Issuu

Fundamentals of Chemistry for Today General, Organic, and Biochemistry 1e Spencer Seager, Tiffiny Rye-McCurdy, Ryan Yoder (Solutions Manual All Chapters, 100% Original Verified, A+ Grade) Chapter 1

Chapter 1: Matter, Measurements, and Calculations CHAPTER OUTLINE 1.1 What is Matter?

1.7 Signiﬁcant Figures

1.2 Physical and Chemical Properties and Changes

1.8 Using Units in Calculations: An Introduction to

1.3 Classifying Matter

Dimensional Analysis

1.4 Measurement Units

1.9 Calculating Percentages

1.5 The Metric System

1.10 Density and Its Applications

1.6 Large and Small Numbers: An Introduction to Scientiﬁc Notation

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Explain what matter is. (Section 1.1; Exercise 1.2)

Explain the diﬀerence between the terms physical and chemical as they apply to the properties of matter and changes in matter. (Section 1.2; Exercises 1.8 and 1.10)

Classify matter as an element, compound, homogenous mixture, or heterogeneous mixture. (Section 1.3; Exercises 1.16, 1.20, and 1.22)

Describe four measurement units used in everyday activities. (Section 1.4; Exercise 1.26))

Convert measurements within the metric system into related units. (Section 1.5; Exercises 1.28 and 1.38))

Convert temperatures measured in Fahrenheit to Celsius and vice versa. (Section 1.5; Exercises 1.41 and 1.42)

Express numbers using scientiﬁc notation. (Section 1.6; Exercises 1.46 and 1.47)

Perform calculations with numbers expressed in scientiﬁc notation. (Section 1.6; Exercises 1.58 and 1.59)

Express measurements and calculations using the correct number of signiﬁcant ﬁgures. (Section 1.7; Exercises 1.62 and 1.64)

10. Use dimensional analysis to solve numerical problems. (Section 1.8; Exercise 1.80) 11. Perform calculations involving percentages. (Section 1.9; Exercise 1.90) 12. Perform calculations involving densities. (Section 1.10; Exercise 1.96)

LECTURE HINTS AND SUGGESTIONS 1.

When describing chemistry as the “central science,” explain how everything around us is somehow related to chemistry. Look around the classroom and point out things which are a result of the study of chemistry; such as the plastic materials which make up part of the furniture, the paint on the walls, the clothing that we have on, the paper that we write on, the ink that we write with, and even the biochemical reactions which take place in our bodies which keep us alive.

Stress that a pure substance contains only one kind of basic building block or one kind of constituent particle. Every constituent particle in a pure substance is the same. If there are two or more kinds of

Chapter 1 constituent particles present, it is a mixture. Sugar has sugar molecules; water has water molecules; and sugar water has both sugar molecules and water molecules. 3.

Emphasize that an important characteristic of a pure substance is a constant composition. Give some simple examples, such as water or salt, which when free of other substances, always have the same composition regardless of source. Simple common solutions such as salt water can be used as examples of mixtures. Also, stress that a mixture may have a varying composition. For example, salt water may contain a very small amount of salt or a lot of salt. Salt water is a mixture. If it is left out in an open dish, the water will evaporate (a physical process) leaving behind the salt.

Students sometimes miss the whole point behind significant figures. The most important point to convey is that all measured data have some uncertainty associated with them that is inherent in the measuring device. A simple demonstration is to have students measure the classroom width using a rope knotted at about one-meter intervals, a meter stick and a tape measure. Note: Since the knots in the rope are not numbered, students need to manually count them. Have three students perform the same counting. The results often differ significantly for a large classroom.

Explain that dimensional analysis is just a way to convert between units and it can really save time when solving complex numerical problems. Begin by showing students how equalities can be written as conversion factors (i.e., fractions) and then move to show how multiplying conversion factors together can eliminate unwanted units and solve for the answer in the unit of interest. Emphasize that learning this method may take some time, however, it can be used to solve quantitative problems presented in not only chemistry but all the natural sciences, and thus it is time well spent learning the method.

Providing a handout with commonly used conversion factors and equations is helpful when introducing unit conversion and dimensional analysis. The example handout titled Chapter 1: Unit Conversion on the next page could be used as a resource for students to reference as they problem solve.

Chapter 1

CHAPTER 1: UNIT CONVERSION TABLE 1.2 Common Preﬁxes of the Metric System

Temperature Scales = °C

= °F

5 ( °F − 32 ) 9

9 ( °C ) + 32° 5

°C =

K − 273

K =

°C + 273

TABLE 1.5 Commonly Used Conversion Factors

A cm3 is commonly abbreviated “cc.” bA foot-pound is the energy it takes to push with one pound-force

(lbf) for a distance of one foot. cA BTU (British thermal unit) is the amount of heat required to increase the temperature of 1 pound of water by 1 °F.

Chapter 1

SOLUTIONS TO ALL END-OF-CHAPTER EXERCISES What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 1.1 WHAT IS MATTER? 1.1

If a heavy steel ball is suspended by a thin wire and hit from the side with a hammer on the moon, the heavy steel ball will hardly move, just like on earth. This experiment depends only on the mass of the ball and the hammer, not their weights.

1.2

All matter occupies space and has mass. Mass is a measurement of the amount of matter in an object. The mass of an object is constant regardless of where the mass is measured. Weight is a measurement of the gravitational force acting on an object. The weight of an object will change with gravity; therefore, the weight of an object will be different at different altitudes and on different planets.

1.3

To prove to a doubter that air is matter, precisely weigh a deflated balloon, then inflate it and weigh it again. The mass of the inflated balloon will be greater than the mass of the deflated balloon because the air in the inflated balloon has mass. The volume of the air is also clearly evident in the increased size of the balloon.

1.4

The distance you can throw a bowling ball will change more than the distance you can roll a bowling ball on a ﬂat, smooth surface. When throwing a ball, gravity pulls the ball towards the ground and air resistance slows its decent. The gravitational force on the moon is approximately 1/6th the gravitational force that is present on the earth; therefore, when throwing a ball on the moon, you should be able to throw it further than you can on earth. The moon does not have air resistance. When rolling a ball, friction helps to slow down the ball. If the ﬂat, smooth surface is the same on the earth and the moon, the amount of friction should remain constant.

1.5

If you were transported from a deep mine to the top of a tall mountain, your mass would not be changed by the move because mass is independent of gravity. b. If you were transported from a deep mine to the top of a tall mountain, your weight would decrease because weight depends on gravity and gravity decreases with distance from the earth’s center. A mountaintop is further from the earth’s center than a deep mine; therefore, your weight will be less on the mountaintop.

1.6

The attractive force of gravity for objects near the earth’s surface increases as you get closer to the center of the earth (Exercise 1.5). If the earth bulges at the equator, the people at the equator are further from the center of the earth than people at the North Pole. If two people with the same mass were weighed at the equator and at the North Pole, the person at the equator would weigh less than the person at the North Pole because the gravitational force at the North Pole is stronger than the gravitational force at the equator.

Chapter 1

SECTION 1.2 PHYSICAL AND CHEMICAL PROPERTIES AND CHANGES 1.7

a. The plum’s color, smell, and taste have changed. This was a change in composition; therefore, it is a chemical change. b. The water vapor can be condensed into liquid water and its properties will not have changed by the boiling. The composition of the water has not changed by boiling; therefore, it is a physical change. c. The glass pieces still have the same chemical composition as the original glass window. This was a change that did not involve composition; therefore, it is a physical change. d. The food is broken down into components that can be used by the body. This is a change that involves composition; therefore, it is a chemical change.

1.8

a. The two pieces of the stick still have the same chemical composition as the original stick. This was a change that did not involve composition; therefore, it is a physical change. b. As the candle burns, it produces carbon dioxide, water, soot, and other products. This is a change that involves composition; therefore, it is a chemical change. c. The pieces of rock salt have the same chemical composition as the original larger piece of rock salt. This was a change that did not involve composition; therefore, it is a physical change. d. Many tree leaves are green in the spring and summer because of the green chlorophyll that is used in photosynthesis to produce energy for the tree. During these seasons, the tree stores the extra energy so that in autumn when the days grow shorter, the chlorophyll is no longer needed. As the leaves in the cell stop producing chlorophyll, the other colors present in the leaves become more visible. This change involves composition; therefore, it is a chemical change.

1.9

a. Physical: a state of matter b. Chemical: binding indicates a change in composition c. Chemical: corrosion indicates a change in composition d. Chemical: neutralizes indicates a change in composition e. Physical: color is easily observed

1.10

a. The phase of matter at room temperature is a physical property because the composition does not change while making this observation. b. The reaction between two substances is a chemical property because the composition of the products diﬀers from the reactants. The products for the reaction between sodium metal and water are sodium hydroxide and hydrogen gas. (Note: Predicting the products for this type of chemical reaction is covered in Section 9.6) c. Freezing point is a physical property because the composition does not change while making this observation. d. The inability of a material to form new products by rusting is a chemical property because rust would have a diﬀerent chemical composition than gold. Attempting to change the chemical composition of a material is a test of chemical property regardless of whether the attempt is

Chapter 1 successful. e. The color of a substance is a physical property because the composition does not change while making this observation. 1.11

a. b.

Dying your hair is a chemical property because a chemical reaction alters the pigments in your hair which result in a color change. When sugar is spun to make cotton candy, it is a physical property because the sugar is only heated to change the state and then spun rapidly in the machine to convert the liquid sugar to thin solid strands. Using hydrogen peroxide to disinfect a wound forms bubbles in a chemical reaction. The hydrogen peroxide comes into contact with an enzyme called catalase, which decomposes into water and oxygen, which forms the bubbles. The use of magnesium chloride as a deicer is a physical property because the compound dissolves in water, which forms a solution that lowers the freezing point of water, which helps prevent ice from forming.

SECTION 1.3 CLASSIFYING MATTER 1.12

Carbon dioxide is heteroatomic. If oxygen and carbon atoms react to form one product, then carbon dioxide must contain these two types of atoms.

1.13

Hydrogen peroxide is heteroatomic. If water (which contains hydrogen and oxygen atoms) and oxygen gas can be produced from hydrogen peroxide, then hydrogen peroxide must contain both hydrogen and oxygen atoms.

1.14

Water is heteroatomic. If breaking water apart into its components produces both hydrogen gas and oxygen gas, then water must contain two types of atoms.

1.15

Heteroatomic: If the products contain hydrogen (in H2) and carbon (in CO2), the hydrogen and carbon must have come from glucose, making it heteroatomic.

1.16

a. b. c.

1.17

a. b. c.

1.18

Substance A is a compound because it is composed of molecules that contain more than one type of atom. Substance D is an element because it is composed of molecules that contain only one type of atom. Substance E is a compound because it is a pure substance that can break down into at least two different materials. Substances G and J cannot be classified because no tests were performed on them. Substance L is a compound. It is formed by combining two elements. Substances M and Q cannot be classified. Without further testing it is impossible to tell if the substances are elements or compounds. Substance X cannot be classified. The absence of a change is not conclusive evidence that a substance is an element or a compound. Substance R might appear to be an element based on the tests performed. It has not decomposed into any simpler substances based on these tests; however, this is not an exhaustive list of tests that could be performed on Substance R. Substance R cannot be

Chapter 1

b. c.

classified as an element or a compound based on the information given. Substance T is a compound. It is composed of at least two different elements because it produced two different substances on heating. The solid left in part b cannot be classified as an element or a compound. No tests have been performed on it.

1.19

Early scientists incorrectly classified calcium oxide (lime) as an element for a number of years. It is possible this mistake in classification was made because calcium oxide was the product of decomposing limestone (calcium carbonate) and it was difficult to further decompose the lime into the elements of calcium and oxygen.

1.20

a. heterogeneous b. homogeneous c. homogeneous d. heterogeneous e. homogeneous f. homogeneous g. heterogeneous

1.21

Muddy ﬂood water

It is heterogeneous because it does not have the same composition throughout (concentration of mud/debris depends on water depth).

Gelatin dessert

Normal urine

Smog-ﬁlled air

An apple

Mouthwash

Petroleum jelly

It is homogeneous because it has the same composition throughout. It is homogeneous because it has the same composition throughout. It is heterogeneous because it does not have the same composition throughout (concentration of smog, oxygen, other gases depend on the altitude.) It is heterogeneous because it does not have the same composition throughout (skin, meat, seeds.) It is homogeneous because it has the same composition throughout. It is homogeneous because it has the same composition throughout.

a. b.

Blood Liquid eye drops

An aspirin tablet

A urine sample with kidney stones Intravenous saline

1.22

This is a pure substance because it contains one substance. This is a solution because it contains many substances (water, salt, glycerin, etc.) This is a solution because it contains many substances (aspirin, starch, preservatives, etc.) This is a solution because it contains urine and kidney stones. This is a solution because it contains water and salt.

Chapter 1

1.23

f. g.

An antibiotic ointment Curdled milk

This is a solution because it contains many substances. This is a pure substance because the curdles are still milk.

a. b. c.

Muddy ﬂood water Gelatin dessert Normal urine

e. f. g.

An apple Mouthwash Petroleum jelly

This is a solution because it contains many substances. This is a solution because it contains many substances. This is a solution because it contains many substances (urea, water, dissolved salts, etc.) This is a pure substance because it contains one substance. This is a solution because it contains many substances. This is a solution because it contains many substances.

SECTION 1.4 MEASUREMENT UNITS 1.24

Modern society is complex and interdependent. Accomplishing projects like building a bridge, constructing a house, or machining an engine may require diﬀerent people to participate. Some people design the project, others supply the necessary materials, and yet another group does the construction. In order for the project to be successful, all of these people need a common language of measurement. Measurement is also important for giving directions, keeping track of the time people work, and keeping indoor environments at a comfortable temperature and pressure.

1.25

In the distant past, 1 in. was deﬁned as the length resulting from laying a speciﬁc number of grain kernels, such as corn, in a row. The size of 1 in. would vary in this system because the size of the individual kernels as well as the tightness of the packing would vary. The unit obtained by this system would not be consistent in all situations.

1.26

The amount of weight that a horse could carry or drag might have been measured in stones. It could also be used to measure people or other items in the 50-500 pound range. It is likely that a large stone was picked as the standard weight for the “stone” unit. Stones may have also been used as counterweights on an old-fashioned set of balances.

SECTION 1.5 THE METRIC SYSTEM 1.27

a. Nonmetric b. Metric c. Nonmetric d. Metric e. Metric f. Nonmetric

1.28

The metric units are (a) degrees Celsius, (b) liters, (d) milligrams, and (f) seconds. The English units are (c) feet and (e) quarts.

1.29

Grams, milligrams Meters Degrees Celsius

Chapter 1 1.30

1.31

1.32

1.33

1.34

1.35

1.36

Meters are a metric unit that could replace the English unit feet in the measurement of the ceiling height. Liters are a metric unit that could replace the English unit quarts in the measurement of the volume of a cooking pot.

Chapter 1

1.37

1.38

1.39

1.40

1.41

1.42

1.43

Chapter 1

SECTION 1.6 LARGE AND SMALL NUMBERS: AN INTRODUCTION TO SCIENTIFIC NOTATION 1.44

1.45

1.46

1.47

a. b. c.

02.7 x 10-3 4.1 x 102 71.9 x 10-6

Improper form because no leading zero is necessary. (2.7 x 10-3) Correct. Improper form because only one digit should be to the left of the decimal point. (7.19 x 10-5) Improper form because a nonexponential term should be written before the exponential term. (1 x 103) Improper form because one nonzero digit should be to the left of the decimal point. (4.05 x 10-4)

103

0.0405 x 10-2

0.119

Improper form because one nonzero digit should be to the left of the decimal point and an exponential term should be to the right of the nonexponential term. (1.19 x 10-1)

a. b. c.

4.2 x 103 6.84 202 x 10-3

0.026 x 10-2

10-2

74.5 x 105

Correct. Improper form because the “x 10” factor is missing. (6.8 x 104) Improper form because only one digit should be to the left of the decimal point. (2.02 x 10-3) Improper form because only one non-zero digit should be to the left of the decimal point. (2.6 x 10-2) Improper form because a nonexponential term should be written before the exponential term. (1 x 10-2) Improper form because only one digit should be to the left of the decimal point (7.45 x 105)

a. 14 thousand b. 365 c. 0.00204 d. 461.8 e. 0.00100 f. 9.11 hundred

= 14,000 =

a. Three hundred b. 4003 c. 0.682 d. 91.86 e. 6000 f. 400

= 300 =

= 9.11 x 100 =

1.4 x 104 3.65 x 102 2.04 x 10-3 4.618 x 102 1.00 x 10-3 9.11 x 102 3 x 102 4.003 x 103 6.82 x 10-1 9.186 x 101 6 x 103 4 x 102

Chapter 1 1.48

a. b.

