Laundry Auxiliaries: Softeners and Bleaches Fabric softeners: quaternary ammonium salts Laundry bleaches: whiter whites Chlorine kills bacteria and whitens clothes.

21.4 All-Purpose and Special-Purpose Cleaning Products Hazards of mixing cleaners Special-purpose cleaners Green cleaners 21.5 Solvents, Paints, and Waxes Solvents Paints Waxes 21.6 Cosmetics: Personal Care Chemicals Skin creams and lotions Moisturizers, sunscreens Antiaging creams and lotions Nanoparticles in cosmetics Lipstick Eye makeup Deodorants and antiperspirants Toothpaste: Soap with grit and flavor It’s the same thing you wash your clothes with!

Perfumes, colognes, and aftershaves Aromatherapy Some hairy chemistry Shampoo, hair coloring, permanent waving, hair sprays, Hair removers, hair restorers The well-informed consumer

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Demonstrations 1. Into a tall glass place some baking soda, some liquid detergent, and a few drops of red food color.

Adding vinegar to this mixture will cause it to bubble up and look somewhat like a strawberry ice cream soda.

2. Add several drops of food coloring to each of two beakers about half full of water. Add a little liquid chlorine bleach to one of the beakers, and watch the color disappear.

Review Questions 1. Soaps are excellent cleansers in soft water, relatively nontoxic, derived from renewable sources, and biodegradable. Soaps do not work well in hard or acidic water.

2. Synthetic detergents are good cleansers in hard water and acidic water. They are more expensive to produce.

3. (1) is (a) CH3COOH is the ingredient in vinegar.

(2) is (f) Na2Al2Si2O8 is a zeolite.

(3) is (b) NaHCO3 is a mild abrasive cleaner.

(4) is (c) NaOCl is used in bleach.

(5) is (d) NaOH is used to make soap.

(6) is (e) Na3PO4 is a water softener.

4. They contain one or more surfactants such as LAS; detergents for automatic dishwashing are strongly alkaline.

5. Diluted ammonia solutions clean mirrors, windows, and other surfaces; they should not be used in a closed room because the vapors are irritating. It should not be mixed with chlorine bleaches.

6. Ammonia is effective for loosening baked-on grease or burned-on food. It should not be used on asphalt tile, wood surfaces, or aluminum; it stains, pits, and erodes those materials surfaces.

7. mild abrasive cleanser that absorbs odors 8. “articles intended to be rubbed, poured, sprinkled or sprayed on, introduced into, or otherwise applied to the human body or any part thereof, for cleansing, beautifying, promoting attractiveness or altering the appearance”; soap is specifically excluded from the definition.

9. It is softer, and it produces a finer lather;

10. Hard water exchanges the sodium ion on the soap for a calcium or magnesium ion in the soap; these compounds are not soluble in water and precipitate out.

11. Abrasives, such as sodium carbonate, silica, pumice, and oatmeal; they scrape soil from hard surfaces.

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12. Hydrocarbon tails dissolve in oils, and the ionic “head” dissolves in water, allowing the oils/dirts to be washed away.

13. Fat is mixed with lye, and the fat is hydrolyzed into a fatty acid (soap) and glycerol.

14. Tripalmitin is mixed with lye (NaOH), and it is hydrolyzed to form sodium palmitate and glycerol.

O 15. (a) CH3(CH2)16C-O

K

(b)

16.

17. It softens the water by binding calcium and magnesium ions to form in soluble molecules, called complexes, and makes the water slightly basic (alkaline).

18. Water runs through the softener and replaces Ca2+ and Mg2+ ions in water with 2 Na+ (ion exchange).

19. Ca2+ and Mg2+ ions 20. CO32-(aq) + Mg2+(aq) →

MgCO3 (s)

21. 2 PO43- + 3 Ca2+ → Ca3(PO4)2(aq) 22. Hard water runs into the softener and through the sodium zeolite where the sodium is released into the water and the calcium ions are bound to the zeolite (ion exchange).