186 thousand mi/s 1100 million km/h

186 x 1000 = 1.86 x 105 mi/s 1100 x 1,000,000 = 1.1 x 109 km/h

1.49

6.51 x 10-2 mm, 2.56 x 10-3 in. 1.52

1.50

The decimal point has been moved 22 places to the left. This places 21 zeros to the right of the decimal point and before the numbers 105 g.

1.51

602 000 000 000 000 000 000 000 hydrogen molecules The decimal point has been moved 23 places to the right.

1.52

a. b. c. d. e.

(8.2 x 10-3)(1.1 x 10-2) (2.7 x 102)(5.1 x 104) (3.3 x 10-4)(2.3 x 102) (9.2 x 10-4)(2.1 x 104) (4.3 x 106)(6.1 x 105)

=9.02 x 10-5 =1.377 x 107 =7.59 x 10-2 =1.932 x 101 =2.623 x 1012

=9.0 x 10-5 with significant figures =1.4 x 107 with significant figures =7.6 x 10-2 with significant figures =1.9 x 101 with significant figures =2.6 x 1012 with significant figures

1.53

a. b. c. d. e.

(6.3 x 105)(4.2 x 10-8) (2.8 x 10-3)(1.4 x 10-4) (8.6 x 102)(6.4 x 10-3) (9.1 x 104)(1.4 x 103) (3.7 x 105)(6.1 x 10-3)

=2.646 x 10-2 =3.92 x 10-7 =5.504 x 100 =1.274 x 103 =2.257 x 103

=2.6 x 10-2 with significant figures =3.9 x 10-7 with significant figures =5.5 x 100 with significant figures =1.3 x 108 with significant figures =2.3 x 103 with significant figures

1.54

a. b. c. d.

(144)(0.0876) =(1.44 x 102)(8.76 x 10-2) =1.26144 x 101 (751)(106) =(7.51 x 102)(1.06 x 102) =7.9606 x 104 (0.0422)(0.00119) =(4.22 x 10-2)(1.19 x 10-3) =5.0218 x 10-5 (128,000)(0.0000316)=(1.28 x 105)(3.16 x 10-5) =4.0448 x 100

=1.26 x 101 with SF =7.96 x 104 with SF =5.02 x 10-5 with SF =4.04 x 100 with SF

1.55

a. b. c. d.

(538)(0.154) (600)(524) (22.8)(341) (23.6)(0.047)

=(5.38 x 102)(1.54 x 10-1) =8.2852 x 101 =3.144 x 105 =(6 x 102)(5.24 x 102) 1 2 =(2.28 x 10 )(3.41 x 10 ) =7.7748 x 103 =(2.36 x 101)(4.7 x 10-2) =1.1092 x 100

=8.29 x 101 with SF =3 x 105 with SF =7.77 x 103 with SF =1.1 x 100 with SF

1.56

3.1 x 103 1.2 x 102

=2.583 x 10-5

=2.6 x 10-5 with SF

7.9 x 102 3.6 x 104

=2.194 x 102

=2.2 x 102 with SF

4.7 x 10-1 7.4 x 102

=6.35135 x 10-4

=6.4 x 10-4 with SF

0.00229 3.16

=7.2468354 x 10-4

=7.25 x 10-4 with SF

119 3.8 x 103

=3.131578947 x 10-2

= 3.1 x 10-2 with SF

0.000 000 000 000 000 000 000 105 g

Chapter 1 1.57

1.58

1.59

233 1.67

=1.33532934142 x 102

=1.34 x 102 with SF

6.7 x 103 4.2 x 104

=1.59523809524 x 10-1

=1.6 x 10-1 with SF

8.7 x 10-4 2.3 x 10-2

=3.78260869565 x 10-2

=3.8 x 10-2 with SF

6.8 x 103 2.7 x 10-4

=2.5185185 x 107

=2.5 x 107 with SF

1.8 x 10-2 6.5 x 104

=2.7692307692 x 10-7

=2.8 x 10-7 with SF

(5.3)(0.22) (6.1)(1.1)

=1.7377 x 10-1

=1.7 x 10-1 with SF

(3.8 x 10-4)(1.7 x 10-2) 6.3 x 103

=1.025 x 10-9

=1.0 x 10-9 with SF

=2.6 x 106 with SF

4.8 x 106 (7.4 x 103)(2.5 x 10-4)

=2.59459 x 106

5.6 (0.022)(109)

=2.335279 x 100 =2.3 x 100 with SF

(4.6 x 10-3)(2.3 x 102) (7.4 x 10-4)(9.4 x 10-5)

=1.520989 x 107 =1.5 x 107 with SF

(7.4 x 10-3)(1.3 x 104) (5.5 x 10-2)

=1.749 x 103

=1.7 x 103 with SF

6.4 x 105 (8.8 x 103)(1.9 x 10-4)

=3.82775 x 103

=3.8 x 105 with SF

(6.4 x 10-2)(1.1 x 10-8) (2.7 x 10-4)(3.4 x 10-4)

=7.668845 x 10-3 =7.7 x 10-3 with SF

(963)(1.03) (0.555)(412)

=4.3378 x 100

1.15 (0.12)(0.73)

=1.312785 x 101 =1.3 x 101 with SF

=4.34 x 100 with SF

Chapter 1

SECTION 1.7 SIGNIFICANT FIGURES 1.60

a. b. c. d.

A ruler with a smallest scale marking of 0.1 cm A measuring telescope with a smallest scale marking of 0.1mm A protractor with a smallest scale marking of 1° A tire pressure with a smallest scale marking of 1 lb/in2

0.01 cm 0.01 mm 0.1° 0.1 lb/in2

1.61

a. b. c. d.

A buret with a smallest scale marking of 0.1 mL A graduated cylinder with a smallest scale marking of 1 mL A thermometer with a smallest scale marking of 0.1°C A barometer with a smallest scale marking of 1 torr

0.01 mL 0.1 mL 0.01°C 0.1 torr

1.62

a. b. c. d.

6.0 mL 37.00°C 9.00 s 15.5°

1.63

A length of two and one-half centimeters measured with a measuring telescope with a smallest scale marking of 0.1 mm. An initial reading of exactly 0 for a buret with a smallest scale marking of 0.1 mL. A length of four and one-half centimeters measured with a ruler that has a smallest scale marking of 0.1 cm. An atmospheric pressure of exactly 690 torr measured with a barometer that has a smallest scale marking of 1 torr.

b. c. d. 1.64

2.500 cm 0.00 mL 4.50 cm 690.0 torr

Measured = 5.06 lbs. Exact = 16 potatoes

5.06 lb. = 0.31625 lb. = 0.316 lb. with SF 16 potatoes potato potato

Measured = percentages Exact = 5 players

71.2 % + 66.9% + 74.1% + 80.9% + 63.6% = 71.34% with SF 5 players

Measured = 1pm, 2pm Exact = 19, 24, 17, 31, 40

19 + 24 + 17 + 31 + 40 people = 26.2 people 5 days day

Measured = heights Exact = 5 players

6’9 + 5’8 + 5’6 + 5’1 + 4’11 = 5’7” with SF 5 players

1.66

a. b. c.

0.0400 309 4.006

3 SF (0.04000) 3 SF 4 SF

d. 4.4 x 10-3 e. 1.002 f. 255.02

2 SF 4 SF 5 SF

1.67

a. b. c.

0.040 11.91 2.48 x 102

2 SF 4 SF 3 SF

d. 149.1 e. 10.003 f. 148.67

4 SF 5 SF 5 SF

1.68

a. b.

(3.71)(1.4) (0.0851)(1.2262)

1.65

5.194 0.10434962

= 5.2 with significant figures = 0.104 with significant figures

Chapter 1 c.

(0.1432)(2.81) (0.7762) (3.3 x 104)(3.09 x 10-3) (760.)(2.00) 6.02 x 1020

0.518412780211

= 0.518 with signiﬁcant ﬁgures

101.97 2.52491694352 x 10-18

= 1.0 x 102 with significant figures = 2.52 x 10-18 with significant figures (assuming 0 in 760 is significant)

a. b. c.

(1.21)(3.2) (6.02 x 1023)(0.220) (0.023)(1.1 x 10-3) 100

3.872 1.3244 x 1023 2.53 x 10-7

= 3.9 with significant figures = 1.32 x 1023 with significant figures = 3 x 10-7 with significant figures

(365)(7.00) 60

42.583333

= 4 x 101 with signiﬁcant ﬁgures

(810)(3.1) 8.632 x 10-1

2908.94347

= 2.9 x 103 with signiﬁcant ﬁgures

a. b. c. d.

0.208 + 4.9 + 1.11 228 + 0.999 + 1.02 8.543 – 7.954 (3.2 x 10-2) + (5.5 x 10-1) (Hint: Write in decimal form ﬁrst, then add.) 336.86 – 309.11 21.66 – 0.02387

= 6.218 = 230.019 = 0.589

= 6.2 with significant figures = 2.30 x 102 with significant figures = 0.589 with significant figures

= 0.582

= 0.58 with signiﬁcant ﬁgures

= 27.75 = 21.63613

= 27.75 with significant figures = 21.64 with significant figures

= 7.289 = 8.181 = 1.126

= 7.3 with significant figures = 8.18 with significant figures = 1.126 with significant figures

= 0.355

= 0.355 with signiﬁcant ﬁgures

= 106.1 = 17.7121

= 106.10 with significant figures = 17.71 with significant figures

= 0.004483037867

= 0.00460 with signiﬁcant ﬁgures

= 2.20817413454

= 2.208 with signiﬁcant ﬁgures

= 2.6453333

= 2.65 with signiﬁcant ﬁgures

= - 12.8633592018

= -13 with signiﬁcant ﬁgures

= 3.3902741324

= 3 with signiﬁcant ﬁgures

d. e. 1.69

1.70

e. f. 1.71

a. b. c. d.

e. f. 1.72

2.1 + 5.07 + 0.119 0.051 + 8.11 + 0.02 4.337 – 3.211 (2.93 x 10-1) + (6.2 x 10-2) (Hint: Write in decimal form ﬁrst, then add.) 471.19 – 365.09 17.76 – 0.0479

(0.0267 + 0.00119)(4.626) 28.7794 b. 212.6 – 21.88 86.37 c. 27.99 – 18.07 4.63 – 0.88 d. 18.87 _ 18.07 2.46 0.88 (Hint: do division ﬁrst, then subtract.) e. (8.46 – 2.09)(0.51 + 0.22)

Chapter 1

f. 1.73

1.74

(3.74 + 0.07)(0.16 + 0.2) 12.06 – 11.84 0.271

= 0.811808118081

= 0.81 with signiﬁcant ﬁgures

132.15 – 32.16 = 1.14209023415 87.55 b. (0.0844 + 0.1021)(7.174) = 0.070012768117 19.1101 c. (2.78 – 0.68)(0.42 + 0.4) = 3.66726296959 (1.058 + 0.06)(0.22 + 0.2) d. 27.635 – 21.71 = 1.2852494577 4.97 – 0.36 e. 12.47 _ 203.4 = 0.781167484035 6.97 201.8 (Hint: Do division ﬁrst, then subtract.) f. 19.37 – 18.49 = 1.07055961071 0.822

= 1.142 with signiﬁcant ﬁgures

= 0.07001 with significant figures = 4 with significant figures = 1.29 with significant figures = 0.78 with significant figures

= 1.1 with signiﬁcant ﬁgures

a. Area (A = 1 x w) Black A = 12.00 cm x 10.40 cm – 124.8 cm2 Red A = 20.20 cm x 2.42 cm – 48.884 cm2 = 48.9 cm2 Green A = 3.18 cm x 2.55 cm = 8.109 cm2 = 8.11 cm2 Orange A = 13.22 cm x 0.68 cm – 8.9896 cm2 = 9.0 cm2

Perimeter (P = 2 (l) + 2 (w)) P = 2(12.00 cm) + 2(10.40 cm) = 44.80 cm P = 2 (20.20 cm) + 2(2.42 cm) = 45.24 cm P = 2 (3.18 cm) + 2(2.55 cm) = 11.46 cm P = 2(13.22 cm) + 2(0.68 cm) = 27.80 cm

Width

Length

Area (A = 1 x w) Perimeter (P = 2 (l) + 2 (w)) 2 P = 2(0.1200 m) + 2(0.1040 m) = 0.4480 m Black A = 0.1200 m x 0.1040 m = 0.01248 m P = 2(0.2020 m) + 2(0.0242 m) = 0.4524 m Red A = 0.2020 m x 0.0242 m = 0.0048884 m2 2 = 0.00489 m P = 2(0.0318 m) + 2(0.0255 m) – 0.1146 m Green A = 0.0318 m x 0.0255 m = 8.109 x 10-4 m2 = 8.11 x 10-4 m2 P = 2(0.1322 m) + 2(0.0068 m) = 0.2780 m Orange A = 0.1322 m x 0.0068 m = 8.9896 x 10-4 m2 -4 2 = 9.0 x 10 m c. No, the number of signiﬁcant ﬁgures in the answers remains constant. The numbers of places

Chapter 1 past the decimal are different; however, that could be fixed by rewriting all of the answers in scientific notation.

SECTION 1.8 USING UNITS IN CALCULATIONS: AN INTRODUCTION TO DIMENSIONAL ANALYSIS 1.75

1 kg

2.20 lbs. c. 1.76

1.094 yd

1g 0.035 oz

0.394 in 1 cm

0.015 grain 1 mg

0.0338 ﬂ oz 1 mL

1L 1.057 qt

1m 1.094 yd

1.77 1.78

1.79

1.80

We’ll use the factor-unit method. Given is the factor 80 mg , which equals 40 mg, or 1 mL 2 mL 1 mL 40 mg

Chapter 1 1.81

Using the factor-unit method, two factors will be employed.

1.82

Using the factor-unit method,

= 286 min.

1.83

Table 1.3 shows that 1 kg = 2.20 lb

1.84

SECTION 1.9 CALCULATING PERCENTAGES 1.85 1.86

55 years

65 years $25.73

$467.80

x 100 = 85%

x 100 = 5.500%

1.87

140 lbs−32 lbs

1.88

1.0 day

1.89 1.90

140 lbs

1.4

mg day

x 100 = 77%

x 100 = 71%

2000 Calories �

45 Calories

100 Calories

� = 900 Calories = 9.0 𝑥𝑥 102 Calories with signi�icant �igures

Total = 987.1 mg + 213.3 mg + 99.7 mg + 14.4 mg + 0.1 mg = 1314.6 mg IgG =

IgD =

987.1 mg

1314.6 mg 14.4 mg

1314.6 mg

x 100 = 75.09 %; IgA = x 100 = 1.10%; IgE =

213.3 mg

1314.6 mg 0.1 mg

1314.6 mg

x 100 = 16.23%; IgM =

x 100 = 0.008%

99.7 mg

1314.6 mg

x 100 = 7.58%

Chapter 1

SECTION 1.10 DENSITY 1.91

1.92

1.93

D = m/v D = 2.10 g = 1.56 g/cm3 1.35 cm3

1.94

Volume = (3.98 cm)3 = 63.0 cm3 Density =

1.95

1.96

mass = 718.3 g = 11.4 g volume (3.98 cm)3 cm3

Chapter 1 1.97

Using the factor-unit method,

ADDITIONAL EXERCISES 1.98

1.99

1.100

1.101

Chapter 1

1.102

1.103

CHEMISTRY FOR THOUGHT 1.104

b. c. d.

1.105

To separate wood sawdust and sand, I would add water. The sawdust will float, while the sand will sink. The top layer of water and sawdust can be poured off into a filter. The water will run through the filter leaving the sawdust in the filter. The sawdust can then be allowed to dry. The remainder of the water and sand can be poured off into a filter and the sand can be allowed to dry. To separate sugar and sand, I would add water to dissolve the sugar. I would then filter the mixture to isolate the sand. I would evaporate the water to isolate the sugar. To separate iron filings and sand, I would use a magnet. The iron filings will be attracted to the magnet, while the sand will not be attracted to the magnet. To separate sand soaked with oil, I would pour the mixture through a filter. The oil will go through the filter and leave the sand behind on the filter.

A bathroom mirror becomes foggy when someone takes a hot shower because the steam from the shower condenses on the cold glass of the mirror. This is a physical change because the water molecules are changing phase of matter, but not composition.

1.106

This student should have used the relationship 2.2 lbs = 1 kg to multiply 44.5 kg by 2.2 lbs/kg to ﬁnd a weight of 97.9 lbs. The mistake she made appears to be that she divided 44.5 kg by 2.2 rather than multiplying by it. Consequently, she found a weight of only 20.2 lbs. Since she knows 2.2 lbs = 1 kg, she was expecting the pound value to be larger than the kilogram value and she determined she had made a calculation error.

Chapter 1 1.107

A mercury thermometer cannot be used to measure a temperature that is -45°C. A thermometer ﬁlled with a liquid that has a freezing point below -45°C could be used to measure this temperature.

1.108 Assuming each of the guests eats one serving of oatmeal, 9 cups of dry oatmeal should be prepared. 1.109

All matter is made up of atoms of the elements and therefore contains chemicals.

1.110

The density of the object is only 8.76 g/mL; therefore, it does not have the same density as silver and is not silver.

Chapter 2

Chapter 2: Atomic Structure and the Periodic Table CHAPTER OUTLINE 2.1 The Periodic Table

2.5 Radioactive Nuclei

2.2 Subatomic Particles

2.6 Where Are the Electrons?

2.3 Atomic Symbols

2.7 Trends within the Periodic Table

2.4 Isotopes and Atomic Weights

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Locate elements in the periodic table on the basis of their group and period designations. (Section 2.1; Exercises 2.1, 2.7, and 2.9)

Describe the charge, relative mass, and location of the three subatomic particles. (Section 2.2; Exercises 2.11 and 2.13)

Write the atomic symbol (AX) for a given set of subatomic particles. (Section 2.3; Exercise 2.15)

Deﬁne the terms isotope and atomic weight. (Section 2.4; Exercises 2.19 and 2.21)

Write balanced equations for radioactive decay and other nuclear processes. (Section 2.5; Exercises 2.41 and 2.43)

Solve problems using the half-life concept. (Section 2.5; Exercises 2.47 and 2.51)

Describe the applications of radiation in health and medicine. (Section 2.5; Exercises 2.57 and 2.59)

Write correct electron conﬁgurations for each element through atomic number 56. (Section 2.6; Exercises 2.77 and 2.79)

Describe periodic trends in atomic radius, ﬁrst ionization energy, and electronegativity. (Section 2.7; Exercises 2.93, 2.95, and 2.97)

LECTURE HINTS AND SUGGESTIONS 1.

The word “element” has two usages: (1) a homoatomic, pure substance; and (2) a kind of atom. This dual usage confuses the beginning students. It often helps the beginning student for the instructor to distinguish the usage intended in a particular statement. e.g. “There are 118 elements, meaning 118 kinds of atoms.” or “Each kind of atom (element) has a name and a symbol.” or “Water contains the element (kind of atom) oxygen.”

The student will memorize the names and symbols for approximately one-third of the 118 elements to be dealt with those commonly encountered in this course or in daily living. Mentioning both the name and the symbol whenever an element is mentioned in the lecture will aid the student’s memorizing.

While memorization of the names and symbols is important, it should not become the major outcome of this class. Avoid reinforcing the mistaken notion that chemistry is merely learning formulas and equations.

If samples of some of the elements are available, showing them may beneﬁt the discussion of the periodic table as it relates to metals, non-metals, similarity within groups, etc.

Chapter 2 5.

Clearly diﬀerentiate between the radioactive substance and the radiation coming from a radioactive substance. Students often think of these as being interchangeable terms because the similar names: alpha radiation and alpha decay, etc.

Many of the students, for whom this text is written, have some basis for understanding the health eﬀects of X-rays. Others have seen ﬁlm badges used by X-ray technicians. This background knowledge can be helpful in discussing the measurement of radiation from nuclear sources.

Extend the concept of chemical change and chemical properties to the atomic level. Chemical changes cause a change in the constituent particles. The atoms are rearranged to make diﬀerent constituent particles. This rearrangement involves the electrons of the atoms. Thus, atoms with similar chemical properties must have similar electronic arrangements.

Relate the arrangement of the periodic table with groups and periods to the electronic structure of the atom, and the ﬁlling order of the orbitals. This facilitates the understanding of the “outer electrons” determining the chemical combinations.

Solutions to All End-of-Chapter Exercises What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 2.1 THE PERIODIC TABLE 2.1

Group a.

IVA (14)

element number 21

III B (3)

zinc

II B (12)

element number 35

VII A (17)

Group

Period

2.2

2.3

2.4

2.5

Period

element number 27

VIII B (9)

IV A (14)

arsenic

V A (15)

II A (2)

How many elements are located in group VIIB (7) of the periodic table?

How many elements are found in period 5 of the periodic table?

How many total elements are in group IVA (14) and IVB (4) of the periodic table?

How many elements are located in group VIIB (7) of the periodic table?

How many total elements are found in periods 1 and 2 of the periodic table?

How many elements are found in period 5 of the periodic table?

This is a vertical arrangement of elements the periodic table

The chemical properties of the elements repeat in a regular way

group periodic law

Chapter 2 as the atomic numbers increase c.

The chemical properties of elements 11, 19, and 37 demonstrate

periodic law

this principle

2.6

Elements 4 and 12 belong to this arrangement

group

This is a horizontal arrangement of elements in the periodic table

period

Element 11 begins this arrangement in the periodic table

period

The element nitrogen is the ﬁrst member of this arrangement

group

elements 9, 17, 35, and 53 belong to this arrangement

group

2.7

a. Transition metal c. Inner-transition metal e. Representative

b. Representative d. Noble gases

2.8

a. Transition metal c. Noble gases e. Representative

b. Inner-transition metal d. Noble gases

2.9

a. c. e.

Nonmetal Metalloid Metal

b. d.

Metal Metalloid

2.10

a. c. e.

Metal Metal Nonmetal

b. d.

Metalloid Nonmetal

SECTION 2.2 SUBATOMIC PARTICLES 2.11

Charge

Mass (u)

8 protons and 9 neutrons

20 protons and 25 neutrons

52 protons and 78 neutrons

130

Charge

Mass (u)

2.12

2.13

9 protons and 10 neutrons

20 protons and 23 neutrons

47 protons and 60 neutrons

107

The number of protons and electrons are equal in a neutral atom. a.

2.14

10 electrons

18 electrons

50 electrons

The number of protons and electrons are equal in a neutral atom. a.

9 electrons

20 electrons

47 electrons

Chapter 2

SECTION 2.3 ATOMIC SYMBOLS 2.15

Symbol

Name

Belongs to group VIA (16) and period 3

Sulfur

The ﬁrst element (reading down) in group VIB (6)

Chromium

The fourth element (reading left to right) in period 3

Silicon

Belongs to group IB (11) and period 5

Silver

Symbol

Name

2.16 a.

The noble gas belonging to period 4

Krypton

The fourth element (reading down) in group IVA (14)

Tin

Belongs to group VIB (6) and period 5

Molybdenum

The sixth element (reading left to right) in period 6

Neodymium

Electrons

Protons

sulfur

element number 24

Electrons

Protons

2.17

2.18 a.

silicon

element number 74

SECTION 2.4 ISOTOPES AND ATOMIC WEIGHTS 2.19

Protons

Neutrons

Electrons

143

Protons

Neutrons

Electrons

3 2

9 4

235 92

34 16

91 40

131 54

cadmium-110

110 4

cobalt-60

60 2

uranium-235

235 9

silicon-28

argon-40

2.20

2.21

2.22

Chapter 2 c.

strontium-88 Mass Number Atomic Number

2.23 a.

6 protons and 6 neutrons

12 6

8 protons and 9 neutrons

17 8

20 protons and 25 neutrons

45 20

2.24

2.25

2.26

2.27

2.28

2.29

Symbol

Mass Number Atomic Number

Symbol

9 protons and 10 neutrons

19 9

20 protons and 23 neutrons

43 20

47 protons and 60 neutrons

107

107 47

contains 20 electrons and 20 neutrons

40 20

contains 1 electron and 2 neutrons

3 1

a magnesium atom that contains 14 neutrons

26 12

contains 17 electrons and 20 neutrons

a copper atom with a mass number of 65

37 17Cl

H M

65 29

a zinc atom that contains 36 neutrons

66 30

the number of neutrons in the nucleus

22.9898 – 11 = 11.9898 ≈ 12 neutrons

the mass (in u) of the nucleus (to three SF)

23.0 u

the number of neutrons in the nucleus

26.982 – 13 = 13.982 ≈ 14 neutrons

the mass (in amu) of the nucleus (to three SF)

27.0 amu

7.42 % x 6.0151 u + 92.58% x 7.0160 u = 0.0742% x 6.0151 u + 0.9258 x 7.0160 u = 6.94173322 u; 6.942 u with SF or (7.42 x 6.0151 u)+(92.58 x 7.0160 u) 100

= 6.94173322 𝑢𝑢; 6.942 u with SF

The atomic weight listed for lithium in the periodic table is 6.941 u. The two values are very close. 2.30

19.78% x 10.0129 amu + 80.22% x 11.0093 amu = 0.1978 x 10.0129 amu + 0.8022 x 11.0093 amu = 10.81221208 amu; 10.812amu with SF or (19.78 x 10.0129 amu)+(80.22 x 11.0093 amu) 100

= 10.81221208 amu; 10.812 𝑎𝑎𝑎𝑎𝑎𝑎 with SF

The atomic weight listed for boron in the periodic table is 10.81 amu. The two values are

Chapter 2 close to one another. 2.31

92.21% x 27.9769 u + 4.70% x 28.9765 u + 3.09% x 29.9738 u = 0.9221 x 27.9769 u + 0.0470 x 28.9765 u + 0.0309 x 29.9738 u = 28.08558541 u; 28.09 u with SF or (92.21 x 27.9769 u)+(4.70 x 28.9765 u)+(3.09 x 29.9738 u) 100

= 28.08558541 u; 28.09 u with SF

The atomic weight listed for silicon in the periodic table is 28.09 u. The two values are the same. 2.32

69.09% x 62.9298 amu + 30.91% x 64.9278 amu = 0.6909 x 62.9298 amu + 0.3091 x 64.9278 amu = 63.5473818 amu; 63.55 amu with SF or (69.09 x 62.9298 𝑎𝑎𝑎𝑎u)+(30.91 x 64.9278 amu) 100

= 63.5473818 amu; 63.55 amu with SF

The atomic weight listed for copper in the periodic table is 63.55 amu. The two values are the same.

SECTION 2.5 RADIOACTIVE NUCLEI 2.33

Radioactive means a material is able to undergo spontaneous nuclear changes and emit energy in the form of radiation. a. Beta rays are not radioactive, they are a form of radiation. b. Radon is not a stable radioactive element, because by deﬁnition a radioactive element is not stable. It undergoes radioactive decay, but as a noble gas, it is also chemically-inert.

2.34

a. b. c.

mass number = 0: beta, gamma, positron. positive charge: alpha, positron charge = 0: gamma, neutron

2.35

a. b. c.

Those with a negative charge = beta Those with a mass number greater than 0 = alpha, neutron Those that consist of particles = alpha, beta, neutron, positron

2.36

a. b. c.

A beta particle = an electron An alpha particle = 2 protons and 2 neutrons A positron = positive electron

2.37

Radiation which has no mass (and therefore no charge) is able to penetrate matter better than radiation which has mass, whether it is charged or neutral. As the charge and mass increase, radiation is less able to penetrate matter.

2.38 a. b. c. d. e.

An alpha particle is emitted. A beta particle is emitted. An electron is captured. A gamma rat is emitted. A positron is emitted.

Atomic number change decrease by 2 increase by 1 decrease by 1 no change decrease by 1

Mass number change decrease by 4 no change no change no change no change

Chapter 2 2.39

2.40

A tin-117 nuclear

117 50

A nucleus of chromium (Cr) isotope containing 26 neutrons

50 24

A nucleus of element number 20 that contains 24 neutrons

44 20

A nucleus of the element in period 5 group VB(5) with a mass number of 96

96 41

A nucleus of element number 37 with a mass number of 80

80 37

A nucleus of the calcium (Ca) isotope that contains 18 neutrons

38 20

10 4

210 83

15 8

204 82

84 35

2.41

2.42

2.43

2.44

Be → ? + 10 B 5

? = 0−1 β

44 22

Bi →42 α + ?

206 ? = 81 Ti

O → ? + 15 N 7

? = 10 β

Pb → ? + 42α

Ti + 0−1e → ?

? = 44 Sc 21

8 4

Be → ? + 24 He

? = 42α

46 23

V → ? + 46 Ti 22

? = 10 β

200 ? = 80 Hg

149 62

Br → ? + 0−1 β

? = 84 Kr 36

34 ? → 15 P + 0−1 β

34 ? = 14 Si

41 ? + 0−1e →19 Br

? = 41 Ca 20

15 8

O + 10 β → ?

? = 15 N 7

121 50

Sm → ? + 145 Nd 60

? = 42α

Sn →0−1 β + 121 Sb 51

Sn (beta emission)

121 50

55 26

Fe (electron capture)

55 26

55 Fe →−01 e + 25 Mn

22 11

Na (daughter= neon-22)

22 11

22 Na →−01 β + 10 Ne

190 78

Pt →42 α + 186 Os 76

Pt (alpha emission)

190 78

67 28

Ni (beta emission)

67 28

Ni →0−1 β + 67 Cun 29

67 31

Ga (daughter= zinc-67)

67 31

Ga →−01 β + 67 Zn 30

157 63

Eu (beta emission)

157 63

Eu →0−1 β + 157 Gd 64

190 78

Pt (daughter= osmium-86)

190 78

Pt → 42α + 186 Os 76

138 62

Sm (electron capture)

138 62

Sm →−01 e + 138 Pm 61

188 80

Hg (daughter= Au-188)

188 80

Hg →−01 β + 188 Au 79

234 90

Th (beta emission)

234 90

234 Th →0−1 β + 91 Pa

218 85

At (alpha emission)

218 85

214 At →42 α + 83 Bi

2.45

Half-life is the amount of time required for half of a sample to undergo a speciﬁc process.

2.46

Half-life is the amount of time required for half of a sample to undergo a speciﬁc process. For example, in the half-life of a cake is one day, then half of a cake will be eaten by the ﬁrst day, the next day half of the remaining cake (1/4 of the original cake) would be eaten, the following day half of the remaining cake (1/8 of the original cake) would be eaten, etc.

Chapter 2 2.47

99 days in three half-lives. After the ﬁrst half-life, 6.00 mg remains. After the second half-life, 3.00 mg remains. After the third half-life, 1.5 mg remains.

2.48

1 Amount remaining = 9.0 ng   2

2.49

1 1 =  128  2  7

 1 half-life  24 hours    6 hours 

= 0.56 ng

 1 half-life  x years    92 years 

1 1   =  2 2

 1 half-life  x years    92 years 

 1 half-life  7 = x years    92 years   92 years  x years = 7   = 644 years  1 half-life  1980 + 644 = 2624 The archaeologist made the discovery in approximately 2624. or 1 Since the fraction remaining is , the number of half-lives which have passed is 7 because 128 7

1 1 . The t 1/ 2 is 92 years; therefore, the amount of time that has passed is   = 128 2 92 x 7 = 644 years. The discovery was made 644 years after 1980, so the year 2624.

2.50

  1  half-lives   ⇒ log (0.0625) = log     2       1  log (0.0625) = half-lives  log     2   half-lives = 4.00

1 100% - 93.75% = 100%   2

half-lives

 5600 years  4 Time elapsed = 4 half-lives   = 22400 years = 2.24 x 10 years 1 half-life   or 1 remains. This means that 4 half-lives have Since 93.75% of the 14C has decayed, 6.25% or 16 4

1 1 . The is t1/ 2 5600 years; therefore, the amount of time that has passed passed, since   = 16 2 is 5600 x 4 = 2.24 x 104 years.

2.51

1 Amount remaining = 50.0 mg   2 or

 1 half-life  10.0 hours    2.5 hours 

= 3.13 mg

Chapter 2

 10.0  The t 1/ 2 is 2.5 hours. After 10.0 hours, 4 half-lives have passed since   = 4.0.  2.5  4

1 1 At this time the fraction remaining = 1/ 2 x 1/ 2 x 1/ 2= =  x 1/ 2 16 2

 1  The amount remaining is 50.0 mg   = 3.13 mg .  16 

2.52

 1  1 = ⇒ log     2  8    1  1 half-lives  log    = log    2  8  half-lives = 3 1   2

half-lives

 1 = log    8 

 5600 years  4 Time elapsed = 3 half-lives   = 16800 years = 1.68 x 10 years 1 half-life   or 3

1 1 The amount remaining is = , which is 1/ 2 x 1/ 2 x 1/ 2 x 1/ 2 =   ; therefore, 3 half-lives have elapsed. 8 2 The t 1/ 2 is 5600 years; therefore, the amount of time that has passed is 5600 x 3 = 1.68 x 104 years.

Sr →−01 β + 90 Y 39

2.53

90 30

2.54

Long-term, low-level exposure to radiation made lead to genetic mutations because ionizing radiation can produce free radicals in exposed tissues. Short-term exposure to intense radiation destroys tissue rapidly and causes radiation sickness. Both forms of exposure have negative eﬀects on health.