23. Any substance added to a surfactant to increase its detergencyA 156 .

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24. By reducing water hardness by the sequestration of calcium and magnesium ions in soluble complexes 25. Zeolites exchange calcium and magnesium ions for sodium ions and keep them soluble.

26. Calcium and magnesium ions are held in suspension rather than being precipitated; they reduce solids forming in the sink.

27. A surfactant molecule that carries both a positive and negative charge.

28. (a) ionic

(b) cationic

29. They would react with each other; the positive end of the cationic surfactant would interact with the negative end of the anionic surfactant, effectively disabling their function.

30. ABS has a highly branched alkyl group on the benzene of the detergent, while LAS has a linear alkyl group on the benzene of the detergent; both are benzenesulfonates.

31. I 32. II 33. They are all biodegradable. I, II, and III. 34. III 35. anionic: I and II

cationic: III

36. Ammonia should not be used on asphalt tile, wood surfaces, or aluminum because it may stain, pit, or erode these materials. Vinegar should not be used on marble because it reacts with it and dissolves the marble, pitting the surface.

37. Quaternary ammonium salts such as hexadecyltrimethylammonium chloride or dioctadecyldimethylammonium chloride 38. They are strongly absorbed by the fabric, forming a flexible monomolecular layer on the surface of the fabric.

39. Trichloroisocyanuric acid and sodium percarbonate.

40. Borates work best with polyester materials but are somewhat toxic and are less active than chlorine bleaches and require higher temperatures, higher alkalinity, and higher concentrations to do an equivalent job. Chlorine bleaches are very inexpensive, work very well with cotton based materials. Due to their reactivity they do not work well with polyesters.

41. They are oxygen-releasing bleaches; they oxidize, with hydrogen peroxide, the stains/chromophores to form colorless compounds.

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42. Sodium hypochlorite; NaOCl 43. Inhaling toxic fumes and flammability 44. Petroleum distillates are flammable, and their fumes are toxic.

45. No. Gasoline is too flammable and has a high vapor pressure; it should never be used as a cleaning solvent.

46. Titanium dioxide; TiO2 47. Emollients are creams and lotions that protect the skin by coating and softening it.

48. 1. Water; 2. an oily substance that forms a protective film; and 3. an emulsifier that binds the polar water to the nonpolar oily component, usually glyceryl stearate.

49. OMC is insoluble in water and will not wash off while swimming. PABA is water soluble and washes off while swimming.

50. Chromium (III) oxide, iron oxides, and carbon 51. Lead (Lead sulfide) 52. ZnO forms a opaque film over your skin that blocks the sun’s rays from your skin, OMC absorbs the sun’s rays.

53. waxes or oils, dyes or pigments, and maybe a perfume and an antioxidant 54. Bromo acid dyes, such as tetrabromofluorescein 55. Bacteria; avoid sugars, and brush and floss after eating.

56. Na2FPO4; it is a source of fluoride ions.

57. (a) glycerol: sweetener. (b) titanium dioxide: appearance of toothpaste—make it white. (c) sodium dodecyl sulfate, detergent to clean your teeth. (d) potassium nitrate: desensitizer. (e) hydrogen peroxide: whitening agent. (f) hydrated silica in toothpaste: abrasive to clean teeth.

58. Fluoride ions convert tooth enamel from hydroxyapatite to fluorapatite, a harder substance.

59. Perfume is a complex mixture of compounds used to make the wearer smell pleasant. Colognes are diluted perfumes.

60. Sex attractant of musk deer; to moderate the odor of the top and middle notes.

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61. Menthol provides a cooling effect when applied to the skin.

62. (a) Aluminum chlorohydrate or aluminum zirconium tetrahydroxy Gly complex These compounds function as astringents, which constrict the openings of the sweat glands and thus restrict the amount of perspiration that can escape. (b) Triclosan, it is a germicide to kill odor-causing bacteria.