2.55

The rem corresponds to the health eﬀects produced by 1 roentgen of gamma or X-rays regardless of the type of radiation involved. If a person is working in an area where exposure to several types of radiation is possible, then a unit that is independent of the type of radiation is the best unit to use.

2.56

Physical units of radiation indicate the activity of a source of radiation; whereas, biological units of radiation indicate the damage caused by radiation in living tissue. Examples of physical units of radiation include the Curie and the Becquerel. Examples of biological units of radiation include the Sievert, the Rad, the Gray, and the Rem.

2.57

A diagnostic tracer is a radioactive isotope that concentrates in a speciﬁc organ. It should have a short, but reasonable, half-life, have a stable daughter isotope, be non-toxic and give oﬀ radiation that can penetrate out of the tissue and be detected.

2.58

Chapter 2 2.59

A therapeutic radioisotope administered internally should be an alpha or beta emitted, have a half-life long enough to perform the desired radiation therapy, be non-toxic (and have non-toxic, non-radioactive daughter isotopes), and be concentrated by the target issue.

2.60

Radioactive isotopes can be used for diagnostic work. When the radioactive isotope concentrates in a tissue under observation, the location is called a hot spot. When the radioactive isotope is excluded or rejected by a tissue under observation, the location is called a cold spot. Both hot spots and cold spots can be used for diagnostic work.

2.61

1 2.2 mCi = 70. mCi   2 2.2 mCi  1    70. mCi  2  5

 1 half-life  x days    2.7 days 

 1 half-life  x days  

 2.7 days    1 1   =  2 2      1 half-life  5 = x days    2.7 days 

 2.7 days  x days = 5   = 14 days  1 half-life  or The percentage of gold-198 that will remain at the lower activity level is

2.2 mCi x 100 = 3.1% . 70. mCi

1 1 This is approximately   = = 3.125% , which is the amount that will remain after 5 half-lives. 32 2 The is 2.7 days; therefore, the amount of time that must pass is 2.7 x 5 = 14 days.

Cr + −01e →51 V; daughter nucleus = vanadium-51 23

2.62

51 24

2.63

After 10 half-lives, the amount of the radioisotope left is less than 1/1000 of the original amount. This would almost be undetectable. Ten half-lives for 14C = 56,000 years. The amount of 14C left after several millions years would be essentially zero.

2.64

By using water that contains a radioactive isotope of oxygen, the oxygen gas produced could be analyzed to see if it contains the radioactive isotope of oxygen from the water or a nonradioactive isotope of oxygen from the hydrogen peroxide.

2.65

Strontium would be likely to be deposited in bones because it is in group IIA (2), just like calcium (a major component of bone).

SECTION 2.6 WHERE ARE THE ELECTRONS? 2.66

Protons are subatomic particles with a positive charge that are located in the nucleus.

Chapter 2 2.67

According to Bohr theory, an electron in an orbit located farther from the nucleus would have higher energy than an electron in an orbit close to the nucleus.

2.68

a. X-rays have a higher amount of energy. b. The red light from a stoplight has a higher amount of energy.

2.69

a. UV radiation b. Gamma-rays

2.70

2.71

A 2p orbital

2 electrons

A 2p subshell

6 electrons

The second shell

8 electrons

A 3d orbital

2 electrons

A 3d subshell

10 electrons

The third shell

18 electrons

2.72

Four (4) orbitals are found in the second shell: one 2s orbital and three 2p orbitals.

2.73

Sixteen (16) orbitals are found in the fourth shell: one 4s orbital, three 4p orbitals, ﬁve 4d orbitals, and seven 4f orbitals.

2.74

Three (p) orbitals are found in a 4p subshell. The maximum number of electrons that can be located in this subshell is 6 because each of the three orbitals can hold two electrons.

2.75

Five (5) orbitals are found in the 3d subshell. The maximum number of electrons that can be located in this subshell is 14 because each of the seven orbitals can hold two electrons.

2.76 a.

element number 37

titanium

2.77

Electron Conﬁguration

Unpaired Electrons

1s22s22p63s23p64s23d104p65s1

1s 2s 2p 3s 3p

1s 2s 2p 3s 3p 4s 3d 2

1s 2s 2p 3s 3p 2

Electron Conﬁguration

Unpaired Electrons 1

1s 2s 2p 3s 3p 4s

1s 2s 2p 3s 3p 4s 3d

chromium

element number 33

s electrons in magnesium

unpaired electrons in nitrogen

ﬁlled subshells in Al

2.78

2 0

1s22s22p63s23p64s23d104p3

1s22s22p63s23p64s23d2

Electron Conﬁguration

Solutions

1s22s22p63s2

1s22s22p3 1s 2s 2p 3s 3p 2

3 1

Chapter 2 2.79 a.

Electron Conﬁguration

Solutions

1s22s22p63s23p64s23d104p2

unpaired electrons

1s22s22p63s23p4

number desig. = 3

3d electrons in tin

1s22s22p63s23p64s23d104p65s24d105p2

Symbol

Name

total electrons in Ge have a number designation (before the letters) of 4

unpaired p electrons in sulfur, number designation of the

2 electrons,

2.80 a.

Contains only two 2p electrons

carbon

Contains an unpaired 3s electron

sodium

Symbol

Name

Contains one unpaired 5p electron

In or I

indium or iodine

Contains a half-ﬁlled 5s subshell

rubidium

arsenic

[Ar] 4s23d104p3

silicon

[Ne] 3s23p2

An element that

[Ar] 4s23d5

element

[Kr] 5s24d105p5

2.81

2.82

contains 25 electrons 2.83

An element that

number 53 [Ar] 4s23d4

iodine

[Kr] 5s24d105p5

[Ar] 4s23d1

copper

[Ar] 4s23d9

phosphorus

[Ne] 3s23p3

contains 24 electrons b.

element number 21

2.84

sodium

[Ne] 3s1

magnesium

[Ne] 3s2

c. d.

sulfur

[Ne] 3s23p4

aluminum

[Ne] 3s 3p

chlorine

[Ne] 3s23p5

silicon

[Ne] 3s23p2

argon

[Ne] 3s23p6

2.85

Eighteen (18) elements have the symbol [Kr] in their abbreviated electronic conﬁgurations.

2.86

Cesium is the period 6 element with chemical properties most like sodium. Cesium has 1 valence-shell electron. Sodium also has only 1 valence-shell electron.

2.87

The period 5 element with chemical properties most like silicon is tin (Sn). Tin has four valence-shell electrons. Silicon also has four valance-shell electrons.

2.88

element number 54

8 electrons

The ﬁrst element (reading down) in group V A (15)

5 electrons

Chapter 2 c.

4 electrons

the fourth element (reading left to right) in period 3

4 electrons

element number 35

7 electrons

2 electrons

strontium

2 electrons

The second element in group VA (15)

5 electrons

2.90

p area

d area

2.91

p area

c. d area

2.89

s area

SECTION 2.7 TRENDS WITHIN THE PERIODIC TABLE 2.92

I would expect to ﬁnd silver and gold in addition to the copper because these elements are all in the same group on the periodic table. Elements that are in the same group have similar chemical properties; therefore, if copper is part of this ore, then the other elements that are most similar to it are also likely to be part of the ore.

2.93

a. b. c. d.

Mg or Sr Rb or Ca S or Te I or Sn

Sr larger radius Rb Te Sn

2.94

a. b. c. d.

Ga or Se N or Sb O or C Te or S

Ga larger radius Sb C Te

2.95

a. b. c. d.

Mg or Al Ca or Be S or Al Te or O

Mg loses e more easily Ca Al Te

2.96

a. b. c. d.

Li or K C or Sn Mg or S Li or N

K loses e more easily Sn Mg Li

2.97

a. Li or Br b. F or Cl c. Sn or C d. Cl or Al

2.98

a. Electronegativity increases from left to right, so Cl is more electronegative. b. Electronegativity decreases as you go down a group, so Cl is more electronegative than I

Br F C Cl

Chapter 2 c. Nitrogen is more electronegative because electronegativity decreases as you go down a group. d. Boron is more electronegative than Be because electronegativity increases as you go from left to right.

Additional Exercises 2.99

The density of the metallic elements increases from left to right across a period of the periodic table because as the mass slowly increases, the volume rapidly decreases across the period; therefore, the density must increase because a larger mass is divided by a smaller volume.

2.100

2.101

Mercury and bromine share the physical property of being liquids at room temperature. They are not in the same group, though, because their chemical properties differ.

2.102

Chemical properties are dependent on the number of valence electrons an atom contains, not the number of neutrons an atom contains; therefore, the chemical properties of isotopes of the same element are the same because all isotopes of the same element contain the same number of electrons, including valence electrons.

2.103

a. Atoms of different elements contain different numbers of protons. b. Atoms of different isotopes contain different numbers of neutrons, but the same number of protons.

Chemistry for Thought 2.104

Aluminum exists as one isotope; therefore, all atoms have the same number of protons and neutrons as well as the same mass. Nickel exists as several isotopes; therefore, the individual atoms do not have the weighted average atomic mass of 58.69 u.

2.105

The ingestion of non-radioactive potassium iodide will satisfy the body’s need or iodine, blocking the absorption of radioactive iodine by the thyroid gland, the part of the body that is most sensitive.

2.106

While radioactive decay does occur naturally, it is unlikely that lead changes into gold naturally because lead has an atomic number of 82 and hold has an atomic number of 79, which makes a difference of 3. None of the common nuclear decay processes change the atomic number of the parent nucleus by 3.

2.107

Because of their mass charge, alpha particles do not travel far enough to be useful in diagnostics.

2.108

In principle, a radioactive isotope never completely disappears by radioactive decay because only half of the sample decays per half-life, so half of the initial sample remains. In reality, all of a sample will decay because eventually, one atom will remain in a sample and when that one atom undergoes decay the entire atom will undergo decay, not half of the atom.

Chapter 2 2.109

While at ﬁrst sending nuclear waste into outer space might seem like an attractive possibility because it would remove the hazardous materials from the earth, it is unlikely to be the best solution for waste disposal. Unfortunately, successfully launching the materials into space would not be assured, and if the spacecraft used were to explode during or shortly after launch, radioactive material would be scattered and the results could be devastating. In addition, we have no idea what the eventual fate of the radioactive waste would be. The presence of pockets of radioactive materials in the cosmos could have a myriad of unintended eﬀects on our universe.

2.110

a. When plutonium-239 undergoes an alpha emission, two protons and two neutrons are released, and uranium-235 will be formed. When Iodine-131 undergoes a beta emission, it releases a beta particle and is transformed into xenon-131. These are the balanced equations: and b. When iodine-131 is ingested, it is readily absorbed by the body’s thyroid gland. The accumulation of iodine-131 in the thyroid can lead to the irradiation of thyroid tissue and the emission of ionizing radiation, which can increase the risk of developing thyroid cancer or other thyroid disorders.

2.111

238 92

234 U → 42α + 90 Th

234 90

234 Th → −01 β + 91 Pa

234 91

234 Pa → −01 β + 92 U

Chapter 3

Chapter 3: Chemical Bonds: Molecule Formation CHAPTER OUTLINE 3.5 Covalent Bonding 3.6 Naming Binary Covalent Compounds 3.7 Lewis Structures of Polyatomic Ions 3.8 Compounds Containing Polyatomic Ions

3.1 An Introduction to Lewis Structures 3.2 The Formation of Ions 3.3 Ionic Compounds 3.4 Naming Binary Ionic Compounds

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1. 2. 3. 4. 5. 6. 7. 8.

Draw correct Lewis dot structures for atoms of the representative elements. (Section 3.1; Exercises 3.1, 3.3, and 3.5) Use electron configurations to determine the number of electrons gained or lost by atoms as they achieve noble gas electron configurations. (Section 3.2; Exercise 3.9) Use electron configurations to determine the series of ions that are isoelectronic with the noble gases. (Section 3.2; Exercise 3.17) Use the octet rule to correctly predict the ions formed during the formation of ionic compounds. (Section 3.3; Exercises 3.21) Write correct formula units for ionic compounds containing a representative metal and a representative nonmetal. (Section 3.3; Exercise 3.23) Determine formula weights for ionic compounds in atomic mass units. (Section 3.3; Exercises 3.27 and 3.28) Correctly name binary ionic compounds. (Section 3.4; Exercise 3.35, 3.39, 3.41) Draw correct Lewis structures for covalent molecules. (Section 3.5; Exercises 3.47, 3.49, and 3.51)

Determine molecular weights for covalent compounds in atomic mass units. (Section 3.5; Exercises 3.53 and 3.55) 10. Correctly name binary covalent compounds. (Section 3.6; Exercise 3.57) 11. Draw correct Lewis structures for polyatomic ions. (Section 3.7; Exercises 3.61 and 3.63) 12. Write correct formulas for ionic compounds containing representative metals and polyatomic ions. (Section 3.8; Exercises 3.65 and 3.69) 13. Correctly name binary ionic compounds containing polyatomic ions. (Section 3.8; Exercises 3.67)

LECTURE HINTS AND SUGGESTIONS 1.

Emphasize how the reactivity of the noble gases is related to the stability of their electronic conﬁgurations. Use several examples of some simple ionic and covalent compounds to show how atoms exchange or share electrons in order to attain noble gas conﬁgurations.

Many students have diﬃculty recognizing the diﬀerence between a covalent molecule such as SO3, sulfur trioxide, and a polyatomic anion, such as SO32-. The importance of putting the charge on the

Chapter 3 ion

should be stressed in this chapter. Also, point out the polyatomic anion is not a compound. It

must be 3.

associated with a positive ion (usually a metal ion) to make a compound.

Be sure that the students understand that there is not really a clear division between covalent and ionic bonds, and that there is a continuous increase in ionic character as the electronegativity diﬀerence between the two bonding atoms increases. Generalizations can be made that ionic compounds are formed from a metal or NH4+ and a non-metal, while covalent compounds are formed from two nonmetals.

Providing a handout summarizing cation and anion charges along with the rules of nomenclature is helpful when introducing ionic and covalent compounds. The example handout titled Naming Ionic and Covalent Compounds on the next two pages could be used as a resource for students to reference.

Chapter 3

Naming Ionic and Covalent Compounds Ionic Compounds Cations 1.

The charge of a cation is the number of electrons lost to achieve a noble gas electron configuration (i.e. excluding transition metals the charge of a cation is the group number of the periodic table.) Group 1 +1 charge Ex. Na+ Group 2 +2 charge Ex. Sr2+

Most transition metals can exist in more than one form. These ions are named with a roman numeral designating the charge.

Name of cations: Element name + ion Examples: Mg2+ Magnesium ion Mn4+ Manganese (IV) ion

There is one polyatomic cation: NH4+ ammonium ion

Anions 1.

The charge of an anion is the number of electrons gained to achieve a noble gas electron configuration Group 16 -2 charge Ex. O2Group 17 -1 charge Ex. Br –

Name of anions: element stem and the suﬃx – ide.

Examples: Cl-

chloride ion

sulﬁde ion

Chapter 3

There are many polyatomic anions:

Naming ionic compounds: 1.

Name = metal + nonmetal stem + -ide Examples:

CaF2

Calcium ﬂuoride

Li2S

Lithium Sulﬁde

Compounds with metals that can form more than one type of ion must indicate which ion is present by a Roman numeral in parentheses following the name of the metal. Examples: NiCl2

Nickel (II) Chloride

Ti2O3

Titanium (III) Oxide

Covalent Compounds Naming Covalent Compounds: 1.

Give the name of the less electronegative element ﬁrst (the element given ﬁrst in the formula)

Give the stem of the name of the more electronegative element next (see Table 3.1) and add the suﬃx -ide.

Indicate the number of each type of atom in the molecule by means of the Greek preﬁxes listed in Table 3.4. a.

the preﬁx mono- is not used when it appears at the beginning of a name.