63. Permanent dyes penetrate the hair and remain there. Temporary dyes are water soluble and can be washed out. New hair will grow in the original shade.

64. Resins are materials that form sticky layers (on the hair). Used in hair sprays.

65. Disulfide linkages are broken and restored when hair is permanently waved or straightened. If the disulfide bridged is restored when the hair is curled, it will remain curled, and if the hair is straight when the disulfide bridge is restored, the hair will remain straight.

66. Detergents such as sodium dodecyl sulfate .

Ammonium lauryl sulfate, or sodium laureth sulfate would also work.

67. Going from dark to light-colored hair is usually done by bleaching the hair, while darkening the hair requires adding a compound that has color or produces color into your hair.

68. Temporary hair dye is a water-soluble dye that is mostly just on the surface of your hair that can be washed out. A semipermanent dye and permanent dyes penetrate the hair and remain; permanent dyes are more strongly bonded to the hair.

69. II 70. I, III, IV 71. I 72. I 73.

O CH3(CH2)14C-O-CH(CH3)2

74. The reaction of interest: Na2CO3 + Ca2+  CaCO3 + 2 Na+ mass of Ca2+ in 10.0 L of water: 295 mg/L x 10.0 L = 2950 mg = 2.95 g moles of Ca2+ in water: 2.95 g/ (40.078 g/mol) = 7.36 x 10-2 mol Ca2+ moles of Na2CO3 = moles of Ca2+ (due to stoichiometry in the first equation) Find mass of Na2CO3: 7.36 x 10-2 mol Na2CO3 x 105.989 g/mol = 7.80 g Na2CO3.

75. 2-phenylethanol is an alcohol and is less volatile due to hydrogen bonding.

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77. Sodium hydroxide reacts with water, generating heat, and reacts with some fats, converting them to soaps and dissolving them.

78. No, they are fluorescent (convert UV light to visible light) which makes fabrics appear whiter.

79. Both are esters of palmitic acid (C16). Carnauba has a much longer carbon chain.

80. Bees wax is the first picture: 81. (a) 2 HCl + CaCO3  CaCl2 + H2O + CO2

(b) 2 NaHSO4 + CaCO3  CaSO4 + H2O + CO2 + Na2SO4 (c) 2 HC6H7O7 + CaCO3  CaC6H7O7 + H2O + CO2

82. (SPF 30 x 15 min)/60 min/hr = 7.5 hrs. You may not have put on the recommend amount, it can rub off, or come off while swimming.

83. (c) Water (H2O), the others are VOCs 84. (c) Molecules that are linear and contain only carbon, hydrogen, and oxygen, like c., degrade more quickly than complicated molecules that contain many rings, branches, or halogen atoms.

85. (a) is 3. NaOCl s used in bleach. (b) is 1. CH3(CH2)14COO- Na+ is a soap. (c) is 4. Na5P3O10 is a detergent builder. (d) is 2. CH3(CH2)14COOCH2(CH2)28CH3 is a wax.

86. Heat = mass x specific heat x temperature change 75 L x (1000 mL/L) x 1.0 g/mL = 75,000 g water Heat = 75,000 g x 1.0 cal/g °C x (65 °C – 10 °C) = 4.1 x 106 calories saved per load.

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Chapter 22 Poisons: Toxicology: What Makes a Poison?

Overview Many compounds are toxic to humans. Chapter 22 discusses many toxic substances that are known, giving priority to those that are more likely to be encountered in everyday life, their effects, detection, identification, and their antidotes.

Lecture Outline 22.1 Natural Poisons Natural poisons and their sources are given, such as oleander, hydrangeas, holly berries, iris, azaleas, and wisteria seeds.

22.2 Poisons and How They Act Understanding corrosive poisons, metabolic poisons, and heavy metal poisons and how each acts.

Antidotes for some common metabolic poisons and heavy metal poisons.