Examples: NBr3 Nitrogen tribromide Cl2O7 Dichlorine heptoxide

Chapter 3

SOLUTIONS TO ALL END-OF-CHAPTER EXERCISES What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 3.1 AN INTRODUCTION TO LEWIS STRUCTURES 3.1

potassium

aluminum

barium

bromine

iodine

tin

strontium

sulfur

3.3

a. b. c. d.

fluorine element number 37 selenium silicon

3.4

a. germanium

b. cesium

c. indium

d. calcium

a. carbon

b. phosphorus

c. sodium

d. rubidium

3.2

3.5

3.6

[He] 2s22p5 [Kr] 5s1 [Ar] 4s23d104p4 [Ne] 2s22p2

a. lithium [He]2s1

b. chlorine [Ne]3s23p5 c. titanium [Ar] 4s23d2 d. [Ar]4s23d104p3 3.7

Any group IIIA (13) element

Any group VI (16) element

Chapter 3

3.8

Any group IIA (2) element

Any group VA (15) element

SECTION 3.2 THE FORMATION OF IONS 3.9 a.

iodine

element 38 (strontium)

Added electrons 1

Removed electrons 17 (7 not including d electrons)

(6 not including d electrons)

15 (5 not including d electrons)

phosphorus

germanium

Added electrons 4

Removed electrons 14

3.10

(4 not including d electrons)

(7 not including d or f electrons)

element number 49

13 (3 not including d electrons)

calcium

(6 not including d electrons)

3.11

Number of electrons lost/gained

Equation

2 electrons lost

Mg → Mg 2+ + 2e −

silicon

4 electrons lost

Si → Si 4+ + 4e −

element 53

1 electron gained

I + e− → I−

sulfur

2 electrons gained

S + 2e − → S 2 −

Number of electrons lost/gained

Equation

1 electron lost

Cs → Cs + + e −

3.12 a.

oxygen

2 electrons gained

O + 2e − → O 2 −

element number 7

3 electrons gained

N + 3e − → N 3 −

iodine

1 electron gained

I + e− → I−

3.13

Equation

Ion Symbol

A bromine atom that has gained one electron

Br + e → Br

A sodium atom that has lost one electron

Na → Na + + e −

Na +

A sulfur atom that has gained two electrons

S + 2e − → S 2 −

S2−

−

Br −

Chapter 3 3.14

3.15

3.16

3.17

3.18

Equation

Ion Symbol 2−

Se 2 −

A selenium atom that has gained two electrons

Se + 2e → Se

A rubidium atom that has lost one electron

Rb → Rb + + e −

Rb +

An aluminum atom that has lost three electrons

Al → Al 3 + + 3e −

Al3+

E−

−

fluorine

E3−

nitrogen

beryllium

lithium

sulfur

E+

sodium

aluminum

chlorine

E2−

Mg 2 +

neon

N3−

neon

Te 2 −

xenon

Be 2 +

helium

Li +

helium

S2−

argon

xenon

−

3.19

a. Kr: Se2-, Br-, Kr, Rb+, Sr2+ b. Xe: Te2-, I-, Xe, Cs+, Ba2+ c. Rn: Ra2+, Fr+, Rn, At-, Po2-

3.20

Cu+

29 protons, 28 electrons

S2-

16 protons, 18 electrons

Sc3+

21 protons, 18 electrons

−

krypton

SECTION 3.3 IONIC COMPOUNDS 3.21

Cation formation a. b. c.

Ca and S

3.24

3.25

+ 2e

→ 2e

Ca → Ca

Mg and N

Mg + Mg

elements num. 19 & 17

3.22

3.23

K + K → 1e

Anion formation

S + 2e → S

−

N + 3e → N −

−

Ca and Cl

lithium and bromine

Ca → Ca

+ 2e

Mg 3 N 2 KCl

−

Anion formation

Formula

Cl + e → Cl

CaCl 2

−

Li → Li + + e − 2+

−

Br + e − → Br −

LiBr

2−

MgS

elements num. 12 & 16

Br −

2−

CaO

S2−

BaSe

I−

BaI 2

P3−

Ba 3 P2

As 3 −

Ba 3 As 2

XO, X = +2

XBr2, X = +2

Mg → Mg + 2e

CaBr2

S + 2e → S

−

Formula CaS

3−

Cl + 1e → Cl

−

Cation formula 2+

2−

S2− 3−

CaS

Ca 3 N 2

Chapter 3 c.

X3P, X = +1

XI, X = +1

3.26

a. c.

BaX, X = -2 Al2X3, X = -2

b. d.

MgX2, X = -1 Na2X, X = -2

3.27

a. c.

Na2O, 61.98 amu PbS2, 271.32 amu

b. d.

FeO, 71.84 amu AlCl3,133.33 amu

3.28

a. c.

NaBr = 102.89 amu Cu2S = 159.15 amu

b. CaF2 = 78.07 amu d. Li3N = 34.83 amu

SECTION 3.4 NAMING BINARY IONIC COMPOUNDS 3.29

3.30

3.31

3.32

3.33

3.34

3.35

3.36

3.37

binary

H2 S

binary

OF2

binary

MgBr2

binary

H 2 SO 4

not binary

PbO 2

binary

Be 3 N 2

binary

CuCl 2

binary

CaCO 3

not binary

KNO 3

not binary

Ca 2 +

calcium ion

Al 3 +

aluminum ion

potassium ion

rubidium ion

Li +

lithium ion

Ba 2 +

barium ion

magnesium ion

cesium ion

chloride ion

S2−

nitride ion

2−

bromide ion

P3−

phosphide ion

oxide ion

fluoride ion

Cl −

3−

Br −

2−

Na 2 O

sodium oxide

LiF

lithium fluoride

CaCl 2

calcium chloride

Ba 3 N 2

barium nitride

Al 2 S 3

aluminum sulfide

SrS

strontium sulfide

Li 2 O

lithium oxide

CaF2

calcium fluoride

MgO

magnesium oxide

BaCl 2

barium chloride

a. magnesium chloride, MgCl2 b. lithium oxide, Li2O

−

sulfide ion selenide ion

Chapter 3 c. potassium fluoride, KF d. sodium nitride, Na3N 3.38

3.39

3.40

3.41

3.42

3.43

3.44

a. cesium iodide CsI b. calcium sulfide CaS c. aluminum chloride AlCl3 d. beryllium phosphide

Be3P2

CrCl 2 and CrCl 3

chromium (II) chloride and chromium (III) chloride

CoS and Co 2 S 3

cobalt (II) sulfide and cobalt (III) sulfide

FeO and Fe 2 O 3

iron (II) oxide and iron (III) oxide

PbCl 2 and PbCl 4

lead (II) chloride and lead (IV) chloride

PbO and PbO 2

lead(II) oxide and lead(IV) oxide

CuCl and CuCl 2

copper(I) chloride and copper(II) chloride

Au 2 S and Au 2 S 3

gold(I) sulfide and gold(III) sulfide

CoO and Co 2 O 3

cobalt(II) oxide and cobalt(III) oxide

CrCl 2 and CrCl 3

chromous chloride and chromic chloride

CoS and Co 2 S 3

cobaltous sulfide and cobaltic sulfide

FeO and Fe 2 O 3

ferrous oxide and ferric oxide

PbCl 2 and PbCl 4

plumbous chloride and plumbic chloride

PbO and PbO 2

plumbous oxide and plumbic oxide

CuCl and CuCl 2

cuprous chloride and cupric chloride

Au 2 S and Au 2 S 3

aurous sulfide and auric sulfide

CoO and Co 2 O 3

cobaltous oxide and cobaltic oxide

manganese (II) chloride

MnCl 2

iron (II) bromide

FeBr2

iron (III) sulfide

Fe 2 S 3

tin (II) chloride

SnCl 2

chromium (II) oxide

CrO

mercury (I) oxide

Hg 2 O

copper (I) nitride

Cu 3 N

lead (II) oxide

PbO

cobalt (II) sulfide

CoS

platinum (IV) iodide

PtI 4

SECTION 3.5 COVALENT BONDING 3.45

Chapter 3 3.46

3.47

IBr

PH 3 (each H atom is bonded to the P atom)

HClO 2 (the O atoms are each bonded to the Cl, and the H is bonded to one of the O atoms)

3.48

3.49

CH 4 (each H atom is bonded to the C atom)

CO 2 (each O atom is bonded to the C atom)

H 2 Se (each H atom is bonded to the Se atom)

NH 3 (each H atom is bonded to the N atom)

a. CCl2O (Each Cl and O atom is bonded to the C atom)

Chapter 3 b. SiF4 (Each F atom is bonded to the Si atom)

c. PF3 (Each F atom is bonded to the P atom)

d. C2H2 (Each H atom is bonded to one C atom)

3.50

a. Boron does not have a full octet

Sulfur has an expanded octet with 12 electrons around the central atom

Sulfur has an expanded octet with ten electrons around the central atom

d. In SeO3, selenium has 12 electrons around the central atom. 3.51

a. O3 (the O atoms are bonded together, like beads on a string) b. CS2 (each S atom is bonded to the C atom) c. SeO2 (each O atom is bonded to the Se atom)

3.52

a. ClO2 = 67.45 amu

Chapter 3 b. N2O = 44.02 amu c. SO2 = 64.06 amu d. CCl4 = 153.81 amu 3.53

a. sulfur hexafluoride, SF6, 146.05 amu b. silicon tetrachloride, SiCl4, 169.89 amu c. dinitrogen trioxide, N2O3, 76.01 amu d. sulfur trioxide, SO3, 80.06 amu

3.54

a. C3H8 = 44.10 amu b. CO = 28.01 amu

3.55

a. Glucose (C6H12O6), 180.16 amu b. The anesthetic Lidocaine (C14H22N2O), 234.34 amu c. The amino acid methionine (C5H11NO2S), 149.21 amu

SECTION 3.6 NAMING BINARY COVALENT COMPOUNDS 3.56

a. b. c.

phosphorus trichloride dinitrogen monoxide carbon tetrachloride

d. e.

boron trifluoride carbon disulfide

3.57

a. b. c.

silicon dioxide silicon tetrafluoride diboron trioxide

d. e.

nitrogen monoxide carbon tetrabromide

3.58

a. SO2, sulfur dioxide, covalent b. MgBr, magnesium bromide, ionic c. BaS, barium sulfide, ionic d. N2O4, dinitrogen tetroxide, covalent

3.59

a. CS2, carbon disulfide, covalent b. SiF4, silicon tetrafluoride, covalent c. CuO, copper(II) oxide, ionic d. NaF, sodium fluoride, ionic

SECTION 3.7 LEWIS STRUCTURES OF POLYATOMIC IONS 3.60

a. SO32– (Each O atom is bonded to the S atom), sulfite

Chapter 3 b. SO42– (Each O atom is bonded to the S atom), sulfate

3.61

a. NO3– (Each O atom is bonded to the N atom), nitrate

b. NO2– (Each O atom is bonded to the N atom), nitrate

3.62

3.63

NH 4 + (each H atom is bonded to the N atom)

PO 4 3 − (each O atom is bonded to the P atom)

SO 3 2 −

ClO 3 − (each O atom is bonded to the Cl atom)

(each O atom is bonded to the S atom)

CN − 13

Chapter 3

CO32- (each O atom is bonded to the C atom)

SECTION 3.8 COMPOUNDS CONTAINING POLYATOMIC IONS 3.64

3.65

3.66

3.67

Any group I A (1) element and SO 3

Any group I A (1) element and C 2 H 3 O −2

Any metal that forms M 2+ ions and Cr2 O7

Any metal that forms M 3+ ions and PO 4

Any metal that forms M 3+ ions and NO −3

M(NO 3 )3

Any metal that forms M + ions and SO 4 2 −

M 2 SO 4

Any metal that forms M 3+ ions and OH −

M(OH)3

Any metal that forms M 2+ ions and HPO 4 2 −

MHPO 4

calcium and the hypochlorite ion

Ca(ClO)2

calcium hypochlorite

cesium and the nitrite ion

CsNO 2

cesium nitrite

Mg and SO 3

2−

MgSO 3

magnesium sulfite

K and Cr2 O7

2−

K 2 Cr2 O7

potassium dichromate

calcium and carbonate ion

CaCO 3

calcium carbonate

sodium and the sulfate ion

Na 2 SO 4

sodium sulfate

K 3 PO 4

potassium phosphate

Mg(NO 3 )2

magnesium nitrate

3.68

3.69

3.70

2−

K and PO 4 2+

M 2 SO 3 MC 2 H 3 O 2 2−

MCr2 O7

3−

and NO 3 −

MPO 4

MgSO 3

(NH 4 )2 SO 4

Ba(OH)2

LiHCO 3

CaCO 3

KOH

Na 3 PO 4

Na 2 CO 3

Ca(NO 3 )2

NH 4 Cl

a. NaNO3 b. Ca2OH c. MgBr2

Chapter 3 d. K2CO3 3.71

a. CsSO3 should be Cs2SO3 b. LiPO4 should be Li3PO4 c. K(MnO4)2 should be KMnO4 d. AlClO3 should be Al3ClO3

3.72

a. FeSO4, ionic, iron(II) sulfate b. NO, covalent, nitrogen monoxide c. CF4, covalent, carbon tetrachloride

3.73

a. NaHCO3, ionic, sodium bicarbonate, or sodium hydrogen carbonate b. Ca2C2H3O2, ionic, calcium acetate c. KI, ionic, potassium iodide

CHEMISTRY FOR THOUGHT 3.74

Hydrogen is the element with an electronic configuration of 1s1. Nitrogen is the element with the electronic configuration of 1s22s22p3. The molecule that contains 3 hydrogen atoms and 1 nitrogen atom is NH 3 . They hydrogen bond to each other as shown here:

3.75

Magnesium is the element in group IIA (2) and period 3. Fluorine is the element with the electron configuration 1s22s22p5. The formula for a compound made from these elements would be MgF2 , magnesium fluoride.

3.76

Potassium dichromate is K 2 Cr2 O7 .

Potassium chromate is K 2 CrO 4 .

Potassium phosphate is K 3 PO 4 .

Potassium permanganate is KMnO 4 .

The colored compounds of potassium ( K 2 Cr2 K 2 O7 and KMnO 4 ) have a transition metal as part of the polyatomic ion. 3.77

Yes, a reaction similar to the one that occurs between sodium metal and chlorine gas would be expected to occur between potassium metal and fluorine gas because potassium is in the same group as sodium and fluorine is in the same group as chlorine. In fact, any combination of metal from group IA (1) (lithium, sodium, potassium, rubidium, cesium, francium) and nonmetal from group VIIA (17) (fluorine, chlorine, bromine, iodine, astatine) should produce a similar reaction.

Chapter 3 3.78

A negatively charged ion will be larger than a nonmetal atom of the same element, because as the electrons are added to the atom, the electron cloud will increase in size as the nucleus is not able to hold onto the electrons as well as their number increases.

3.79

When a metal changes to form a positively charged metal ion, it loses electrons. The remaining electrons are then pulled closer to the positively charged nucleus, which makes the size of the metal ion smaller than the size of the original metal atom.

3.80

There is one pair of nonbonding electrons present on the nitrogen atom in methamphetamine.

3.81

Neon atoms do not combine to form Ne 2 molecules because each neon atom has a completed valence shell and does not need to form bonds with another atom in order to satisfy its valence.

3.82

There are five nonbonding electron pairs on glycine that are located on the nitrogen and oxygen atoms: two are on the C=O oxygen atom, two are on the OH oxygen atom, and one pair is on the NH2 nitrogen atom.

3.83 3.84

There are four nonbonding electron pairs on propanoic acid that are located on the two oxygen atoms: two are on the C=O oxygen, and two are on the OH oxygen.

Chapter 4

Chapter 4: The Mole and Chemical Reactions CHAPTER OUTLINE 4.1 Avogadro’s Number: The Mole 4.2 The Mole and Chemical Formulas 4.3 Chemical Equations

4.4 The Mole and Chemical Equations 4.5 Reaction Yields 4.6 Types of Reactions

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Convert between the number of moles, number of grams, and number of atoms of elements. (Section 4.1; Exercises 4.1 and 4.3)

Convert between the number of moles, number of grams, number of compounds, and number of atoms in compounds. (Section 4.2; Exercise 4.9)

Calculate the percent by mass of atoms in compounds. (Section 4.2; Exercise 4.11)

Write a balanced chemical equation when provided with a chemical reaction. (Section 4.3; Exercises 4.31 and 4.32)

Use the mole concept to perform calculations based on the stoichiometry of balanced chemical equations. (Section 4.4; Exercises 4.35 and 4.37)

Use the mole concept to calculate percent yields. (Section 4.5; Exercises 4.45 and 4.47)

Identify the oxidizing and reducing agents in oxidation–reduction (redox) reactions when given the oxidation states of each element in the reactants and products. (Section 4.6; Exercises 4.50 and 4.51)

Classify reactions ﬁrst as redox or nonredox, then as decomposition, combination, single replacement, double replacement, or combustion. (Section 4.6; Exercises 4.54 and 4.55)

LECTURE HINTS AND SUGGESTIONS 1.

It should be emphasized that the mole is a convenient way of measuring out needed numbers of atoms and molecules in the correct ratios for chemical reactions. Explain that the term “mole” is the same type of term as “dozen,” “pair,” or “gross,” except that it speciﬁes a much larger number of items.

Emphasize the main roles of oxidizing numbers are in recognizing redox reactions and in balancing redox equations. To simplify the problems oxidation numbers are provided and students are simply asked to identify oxidizing and reducing agents, rather than memorize the rules of determining oxidation numbers. 3.

Use some simple redox reactions and relate the oxidation number change with the loss and gain of electrons. For example, in the reaction 2Na + Cl2 → 2NaCl, each sodium atom loses one electron, and its oxidation number goes from 0 to + 1; and each chlorine atom gains an electron, and its oxidation number goes from 0 to -1.