22.3 More Chemistry of the Nervous System Nerve poisons: stopping the acetylcholine cycle (various substances interfere with the action of neurotransmitters).

Nerve poisons are the most deadly killing agents developed for military use.

22.4 The Lethal Dose How much does it take to kill? The concept of the LD50 value.

22.5 The Liver as a Detox Facility How the body is able to detoxify some substances.

22.6 Carcinogens and Teratogens Substances shown to cause cancer are known as carcinogens. Cancers and how they develop are discussed.

Three ways to test for carcinogens: The Ames test uses bacteria; animal testing uses test animals; epidemiological studies are a statistical approach.

A teratogen is a substance that causes birth defects.

22.7 Hazardous Wastes.

Industrial waste, reactive wastes, flammable wastes, toxic wastes, corrosive wastes, and what is the price we pay for these wastes/poisons.

Demonstrations [None] Review Questions 1. Although not thought of as a poison, yes it can be; in large amounts and to a diabetic.

2. Individuals can have different responses to compounds; sugar is more harmful to a diabetic than to a nondiabetic.

3. The toxicity will change based on the route of administration. Water inhaled can be deadly; nicotine is 50 times more toxic when taken intravenously than when taken orally.

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4. When fluoroacetic acid is ingested, it is incorporated into fluorocitric acid, which in turn blocks the Krebs cycle. Fluorocitric acid inhibits the citric acid cycle by tying up (deactivating) the enzyme that acts on citric acid.

5. Mercury: broken thermometers, tooth fillings, compact fluorescent light bulbs, agricultural uses.

Lead: lead pipes, paint, pencils, food containers.

6. Cigarette smoking (30%) and diet or lack of exercise leading to obesity (30%).

7. Many answers apply; a few more common ones are: sunlight, radon, safrole in sassafras. Many carcinogenic compounds are found in a variety of substances such as mushrooms, basil, celery, figs, mustard, pepper, fennel, parsnips, and citrus oils.

8. Benzopyrene, a carcinogen found in tobacco smoke, is oxidized into an active carcinogen that binds to the P53 gene, which causes the mutation. The P53 gene is mutated in 60% of all lung cancers.

9. a. Sulfuric acid (H2SO4), d. sodium hydroxide (NaOH) 10. Dilute solutions of acids and bases catalyze the hydrolysis of protein molecules in living cells by breaking peptide linkages.

11. Hydrolysis of proteins causes them to fail to carry out their functions, such as the deactivation of an enzyme. Extreme exposure can cause complete destruction of the proteins.

12. Ozone is a very strong oxidizer, it deactivates enzymes through oxidation, especially enzymes that have sulfur-containing amino acids such as cysteine and methionine (which most do).

13. A metabolic poison is a chemical substance that prevents cellular oxidation of metabolites by blocking the transport of oxygen in the bloodstream or by interfering with oxidative processes in the cells. Substances such as carbon monoxide and nitrates.

14. Methemoglobin is when the Fe2+ in hemoglobin is oxidized to Fe3+ , Hemoglobin is bright red and is responsible for the red color of blood. Methemoglobin is brown. During cooking, red meat turns brown because hemoglobin is oxidized to methemoglobin. Dried bloodstains turn brown for the same reason; Methylene blue is capable of reducing the methemoglobin back to hemoglobin.

15. Nitrate ions, found in dangerous amounts in the groundwater in some agricultural areas diminish the ability of hemoglobin to carry oxygen. Microorganisms in the digestive tract reduce nitrates to nitrites. The nitrite ions oxidize the Fe+2 in hemoglobin to Fe+3 forming methemoglobin (see previous question).

16. by converting the toxic cyanide to a nontoxic thiocyanate ion 17. c. Molybdenum and d. Tellurium. (Beryllium is very toxic metal, but not a heavy metal.)