Many double replacement reactions are easy to demonstrate in class. Simply mix two solutions, which will provide two ions that form a precipitate: for example, sodium carbonate and calcium 1

Chapter 4 chloride or barium chloride and sodium sulfate or silver nitrate and potassium chromate. Try to pick examples that have some relevance such as in these cases where the formation of calcium carbonate would illustrate what is meant by hard water and where the insolubility of barium sulfate is related to its usefulness as a radiopaque diagnostic material. The silver chromate is a chemical precursor to modern photography that has been important in neuroscience, as it is used in the “Golgi method” of staining neurons for microscopy. The reaction that produces silver chromate is also highly visible in a large lecture theater because the reaction involves a color change in addition to the formation of a precipitate.

Chapter 4

Solutions To All End-Of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 4.1 AVOGADRO’S NUMBER: THE MOLE 4.1

a. 1.003 x 1023 atoms are in 3.10 grams of phosphorus.

b. 3.207 g of sulfur would contain the same number of atoms as 3.10 grams of phosphorus.

4.2

a. 6.022 x 1022 atoms are in 1.60 grams of oxygen.

b. 1.90 grams of fluorine would contain the same number of atoms as 1.60 grams of oxygen.

4.3

The number of moles of beryllium atoms in a 10.0 g sample of beryllium

The number of lead atoms in a 2.0 mol sample of lead

Chapter 4

4.4

The number of sodium atoms in a 50 g sample of sodium

The mass of grams of one phosphorous atom

The number of grams of aluminum in 1.65 mol of aluminum

The total mass in grams of one-fourth Avogadro’s number of krypton atoms

SECTION 4.2 THE MOLE AND CHEMICAL FORMULAS 4.5

Fructose, C6H12O6 = (6 ´ 12.01 g/mol C) + (12 ´ 1.008 g/mol H) + (6 ´ 16.00 g/mol O) = 180.2 g/mol

4.6

Creatine, C4H9N3O2, (4 x 12.01 g/mol C) + (9 x 1.008 g/mol H) + (3 x 14.00 g/mol N) + (2 x 16.00 g/mol O) = 131.1 g/mol

4.7

Olinvyk, C22H30N2O2S, (22 ´ 12.01 g/mol C) + (30 ´ 1.008 g/mol H) + (2 ´ 14.00 g/mol N) + (2 ´ 16.00 g/mol O) + (1 ´ 32.06 g/mol O) = 386.5 g/mol

4.8

Camzyos, C15H19N3O2, (15 x 12.01 g/mol C) + (19 x 1.008 g/mol H) + (3 x 14.00 g/mol N) + (2 x 16.00 g/mol O) = 273.3 g/mol

4.9

(1 x 31.0 u) + (3 x 1.01 u) = 34.u; 1 mole PH3 = 34.0 g PH3 (1 x 32.1 u) + (2 x 16.0 u) = 1 mole SO2 = 64.1 g SO2

Chapter 4 4.10

(1 x 10.8 u) + (3 x 19.0 u) = 67.8 u; 1 mole BF3 = 67.8 g BF3 (2 x 1.01 u) + (1 x 32.1 u) = 34.1 u; 1 mole H2S = 34.1 g H2S

4.11

a. 2 mol of O atoms

4.12

4.13

4.14

First, let’s use the formula for percent to ﬁnd the mass of iron present.

4.15

72.0 g C x 100 = 40.0% of C in C6 H12 O6 180.0 g C6 H12 O6 24.0 g C x 100 = 52.5% of C in C 2 H6 O 46.0 g C 2 H6 O

4.16

4.04 g H x 100 = 25.2% of H in CH 4 16.04 g CH 4

Chapter 4

6.06 g H x 100 = 20.1% of H in C 2 H6 30.01 g C 2 H6 4.17

Statement 4.

6.02 x 1023 C6H5NO3 molecules contain 36.12 x 1023 C atoms, 30.01 x 1023 H atoms, 6.02 x 1023 N atoms, and 18.06 x 1023 O atoms.

Statement 5.

1 mol C6H5NO3 molecules contain 6 moles of C atoms, 5 moles of H atoms, 1 mole of N atoms, and 3 moles of O atoms.

Statement 6.

139 g of nitrophenol contains 72.0 g of C, 5.05 g of H, 14.0 g of N, and 48.0 g of O.

a. (Statement 6) 139 g of nitrophenol contains 72.0 g of C, 5.05 g of H, 14.0 g of N, and 48.0 g of O

b. (Statement 5) 1 mol of C6H5NO3 molecules contain 6 moles of C atoms, 6 moles of H atoms, 1 mole of N atoms, and 3 mol of O atoms.

c. (Statement 4) 6.02 x 1023 C6H5NO3 molecules contain 36.12 x 1023 C atoms, 30.1 x 1023 H atoms, 6.02 x 1023 N atoms, and 18.06 x 1023 O atoms. 4.18

a. 180 g of fructose contains 72.0 g of C, 12.1 g of H, and 96.0 g of O.

b. 1 mol of C6H12O6 molecules contains 6 mol of C atoms, 12 mol of H atoms, and 6 mol of O atoms.

c. 6.02 x 1023 C6H12O6 molecules contain 36.12 x 1023 C atoms, 72.24 x 1023 H atoms, and 36.12 x 1023 O atoms.

Chapter 4

4.19

Urea (CH4N2O) contains the higher mass percentage of nitrogen as shown below:

4.20

Using the periodic table, the mass of 1 mol of epinepherine can be calculated: 9 C’s = 9 x 12.01 = 108.9 13 H’s = 13 x 1.008 = 13.104 1 N = 1 x 14.01

4.21

Using the periodic table, the mass of 1 mol of cortisone can be calculated: 21 C’s = 21 x 12.01 = 252.21 28 H’s = 28 x 1.008 = 28.224

80.00 or 360.4 g/mol 360.434 Now let’s use the factor-unit method. Step 1: 10.0 mg Step 2: 10.0 mg =

5 O’s = 5 x 16.00 =

Step 3: 10.0 mg x Step 4:

1 g 1000 mg

1g 1000 mg x = mol 1000 mg 360.4 g

1000 mg 360.4 g

= 2.77 x 10 −5 mol

4.22

There are 1759.34 grams in 5.5 moles of chloroquine.

4.23

a. C6H14N2O2

Chapter 4 b. Molar Mass of C6H14N2O2, (6 ´ 12.01 g/mol C) + (14 ´ 1.008 g/mol H) + (2 ´ 14.00 g/mol N) + (2 ´ 16.00 g/mol O) =146.2 g/mol c. 19.2%

d. 0.0171 moles of lysine are in 2.5 g.

4.24

a. C5H8NO4 b. Molar mass of C5H8NO4 = (5 x 12.01 g/mol C) + (8 x 1.008 g/mol H) + (1 x 14.01 g/mol N) + (4 x 16.00 g/mol O) = 146.1 g/mol c. 43.8%

d. Yes, the risk value is 80 mg/kg and if a 9 kg the toddler ingested 5.4 g of glutamate, they will have ingested over the risk value

SECTION 4.3 CHEMICAL EQUATIONS 4.25

Reactants

Products

CH 4 ,O 2

CO 2 ,H 2 O

Al,Cl 2

AlCl 3

CH 4 (g) + 2O 2 (g) → CO 2 (g) + H 2 O(g)

2Al(g) + 3Cl 2 (g) → 2AlCl 2 (g)

methane + water → carbon monoxide + hydrogen

methane, water

copper(II) oxide + hydrogen → copper + water

copper(II) oxide

4.26 a. b.

carbon monoxide, hydrogen copper, water

hydrogen Reactants

Products

H 2 (g) + Cl 2 → (g) 2HCl (g)

H 2 , Cl 2

HCl

2 KClO 3 (s) → 2 KCl (s) + 3 O 2 (g)

KClO 3

KCl, O 2

Chapter 4

magnesium oxide + carbon → magnesium + carbon monoxide

4.27

magnesium oxide, carbon

magnesium carbon monoxide

ethane, oxygen

carbon dioxide,

ethane + oxygen → carbon dioxide + water

Fe (s) + O 2 (g) → Fe 2 O 3 (s) is not consistent with the law of conservation of matter because neither the number of iron atoms nor the number of oxygen atoms are the same on both sides

water

of the reaction. b.

2Na 3 PO 4 (aq) + 3MgCl 2 (aq) Mg 3 (PO 4 )2 (s) + 6NaCl (aq) is consistent with the law of conservation of matter.

3.20 g oxygen + 3.21 g sulfur → 6.41 g sulfur dioxide is consistent with the law of conservation of matter.

Notice, the number of moles of oxygen and sulfur are equal on both sides of the equation.

 1 mole O 2  Reactants: 3.20 g O 2   = 0.100 moles O 2 ;  32.0 g O 2   1 mole S  3.21 g S   = 0.100 moles S  32.1 g S   1 mole SO 2   2 moles O  Products: 6.41 g SO 2    = 0.200 moles O;  64.1 g SO 2   1 mole SO 2   1 mole SO 2   1 mole S  6.41 g SO 2    = 0.100 moles S  64.1 g SO 2   1 mole SO 2  d.

4.28

CH 4 (g) + 2 O 2 (g) → CO 2 (s) + 2 H 2 O (g) is consistent with the law of conservation of matter. ZnS (s) + O 2 (g) → ZnO (s) + SO 2 (g) is not consistent with the law of conservation of matter

because the reactant (left) side of the equation has two moles of oxygen atoms, while the product (right) side of the equation has three moles of oxygen atoms. b.

Cl 2 (aq) + 2 I − (aq) → I 2 (aq) + 2Cl − (aq) is consistent with the law of conservation of matter.

1.50 g oxygen + 1.50 g carbon → 2.80 g carbon monoxide is not consistent with the law of conservation of matter because the mass of the reactants is 3.00 g, while the mass of the products is only 2.80 g.

Notice the number of moles of oxygen and carbon are not equal on both sides of the equation either.

Chapter 4

 1 mole O 2  Reactants: 1.50 g O 2   = 0.0469 moles O 2 ;  32.0 g O 2   1 mole S  1.50 g C   = 0.125 moles C  12.0 g S   1 mole CO   1 mole O  Products: 2.80 g CO    = 0.100 moles O;  28.0 g CO   1 mole CO   1 mole CO   1 mole C  2.80 g CO    = 0.100 moles C  28.0 g CO   1 mole CO  d.

2 C 2 H6 (g) + 7 O 2 → 4 CO 2 (g) + 6 H 2 O (g) is consistent with the law of conversation

of matter.

4.29

4.30

3 Pb(NO 3 )2 (aq) + 2 AlBr3 (aq) → 3 PbBr2( 3) + 2 Al(NO 3 )3 (aq)

K (s) + H 2 O (l) → KOH (aq) + H 2 (g)

CaCO 3 (s) → CaO (s) + CO 2 (g)

Ba(ClO 3 )2 (aq) + H 2 SO 4 (aq) → 2HClO 3 (aq) + BaSO 4 (s)

Ag(s) + Cu(NO 3 )2 (aq) → Cu (s) + AgNO 3 (aq)

2 N 2 O(g) + 3 O 2 (g) → 4 NO 2 (g)

Chapter 4

Mg (s) + O 2 (g) → 2 MgO (s)

H 2 SO 4 (aq) + Ca(OH)2 (aq) → CaSO 4 (s) +2 H 2 O (l)

4.31

Li 3 N (s) → Li (s) + N 2 (g)

2Li 3 N (s) → 6Li (s) + N 2 (g)

4.32

KClO 3 (s) → KCl (s) + O 2 (g)

2 KClO 3 (s) → 2 KCl (s) + 3 O 2 (g)

Chapter 4

SECTION 4.4 THE MOLE AND CHEMICAL EQUATIONS 4.33

4.34

2SO 2 (g) + O 2 (g) → 2SO 3 (g)

4HCl (g) + O 2 (g) → 2Cl 2 (g) + 2 H 2 O (g)

Fe 2 O 3 (s) + 3 C (s) → 2Fe (s) + 3CO (g)

2H 2 O 2 (aq) → 2 H 2 O (l) + O 2 (g)

2 C 3 H6 (g) + 9 O 2 (g) → 6 CO 2 (g) + 6 H 2 O (g)

S (s) + O 2 (g) → SO 2 (g)

Chapter 4 b.

Sr (s) + 2 H 2 O (l) → Sr(OH)2 (s) + H 2 (g)

2 H 2 S (g) + 3 O 2 (g) → 2 H 2 O (g) + 2 SO 2 (g)

4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O (g)

CaO (s) + 3 C (s) → CaC 2 (s) + CO (g)

4.35

For the following equation: 4H2O(l) + SnCl4(s) ® Sn(OH)4(s) + 4HCl(aq) a. How many moles of H2O(l) will react with 1.5 moles of SnCl4(s):

b. 292 grams of HCl will be produced.

Chapter 4 4.36

2 SO 2 + O 2 → 2 SO 3 (g)

Factors:

12.0 x 10 23 SO 2 molecules 6.02 x 10 23 O 2 molecules 12.0 x 10 23 SO 3 molecules 12.0 x 10

; ;

6.02 x 10 23 O 2 molecules 12.0 x 10 23 SO 2 molecules 6.02 x 10 23 O 2 molecules

SO 2 molecules 12.0 x 10

SO 3 molecules

; ;

12.0 x 10 23 SO 2 molecules 12.0 x 10 23 O 2 molecules 12.0 x 10 23 SO 3 molecules 6.02 x 10 23 O 2 molecules

; ;

2 moles SO 3 2 moles SO 2 1 mole O 2 2 moles SO 2 2 moles SO 3 1 mole O 2 ; ; ; ; ; 1 mole O 2 2 moles SO 2 2 moles SO 3 2 moles SO 2 2 moles SO 3 1 mole O 2 160.0 g SO 3 160.0 g SO 3 128 g SO 2 32.0 g O 2 128 g SO 2 32.0 O 2 ; ; ; ; ; 32.0 O 2 128 g SO 2 160.0 g SO 3 128 g SO 2 160.0 g SO 3 32.0 g O 2

This list does not include all possible factors. 4.37

2 SO 2 + O 2 → 2 SO 3 (g)

 128 g SO 2  350. g SO 3   = 280. g SO 2  160. g SO 3  280. g SO 2 must react to produce 350. g SO 2 4.38

CaCO 3 (s) → CaO (s) + CO 2 (g)

 100 g CaCO 3  500. g CaO   = 891.265597148 g CaCO 3  56.1 g CaO  ≈ 891 g CaCO 3 4.39

CaCO 3 (s) → CaO (s) + CO 2 (g)  1 mole CO 2  500. g CaO   = 8.91265597 148 moles CO 2  56.1 g CaO  ≈ 8.91 moles CO 2

Chapter 4 4.40

2 Al (s) + 3 Br2 (l) → AlBr3 (s)

 479 g Br2  50.1 g Al   = 444.4055556 g Br2  54.0 g Al  ≈ 444 g Br2 with SF

4.41

3 Ag 2 S (s) + 2 Al (s) → 6 Ag (s) + Al 2 S 3 (s) a. b.

4.42

 54.0 g Al  0.250 g Ag 2 S   = 0.01814516129 g Al  744 g Ag 2 S  -2 SF  1 mole Al 2 S 3≈ 1.81 x 10 g Al with 0.250 g Ag 2 S  3.36021505 x 10 -4 moles Al 2 S 3 =   744 g Ag 2 S  ≈ 3.36 x 10 -4 moles Al 2 S 3 with SF

TiCl 4 (s) + 2Mg (s) → Ti (s) 2 MgCl 2 (s)  1000 g Ti   48.6 g Mg  1.00 kg Ti    = 1014.61377871 g Mg  1 kg Ti   47.9 g Ti  ≈ 1.01 x 10 3 g Mg with SF

4.43

C6 H12 O6 (aq) + 6 O 2 (aq) → 6 CO 2 (aq) + 6 H 2 O (l) a.

4.44

 108 g H 2 O  1.00 mole glucose   = 108 g H 2 O  1 mole glucose  = 1.08 x 10 2 g H 2 O   192 g O 2 1.00 mole glucose   = 192 g O 2  1 mole glucose  = 1.92 x 10 2 g O 2

C6 H12 O6 (aq) + 8 O 2 (aq) → 6 CO 2 (aq) + 6 H 2 O (l)   256 g O 2 1.00 mol caproic acid   = 256 g O 2  1 mol caproic acid 

SECTION 4.5 REACTION YIELDS 4.45 4.46

14.37 g x100 = 81.88% yield 17.55 g 2 HgO (s) → 2 Hg (l) + O 2 (g)

 402 g Hg  7.22 g HgO   = 6.6876 g Hg  434 g HgO 

5.95 g  402 g Hg  7.22 g Hg    434 g HgO 

x 100 = 89.0% yield

Chapter 4 5.95 g x 100 = 89.0% yield 6.6876 g Hg

4.47

2 Ca (s) + O 2 (g) → + 2 CaO (s)  112 g CaO  2.00 g Ca   = 2.7930 g CaO  80.2 g Ca 

2.26 g

 112 g CaO  2.00 g Ca    80.2 g Ca 

x 100 = 80.9% yield

2.26 g x 100 = 80.9% yield 2.7930 g Ca

SECTION 4.6 TYPES OF REACTIONS 4.48

O.N.