18. Too little: anemia; Too much: vomiting, shock, coma, and death.

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19. In fish, mercury is converted to a very toxic Hg+ ion (in the form of CH3Hg+). Ordinary exposure to mercury results in the mercury being converted to Hg+2 ions, still a poison, but not as toxic as Hg+.

20. BAL acts by chelating (binding to) the mercury ion, thus allowing the mercury to be excreted.

21. EDTA chelates lead ions, thus enhancing their excretion.

22. Too much cadmium leads to a loss of calcium ions from the bones, severe abdominal pain, vomiting, diarrhea, and choking sensation; itai-itai is “it hurts, it hurts” disease.

23. Tabun, Sarin, and Soman 24. Tabun has a fruity odor and is quickly detected.

25. Anticholinesterases are nerve poisons that inhibit the enzyme cholinesterase and are generally organophosphate compounds; The organophosphate ties up acetylcholinesterase, preventing it from breaking down acetylcholine. Acetylcholine therefore builds up, causing postsynaptic nerve cells to fire repeatedly. This overstimulates the muscles, glands, and organs. The heart beats wildly and irregularly, and the victim goes into convulsions and dies quickly.

26. Although botulin is the most toxic substance known, it has a commercial formulation called Botox®. It is used in medicine to treat intractable muscle spasms and is widely used in cosmetic surgery to remove wrinkles.

27. Nicotine is detoxified by oxidation to cotinine, which is less toxic than nicotine, and cotinine is more water soluble than nicotine, and thus more readily excreted in the urine.

28. The P-450 system is a group of enzymes in the liver. The P-450 system oxidizes fat-soluble substances into water-soluble substances so that they are readily excreted.

29. No, some are made more toxic by the oxidation—such as the oxidation of benzene.

30. 1. Toluene is oxidized to benzoic acid. 2. Benzoic acid reacts with glycine forming hippuric acid, which is water soluble and can be excreted.

31. The liver enzymes that oxidize alcohols also react with the male hormone testosterone. More liver enzymes means more hormones also reacting.

32. Fomepizole is a safer antidote than ethanol because it inhibits the liver enzyme that catalyzes the oxidation of alcohols, not blocking it with an excess of an ethanol, which can be dangerous.

33. 125 lb x 0.454 kg/lb = 56.8 kg 140 mg/kg x 56.8 kg = 7950 mg methyl isocyanate (7.95 g) 34. 307 mg/kg x 18 kg = 5526 mg of carbaryl (5.5 g) 163 .

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35. Mutagens are substances that create mutations; The Ames test: Bacteria are incubated in the presence of a mutagenic substance; it assumes that carcinogens are also mutagens.

36. Teratogens are a chemical substance that causes birth defects. The defect may not be noticed for several years after birth. It is often difficult to relate the defect back to the mother or the chemical after all of the years (time delay).

37. PAHs are formed primarily during the incomplete burning of nearly any organic material, such as grilled meats, cigarette smoke, coffee, burnt sugar, and most any burnt food.

38. a. β-naphthylamine and benzidine

b. dimethylnitrosamine and vinyl chloride

39. Humans are not exposed to comparable doses; human metabolism is different from test-animal metabolism.

40. A study of a human population to try to link human health effects (e.g., cancer) to a cause (e.g., exposure to a specific chemical); no 41. Oncogenes are genes that regulate cell growth, and they sustain the abnormal growth characteristic of cancer, while suppressor genes are genes that prevent the development of cancer and are deactivated before a cancer can develop.

42. The Ames test is conducted to find out whether or not a substance may be carcinogenic. It does this by determining whether bacteria mutate in the presence of a chemical. If the bacteria mutate, it is assumed they are carcinogenic. The assumption that because they are mutagens means they are carcinogenic may not be true. Animal tests may also be done, but they may not lead to conclusive correlations to humans either.