Change

Classiﬁcation

4.49

O.N.

Change

Classiﬁcation

4.50

2 Cu (s) + O 2 (g) → 2 CuO (s)

Cl 2 (aq) + 2 KI (aq) → 2 KCl (aq) + I 2 (aq)

3 MnO 2 (s) + 4 Al (s) → 2 Al 2 O 3 (s) + 3 Mn (s)

Chapter 4

4.51

2 H + (aq) + 3 SO 3 2 − (aq) → 2 NO 3 − (g) + H 2 O (l) + 3 SO 4 2- (aq)

Mg (s) + 2 HCl (aq) → MgCl 2 (aq) + H 2 (g)

4 NO 2 (g) + O 2 (g) → 2 N 2 O 5 (g)

H 2 (g) + Cl 2 (g) → 2 HCl (g)

H 2 O(g) + CH 4 (g) → CO (g) + 3 H 2 (g)

CuO (s) + H 2 (g) → Cu (s) + H 2 O (g)

B2 O 3 (s) + 3 Mg (s) → 2 B (s) + 3 MgO (g)

Fe 2 O 3 (s) + CO (g) → 2 FeO (s) + CO 2 (g)

Chapter 4 f.

4.52

Cr2 O7 2 − (aq) + 2 H + (aq) → 3 Mn 2 + (aq) + Cr 3 + (aq) + 3 MnO 2 (s) + H 2 O (l)

6 NaOH (aq) + 2 Al (s) → 3 H 2 (g) + Na 3 AlO 3 (aq) + heat

The oxidizing agent is NaOH because the oxidation number for the hydrogen changes from +1 to 0. The reducing agent is Al because the oxidation number for the aluminum changes from 0 to +3. 4.53

Yes, using the rules for oxidation numbers, assign oxidation numbers. Rule 1: The O.N. for O2 is O. Rule 4: The O.N. for combined H is +1. Rule 5: The O.N. for combined oxygen is -2. Rule 6: Allows us to calculate the O.N. of C on each side of the equation. In CO2, (O.N. of C) + 2 (O.N. of O) = 0 (O.N. of C) + 2 (-2) = 0 (O.N. of C) + (-4) = 0 O.N. of C - +4 In C6 H12 O6 , 6 (O.N. of C) + 12 (O.N. of H) + 6 (O.N. of O) = 0 Thus, carbon on the left side of the equation goes from O.N. of +4 to 0 (right side). Therefore, carbon has been reduced and is the oxidizing agent. The oxygen on the left side of the equation is going from O.N. of -2 to 0 (in the O2) and stays of -2 (in the C6 H12 O6 ). Therefore, some of the oxygen is being oxidized and is the reducing agent.

4.54 a. nonredox:

decomposition

b. redox: single-replacement c. nonredox: double-replacement d. nonredox: combination e. redox: combination f. redox: combustion g. redox: combustion

4.55

N 2 O 5 (g) + H 2 O (l) → 2HNO 3 (aq)

Chapter 4

4.56

Nonredox decomposition

4.57

Na H CO 3 (aq) + H (aq) → Na (aq) + H 2 O (l) + CO 2 (g) ; This reaction is a nonredox reaction.

+4 − 2

+1 − 2

+4 − 2

4.58

Redox.

4.59

Redox.

4.60

Cl 2 (aq) + H 2 O (l) HOCl (aq) + HCl (aq) ; This is a redox reaction.

4.61

Ca 3 (PO 4 )2 (s) + 4 H 3 PO 4 (aq) → 3 Ca(H 2 PO 4 )2 (s) ; This is a nonredox combination reaction.

4.62

Oxidizing agent: N2

+1 − 2

+5 − 2

+1 − 2 + 2

+1 − 1

+2 + 1 + 5 − 2

+1 + 5 − 2

Reducing agent: H2

ADDITIONAL EXERCISES 4.63

1.0 x 10 9 x 100 = 1.66 x 10 −13 % 6.02 x 10 23

4.64

4.65

D2O : (2 x 2u) + (1 x 16.00 u) = 20 u

4.66

Due to its highly reactive nature and tendency to donate electrons, lithium can participate in redox reactions as a reducing agent.

4.67

I would not expect argon to be involved in a redox reaction because it is a noble gas. Noble gases do not tend to form ions, and in order to participate in a redox reaction, argon would have to form ions.

4.68

    60 60.0 g Fe  = 7.5 g 60 Fe 6.983 g naturally occuring elemental iron   55.9 g naturally occuring    elemental iron  

Chapter 4 4.69

Sodium is the element with an electron configuration of 1s22s22p63s1. Phosphorus is the element with 15 protons in its nucleus.

3 Na (s) + P (s) → Na 3 P (s) 4.70

a. There are approximately 1.479 x 10-5 moles of tetrodotoxin in 2.4 mg of (C11H17N3O8).

b. There are 8.893 x 1018 molecules of tetrodotoxin in 2.4 mg of C11H17N3O8.

CHEMISTRY FOR THOUGHT 4.71

*Many vitamin supplements are candy-like in appearance, which can lead to iron poisoning in small children. Symptoms will appear in a 3-year-old at doses greater than 10 mg/kg. a.

No. If a 10 kg patient ingested 5 multivitamin gummies at 12 mg each, 60 mg of iron was ingested. The threshold for iron poisoning for a 10 kg patient would be 100 mg.

b. A 10 kg pediatric patient would receive 36 g of deferoxamine in 24 hours for the treatment for iron poisoning.

4.72

When a yield of more than 100% occurs for a compound prepared by precipitation from water solutions, it is likely that the “dry” compound is still moist and contains extra water mass.

4.73

The formula of the sample that was decomposed was N2O5, dinitrogen pentoxide.  1 mole O  80.0 g O   = 5.00 moles O  16.0 g O 

4.74

  1 mole N 1.20 x 10 24 atoms N   = 1.99 moles N 23  6.02 x 10 atoms g N  Zinc changing from zinc metal into zinc ions is an oxidation reaction because the zinc atoms are

losing electrons to become cations. This reaction is the source of electrons as shown in the following equation: Zn (s) → Zn2+ (aq) + 2 e-

Chapter 5

Chapter 5: Heating and Changes of State: A Story of Polarity and Intermolecular Forces CHAPTER OUTLINE 5.1 Geometries of Molecules and Polyatomic Ions

5.4 Energy and Properties of Matter

5.2 The Polarity of Covalent Molecules

5.5 Changes of State

5.3 Intermolecular Forces

5.6 Energetics: Changes of State vs. Speciﬁc Heat

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Use VSEPR theory to predict the geometries of molecules and polyatomic ions. (Section 5.1; Exercises 5.3 and 5.5)

Use electronegativities to classify the covalent bonds in molecules. (Section 5.2; Exercises 5.20 and 5.21)

Determine whether covalent molecules are polar or nonpolar. (Section 5.2; Exercises 5.22 and 5.23)

Identify the major attractive force between molecules in a given substance. (Section 5.3; Exercises 5.30 and 5.31)

Distinguish phenomena that demonstrate kinetic energy from those that demonstrate potential energy. (Section 5.4; Exercise 5.42)

Identify the states of matter using ﬁve properties: density, shape, compressibility, particle interaction, and molecular movement. (Section 5.4; Exercise 5.41)

Classify changes of state as exothermic or endothermic. (Section 5.5; Exercises 5.45 and 5.47)

Use the factors that aﬀect evaporation and condensation to rank substances in order of increasing vapor pressure. (Section 5.5; Exercise 5.53 and 5.55)

Use the factors that aﬀect boiling and melting to rank substances in order of increasing boiling point and melting point. (Section 5.5; Exercises 5.59 and 5.62)

10. Calculate energy changes that accompany heating, cooling, or changing the state of a substance. (Section 5.6; Exercises 5.69 and 5.71)

LECTURE HINTS AND SUGGESTIONS 1.

Provide students Table 5.1 as a full-page handout when introducing VSEPR theory (see next page.)

Have the students make molecular models using balloons as a 3D visual aid. Stress the idea that the molecular shape is the result of electron pairs spreading out as far from each other as possible. The large magnitude of the heat of fusion of water can be demonstrated by taking equal masses of ice and warm water and stirring them together in a beaker. Show that the temperature produced is 0 °C rather than a temperature which is midway between that of the ice and water as the students might initially guess.

Many of the students, for whom this text is written, have a strong interest in health and medicine. To better understand speciﬁc heat, emphasize the real-world application of this concept with a discussion of therapeutic hypothermia (Example 5.11.)

Chapter 5

VSEPR Theory

Chapter 5

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 5.1 GEOMETRIES OF MOLECULAR AND POLYATOMIC IONS 5.1

a. CH4 (each H atom is bonded to the C atom) four bonding domains

b. SO2 (each O atom is bonded to the S atom) two bonding domains and one nonbonding domain

c. AlCl3 (each Cl atom is bonded to the Al atom) three bonding domains

5.2

a. NH3 (each H atom is bonded to the N atom) has three bonding domains and one nonbonding domains

b. BeCl2 (each Cl atom is bonded to the Be atom) has two bonding domains

c. ClCN (the Cl and N atoms are bonded to the C atom) has two bonding domains

Chapter 5 5.3

a. BF3 (each F atom is around the B atom) trigonal planar

b. PH3 (each H atom is around the P atom) trigonal pyramidal

c. SCl2 (each Cl atom is bonded to the S atom) bent

5.4

H 2 S (each H atom is bonded to the S atom) Lewis structure

VSEPR

PCl 3 (each Cl atom is bonded to the P atom) Lewis structure

3D Structure

VSEPR

OF2 (each F atom is bonded to the O atom) Lewis structure

3D Structure

VSEPR

Chapter 5 5.5

O 3 (the O atoms are bonded together, like beads on a string) Lewis structure

3D Structure

VSEPR

3D Structure

VSEPR

NO 2 − (each O is bonded to N) Lewis structure

VSEPR

SO 3 (each O atom is bonded to the S atom) Lewis structure

5.6

3D Structure

PH 3 (each H atom is bonded to the P atom) Lewis structure

VSEPR

SeO 2 (each O atom is bonded to the Se atom) Lewis structure

3D Structure

VSEPR

ClO 3 − (each O is bonded to Cl) Lewis structure

3D Structure

VSEPR

Chapter 5

CO 3 2 − (each O is bonded to C) Lewis structure

3D Structure

VSEPR trigonal planar with C in the middle

H 3 O + (each H is bonded to O. Note the positive charge; compare with NH 4 + ) Lewis structure

5.7

3D Structure

VSEPR

PO 3 3 − (each O is bonded to P) Lewis structure

VSEPR

NH 2 − (each H is bonded to N) Lewis structure

3D Structure

VSEPR

BeCl 4 2 − (each Cl is bonded to Be) Lewis structure

3D Structure

VSEPR

Chapter 5

ClO 4 − (each O is bonded to Cl) Lewis structure

5.8

Trigonal pyramidal

5.9

Trigonal planar

5.10

Bent

5.11

Trigonal planar

3D Structure

VSEPR

Chapter 5

SECTION 5.2 THE POLARITY OF COVALENT MOLECULES 5.12

a. O (EN = 3.5) and S (EN = 2.5) are in the same group, but O is above S, so O is more electronegative b. Electronegativity increases as one goes from left to right and top to bottom in the periodic table, which means F (EN = 4.0) is more electronegative than Si (EN = 2.1).

5.13

a. F and Cl, F is more electronegative b. C and N, N is more electronegative

5.14

a. N-F is more polar than N-C as F is more electronegative than C, which makes the bond more polar (ENC = 2.5, ENN = 3.0, ENF = 4.0). b. Si-Cl is more polar than Si-S because Cl is more electronegative than S, which makes the bond more polar (ENSi = 2.1, ENCl = 3.0, ENS = 2.5)

5.15

a. C-O or C-S, C-O is more polar b. Br-Cl or Br-F, Br-F is more polar

5.16

because N (EN = 3.0) is more electronegative than H (EN = 2.1)

because F (EN = 4.0) is more electronegative than Cl (EN = 3.0)

because F (EN = 4.0) is more electronegative than C (EN = 2.5)

5.17

a. b. c.

5.18

5.19

Chapter 5

5.20 a.

LiBr

Calculation 2.8 – 1.0 = 1.8

Classiﬁcation ionic

HCl

3.0 – 2.1 = 0.9

polar covalent

PH 3 (each H is bonded to P)

2.1 – 2.1 = 0.0

nonpolar covalent

SO 2 (each O is bonded to S)

3.5 – 2.5 = 1.0

polar covalent

CsF

4.0 – 0.7 = 3.3

ionic

Calculation

Classiﬁcation

5.21 a.

MgI 2 (each I is bonded to Mg)

2.5 – 1.2 = 1.4

polar covalent

NCl 3 (each Cl is bonded to N)

3.0 – 3.0 = 0.0

nonpolar covalent

H 2 S (each H is bonded to S)

2.5 – 2.1 = 0.4

polar covalent

d. e.

RbF SrO

4.0 – 0.8 = 3.2 3.5 – 1.0 = 2.5

ionic ionic

magnesium and chlorine

Calculation 3.0 – 1.2 = 1.8

Classiﬁcation polar covalent

5.22

5.23

5.24 a.

Chapter 5 b. c.

carbon and hydrogen phosphorus and hydrogen

2.5 – 2.1 = 0.4 2.1 – 2.1 = 0.0

nonpolar covalent nonpolar covalent

C and Br aluminum and chlorine nitrogen and oxygen

Calculation 2.8 – 2.5 = 0.3 3.0 – 1.5 = 1.5 3.5 – 3.0 = 0.5

Classification polar covalent polar covalent polar covalent

5.25 a. b. c.

5.26

5.27

5.28

a. SOCl2 (S atom in the middle) SOCl2 is polar

Chapter 5

b. NF3 (N atom in the middle) NF3 is polar

c. CH4 (C atom in the middle) CH4 is nonpolar

5.29

a. AlCl3 (Al atom in the middle) AlCl3 is nonpolar

b. OF2 (O atom in the middle) OF2 is polar

c. H2S (S atom in the middle) H2S is nonpolar

SECTION 5.3 INTERMOLECULAR FORCES 5.30

a. CF4 London dispersion forces b. CS2 London dispersion forces c. HCl Dipole-dipole

Chapter 5 5.31

a. CH3OH hydrogen bonding b. CH3CH2CH2CH3 London dispersion forces c. CH2Cl2 London dispersion forces

5.32

The alcohol has higher melting and boiling points than the ether. The forces that hold the alcohol molecules together must be stronger and harder to break than the forces that hold the ether molecules together.

5.33

CH4 is nonpolar because its symmetric tetrahedral shape cancels out the weak individual bond dipoles, while CH3Cl is polar due to the presence of a polar C-Cl bond that creates an imbalance in electron distribution in the tetrahedral molecular shape.

5.34

London dispersion forces < Dipole-dipole forces < Hydrogen bonding < Ion-dipole interactions.

5.35

Hydrogen bonding is the most important intermolecular force vitamin C experiences with water. In addition to hydrogen bonding, dipole-dipole interactions are present due to the polar functional groups as well as London dispersion forces.

5.36

A single water molecule can form a maximum of four hydrogen bonds. This is because each water molecule has two hydrogen atoms that can donate a hydrogen bond and two lone pairs of electrons that can accept a hydrogen bond.

5.37

In methylamine, the nitrogen atom has a lone pair of electrons available for hydrogen bonding. This lone pair can form a hydrogen bond with the hydrogen atom of another methylamine molecule. Since methylamine has two hydrogen atoms bonded to the nitrogen atom, it can potentially form two hydrogen bonds with neighboring methylamine molecules.

SECTION 5.4 ENERGY AND PROPERTIES OF MATTER 5.38

The molecules of a liquid possess kinetic and potential energy. The kinetic energy is not great enough to overcome the attractive forces between the molecules; therefore, the molecules are able to ﬂow together into the shape of the container, but the volume of the liquid remains constant.

Solid and liquids are composed of molecules with considerable attractive forces between the molecules. These attractive forces bring the molecules close together and cause them to be diﬃcult to compress further.

Gases are composed of molecules with high kinetic energy and low potential energy. The molecules are small compared to the amount of space occupied by the gas. The molecules strike each other and the walls of the container; however, all of the collisions are elastic and no net energy is lost from the system. Since the molecules are in constant motion, the same average number of gas molecules will strike the walls of the container at any given time and the elastic collisions result in uniform pressure on the walls of its container.