43. Toxic wastes are substances released into the environment in quantities sufficient to pose a hazard to human health or to the environment. Many apply; some household toxic wastes are paint, pesticides, motor oil, medicines, battery acid, drain cleaners, and mercury from CFLs. 44. (a) sodium hydroxide and d. phosphoric acid.

45. LD50 = 200 pg/kg x 66 kg = 1.32x 104 pg 1.32 x 104 pg x 1.0 x 10-12 g/pg = 1.32 x 10-8 g 1.32 x 10-8 g/(150,000 g/mol) = 8.8 x 10-14 mol 8.8 x 10-14 mol x 6.02 x 1023 = 5.3 x 1010 molecules of botulin 46. (a) LD50 = 200pg/kg x 60 kg = 12,000 pg per person (b) 12,000 pg x 7.0 x 109 (1.0 g/1012pg)= 84 g 47. (a) C10H16O (b) 87.5 mg/kg x 55 kg = 4800 mg = 4.8 g (c) 10 mg/L x 0.50 L = 5.0 mg

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48. (a) LD50 = 70 mg/55kg = 1.3 mg/kg (b) between ketamine and ethylene glycol

1350 x100% = 99.48% 1357 1296 % smokers without cancer in control group = x100% = 95.50% 1357 You can infer that you are studying similar populations and you are more likely to get cancer if you smoke (b) 0.9948:1 for cancer cases; 0.9550:1 for controls (c) 0.9550/0.9948 = 0.96:1 – about the same

49. (a) % smokers with cancer =

50. Using density greater than 5 g/mL. Iron, copper, zinc, manganese, molybdenum, chromium, cobalt, vanadium, nickel, cadmium, tin, lead.

51. Depending on the poison, the amount you intake (the dose) will determine whether it poisons you.

Some substances may be required at low levels but are toxic at high levels.

52. 200 μg = 0.000020 g, between sarin/Agent VX and tetanus toxin 53. (a) 200 μg/kg x (1 mg/1 x 103μg) = 0.20 mg/kg LD50 1.5 mg/55 kg = 0.027 mg/kg (27 μg/kg); no, it is below 200 μg/kg. (b) 200 μg/kg x (1 mg/1 x 103μg) = 0.20 mg/kg LD50 20 mg/55 kg = 0.364 mg/kg (364 μg/kg); yes, it is plenty to be lethal.

54. From Table 22.1: LD50 for arsenic trioxide = 15 mg/kg (a) (15 mg/kg) x 60 kg = 900 mg needed to be lethal; no. 10 mg is not enough.

From Table 22.1: LD50 for ketamine = 0.229 mg/kg (b) (0.229 mg/kg) x 60 kg = 13.7 mg needed to be lethal, yes, 20 mg is more than enough.

55. Benzene, because of its general inertness in the body, is not acted upon until it reaches the liver.

There it is slowly oxidized to an epoxide. The epoxide is a highly reactive molecule that can attack certain key proteins. The damage done by this epoxide sometimes results in leukemia. If the body could extract it without reacting with it, it would not be toxic.

56. It becomes more polar and allows it to hydrogen bond to water, stay dissolved, and be excreted.

57. (a) 10.3 g/kg x 0.395 kg = 4.1 g ethanol, a density of 0.92g/ml for a 50% water/ethanol mix (100 proof) = 4.1 g / (0.92g/mL) = 4.46 mL ethanol. 100 Proof is only half ethanol, so 4.46 mL x 2 = 8.9 mL vodka. (b) 395 g rat x 0.67 g water/g rat = 265 g water = 265 mL water 4.1 g ethanol x (1000 mg/g) = 4100 mg ethanol/265 mL water = 15.5 mg/mL

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58. False 59. (b) Rodent testing 60. (b) A chemist identifies structural features of a chemical that are responsible for hazardous properties and replaces those portions of the molecule with safer alternatives.

61. (b) By linking specific structural features to hazards, chemists can design those features out of new chemicals.

62. Diquat is as effective as paraquat but is much less toxic (Principle 3).

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