Chapter 5 5.39

The distance between gas molecules is much greater than between molecules of solids and liquids; hence, the mass of a volume of gas is much less than the mass of an equal volume of either a solid or liquid. Consequently, gases have low densities.

The diﬀerence between a solid and a liquid is that the molecules of a solid are organized, whereas the molecules of a liquid are random. The intermolecular distances are similar, and the amount of space occupied by a particular mass of the substance remains very much the same; hence, the densities of liquids and solids are very similar.

The average molecular speed increases as the temperature rises resulting in the tendency to overcome the intermolecular attractive forces; hence, the molecules tend to be further apart with an increase in temperature. Consequently, solids, liquids, and gases all expand when heated.

5.40

Liquids and gases are similar in their fluidity, ability to mix, and response to temperature and pressure changes. However, liquids have stronger intermolecular forces, a fixed volume, and are denser, while gases have weaker intermolecular forces, are more compressible, and have greater diffusion rates.

5.41

a. Phosgene is a gas that is shipped in a compressed steel cylinder. This is because gases have a low density and are easily compressed, making them ideal for shipping in cylinders. It is a colorless gas that is heavier than air. b. Benzocaine is a solid that has a high boiling point, low compressibility, and a density heavier than water. This means that it is difficult to turn into a gas, and it is not easily compressed.

5.42

Kinetic energy is the energy of motion. The particles in a solid have the lowest kinetic energy because the particles are moving the least in this phase of matter. The particles in a liquid have higher kinetic energy than a solid, as well as lower kinetic energy than a gas. The particles in a gas have the highest kinetic energy because the particles are moving the most in this phase of matter. The potential energy of the three phases of matter is easiest to compare during a phase transition. When a solid melts into a liquid, the temperature does not change, even though energy is added to the solid; therefore, both the liquid and the solid particles have the same average kinetic energy at the melting point and the liquid must have higher potential energy than the solid. Similarly, when a liquid evaporates into a gas, the temperature does not change, even though energy is added to the liquid; therefore, both the liquid and the gas particles have the same average kinetic energy at the melting point and the gas must have higher potential energy than the liquid.

SECTION 5.5 CHANGES OF STATE 5.43

a. b.

This state is characterized by the lowest density of the three. This state is characterized by an indeﬁnite shape and a high density. 13

gaseous liquid

Chapter 5

5.44

c. d.

In this state, disruptive forces prevail over cohesive forces. In this state, cohesive forces are most dominant.

a. b. c.

Temperature changes influence the volume of this state substantially. gaseous In this state, constituent particles are less free to move about than in other states. solid Pressure changes influence the volume of this state more than that of the other gaseous two states. This state is characterized by an indefinite shape and a low density. gaseous

(l) → (s); remove heat (s) → (g); add heat (l) → (g); add heat

gaseous solid

5.45

a. b. c.

Freezing Sublimation Vaporization

5.46

a. b. c.

Condensation Liquefaction Boiling

5.47

a. Frost appearing on a car window after a cold evening is an exothermic process. This is because the water vapor in the air condenses to form ice, releasing heat in the process. The heat is released because the molecules in the water vapor lose energy as they come closer together to form ice.

(g) → (l); remove heat (g) → (l); remove heat (l) → (g); add heat

exothermic endothermic endothermic exothermic exothermic endothermic

b. Coffee is freeze-dried before shipping is an endothermic process. This is because the water in the coffee is removed by sublimation, which requires energy. Sublimation is the process of going directly from a solid to a gas without going through the liquid state. The energy required for sublimation comes from the coffee, which cools down as the water is removed. 5.48

a. Rubbing alcohol on skin before an injection quickly disappears due to an endothermic process. Evaporation requires energy, and breaking intermolecular forces needs energy. The rubbing alcohol evaporates, cooling down the skin. b. Frost on food in the freezer is an exothermic process. Water vapor in the air condenses and turns into ice, releasing heat as molecules lose energy and come closer together.

5.49

a. A cold pack is used to reduce swelling on a muscle from an injury. This is an endothermic process because the cold pack absorbs heat from the surrounding tissue, which helps to reduce the swelling. b. A heat pack is used to relax a muscle cramp. This is an exothermic process. This is because the heat pack releases heat to the surrounding tissue, which helps to relax the muscle.

5.50

a. An ice cube left melts after being left out on the table. This is an endothermic process. This is because the ice cube absorbs heat from the surrounding environment, which melts it.

Chapter 5 b. Cellular metabolism breaks chemical bonds and generates heat. This is an exothermic process. This is because the breaking of chemical bonds releases energy, which is then used to power the cell's activities. 5.51

(lowest vapor pressure) heptane < hexane < pentane < butane (highest vapor pressure) All of the compounds are nonpolar liquid hydrocarbons; thus, they all experience the same type of intermolecular forces (dispersion forces) which increase as the molecular weight increases. Vapor pressure is low for substances with high intermolecular forces and vapor pressure is high for substances with low intermolecular forces. Therefore, the substance with the lowest vapor pressure will have the highest molecular weight and the substance with the lowest vapor pressure will have the lowest molecular weight.

5.52

Liquid water was heated to become steam (endothermic process), and then the steam was cooled to become liquid water again (exothermic process).

5.53

CH3Cl < CH2Cl2 < CCl4 < CH4

5.54

Methylene chloride is a volatile liquid. When it was sprayed in the mouth, the methylene chloride absorbed heat from the tissue and evaporated. The tissue became cold and was anesthetized.

5.55

Water has a higher vapor pressure than hydrogen sulﬁde. This is because water molecules have stronger intermolecular forces compared to hydrogen sulﬁde molecules, leading to a higher vapor pressure for water at the same temperature.

5.56

Water and ethylene glycol diﬀer in their boiling points. Water has a boiling point of 100°C, while ethylene glycol has a boiling point that is higher than 100°C. To determine the identity of the two boiling liquids, use a thermometer to measure the boiling point.

5.57

As pressure on a liquid increases, the boiling point also increases. The temperature inside an autoclave would be 121°C.

5.58

Water (H2O) has a lower vapor pressure than 1-propanol (CH3CHOHCH3). This is because the hydrogen bonding intermolecular forces in water are stronger than in 1-propanol, which makes it more difficult for the water molecules to escape from the liquid state and enter the gas state.

5.59

Octane (CH3CH2CH2CH2CH2CH2CH2CH3) will have a higher boiling point than pentane (CH3CH2CH2CH2CH3). This is because octane has a longer carbon chain than pentane, which means that the octane molecules have stronger London dispersion forces. Stronger intermolecular forces make it more difficult for the octane molecules to escape from the liquid state and enter the gas state, which means that octane has a higher boiling point.

5.60

C5H12 < C6H14 < C8H18 < C10H22 This is because the boiling point increases as the number of carbon atoms in the molecule increases. This is because the longer carbon chain results in stronger London dispersion forces, which leads to a higher boiling point.

Chapter 5 5.61

CF4 < CCl4 < CBr4 < CI4 The boiling point of substances is determined by intermolecular forces. CF4, CCl4, CBr4, and CI4 are all nonpolar, but CCl4, CBr4, and CI4 have more electrons and stronger London dispersion forces. CI4 has the most electrons and the strongest dispersion forces among the four compounds.

5.62

Mg(OH)2 will have a higher melting point than palmitic acid CH3(CH2)14COOH because Mg(OH)2 is an ionic compound, while palmitic acid is a molecular compound. Ionic compounds have stronger intermolecular forces than molecular compounds and therefore have higher melting points.

5.63

Calcium carbonate (CaCO3) will have the higher melting point than cetyl palmitate (C32H64O2) because CaCO3 is an ionic compound, while cetyl palmitate is a molecular compound. Ionic compounds have stronger intermolecular forces than molecular compounds, and therefore have higher melting points.

5.64

To obtain pure solid iodine from a mixture of solid iodine and sand, heat the mixture until the iodine sublimes and provide a cold surface above the mixture so the iodine can be deposited as a solid. The sand will not sublime and the solid on the cold surface will be pure iodine.

5.65

Dry ice (solid CO2) will sublime gaseous CO2 without leaving any liquid behind, keeping the samples dry. Water ice will melt to leave behind liquid water, which may destroy the samples.

5.66

When the ice and water mixture reached a constant temperature of 0.0°C, the system shared not only the same temperature throughout, but also the same vapor pressure. The vapor pressure of both the water and the ice is 4.58 torr.

SECTION 5.6 ENERGETICS: CHANGES OF STATE VS SPECIFIC HEAT 5.67

a. 1 kilocalorie is equal to 4.184 kJ. So, a protein bar that produces 280 kilocalories when burned produces 280 x 4.184 = 1171.5 kJ. b. Two apples that produce 144 kilocalories when burned produce 144 x 4.184 = 602.5 kJ.

5.68

a. An IV of dextrose that produces 170 kilocalories when burned produces 170 x 4.184 kJ/kcal = 711.3 kJ = 710 jK. b. One packet of electrolyte powder that produces 25 kilocalories when burned produces 25 x 4.184 = 104.6 kJ = 1.0 x 102 kJ.

5.69

Heat needed = 2.5 g × 80 cal/g = 200 cal

5.70

Heat needed = 66 g x 540 cal/g = 35640 cal = 36 kcal

Chapter 5 5.71

210. g of copper from 40°C to 95°C

heat = (210. g)(0.093

cal )(95 C − 40 C)  g C

= 1.1 x 10 3 cal b.

150. g of mercury from 120°C to 300°C

heat = (150. g)(0.033

cal )(300 C − 120 C) g C

= 8.9 x 10 2 cal c.

2.50 x 103 g of helium gas from 250°C to 900°C

heat = (2.50 x 10 3 g)(1.25

cal )(900 C − 250 C) g C

= 2.0 x 10 6 cal 5.72

50. g of aluminum from 25°C to 55°C

heat = (50. g)(0.24

cal )(55 C − 25 C)  g C

= 3.6 x 10 2 cal b.

2.50 x 103 g of ethylene glycol from 80°C to 85°C

heat = (2.50 x 10 3 g)(0.57

cal )(85 C − 80 C) g C

= 7 x 10 3 cal c.

500. g of steam from 110°C to 120°C

heat = (500. g)(0.48

cal )(120 C − 110 C) g C

= 2.4 x 10 3 cal

5.73

CaCl 2 6H 2 O: melting point = 30.2 C ‫ ﻠ‬heat of fusion = 40.7 cal/g

 1000 g   40.7 cal  heat = (1000. kg)    ‫ككككككككككككككككككك‬ g  1 kg    = 4.07 x 107 cal

b.‫ك‬LiNO 3 3H 2 O: melting point = 29.9 C, heat of fusion = 70.7 cal/g  1000 g   70.7 cal  heat = (1000. kg)    ‫ككككككككككككككككككك‬ g  1 kg    = 7.07 x 107 cal

Chapter 5

c. Na 2 SO 4 10H 2 O: melting point = 32.4 C, heat of fusion – 57.1 cal/g  1000 g   57.1 cal  heat = (1000. kg)    ‫ككككككككككككككككككك‬ g  1 kg    7 = 5.71 x 10 cal

5.74

In a solar heat storage system, solids are melted, and the energy is stored in the liquid form until the heat is released upon re-solidiﬁcation. The melting point of K2SO4 is 1069˚C which is much too high to be useful. The melting temperature must be low enough to easily maintain the compound in the liquid state for storage on cloudy days.

5.75

2550. cal + 40000. cal + 50000. cal + 270000. cal + 4800. cal = 367350. cal ⇒ 3.67 x 105 cal

5.76

 1000 g   38.6 cal  4 (2.00 kg)    = 7.72 x 10 cal 1 kg 1 g   

ADDITIONAL EXERCISES 5.77

As the sevoflurane is heated, the kinetic energy of the molecules increases, thus increasing the disruptive forces. At the boiling point, the disruptive forces overcome the cohesive forces, and the molecules become gaseous, escaping the liquid.

5.78

If one atom of oxygen reacted with two atoms of nitrogen to form a molecule, the formula of the molecule would be N 2 O . The electronegativity difference between nitrogen and oxygen is 0.5; therefore, the bond between nitrogen and oxygen is covalent. The name of N 2 O is dinitrogen monoxide.

5.79

Boiling water on the summit of Mount Everest might not make it safe to drink because the boiling point of water on the summit of Mount Everest is only 76.5°C. This might not be a high enough temperature to sterilize the water.

Chapter 5 5.80

The heat from the burner does not increase the temperature of the water-containing cup to the ignition temperature because the cup is the same temperature as the water. The energy from the burner must ﬁrst heat the water to the boiling point and completely boil the water to dryness before the cup reaches a temperature above the boiling point of water. If a sample of water were heated to the boiling point in a glass beaker using a single burner and then a second burner were added, the temperature of the boiling water would stay the same. The boiling point of water is not dependent on the amount of energy added to the water. The time required to reach the boiling point of water may decrease when using two burners rather than just one.

5.81

If four 250-mL bottles of water were placed in an ice chest ﬁlled with crushed ice, 2.1 x 102 g of ice would be needed to cool the bottles of water form 23°C to 5°C, as shown by the calculation below:

 250. mL   1 g   1 cal   4 bottles  − 5 C)     (23 C =   1 bottle  1 mL   g ⋅ C 

 1 cal    80. cal   (x)   + (x)    (5 C − 0. C) g g C ⋅    

 85. cal    g 

‫كككككككككككككككككككككككككككككككككككككككككككككككككككك‬18000. cal = (x) 

‫ككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككك‬ ‫ك‬ ‫كككك‬x = 211.76 g ⇒ 2.1 x 10 2 g ice 5.82

Lead has the smallest speciﬁc heat value, which indicates a greater change in temperature when equal amounts of energy are added to metals at the same mass.

5.83

If 43.4 kJ of heat is released when all ice melts at 0°C, 43.4 kJ = (mass of water ) x 334 J/g is rearranged to

5.84

The heat required = 0.0688 kg x 3.32 kJ/kg•˚C x (183˚C-175˚C) = 1.83 kJ

5.85

The heat released from 1 mg of protein can be calculated by the following:

Chapter 6

Chapter 6: Gases and Solutions CHAPTER OUTLINE 6.1 Properties of Gases

6.5 Solutions and Solubility

6.2 Pressure, Temperature and Volume Relationships

6.6 Electrolytes and Net Ionic Equations

6.3 The Ideal Gas Law

6.8 Solution Preparation

6.4 Dalton’s Law

6.9 Osmosis and Dialysis

6.7 Solution Concentrations

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Calculate the kinetic energy of moving particles. (Section 6.1; Exercise 6.2)

Convert pressure and temperature values into various units. (Section 6.1; Exercise 6.5)

Do calculations based on Boyle’s law, Charles’s law, and the combined gas law. (Section 6.2; Exercises 6.11, 6.13, and 6.15)

Do calculations based on the ideal gas law. (Section 6.3; Exercises 6.31, 6.33, and 6.35)

Do calculations based on Dalton’s law. (Section 6.4; Exercises 6.44 and 6.45)

Apply your knowledge of intermolecular forces to determine the solubility of substances in a given solvent. (Section 6.5; Exercises 6.49)

Write molecular equations in total ionic and net ionic forms. (Section 6.6; Exercises 6.61 and 6.62)

Calculate solution concentrations in units of molarity, weight/weight percent, weight/volume percent, and volume/volume percent. (Section 6.7; Exercises 6.65, 6.73, and 6.79)

Describe how to prepare solutions of speciﬁc concentration by diluting concentrated solutions. (Section 6.8; Exercises 6.91 and 6.93)

10. Describe the process of osmosis. (Section 6.9; Exercise 6.95) 11. Describe the role of osmosis and diﬀusion in dialysis. (Section 6.9; Exercises 6.98)

LECTURE HINTS AND SUGGESTIONS 1.

The individual gas laws are important from a historical point of view, however, the ideal gas law PV=nRT can be used to solve most problems, thus your emphasis should be on this equation more than the individual laws. Moreover the major focus on the gas laws should be relating the kinetic molecular theory to the physical behavior of gases (how the laws relate to common and everyday phenomenon)

Combining Table 6.2 and Study Tools 6.1 together on a one-page handout (see next page) creates a

Treat Avogadro's volume, 22.4 L/mol at STP, as another conversion factor that is always available for

useful resource for students to reference when solving gas law problems. use when using dimensional analysis to solve problems. 4.

The solubility of sugar, cooking oil and rubbing alcohol in water can be demonstrated. The diﬀerence between the terms soluble, insoluble and miscible should be illustrated. Explain the general statement "like dissolves like" and explain in terms of chemical structure and the attractive forces between molecules. 1

Chapter 6

Units of Pressure and Gas Laws

Chapter 6

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

SECTION 6.1 PROPERTIES OF GASES 6.1

The kinetic energy of the halfback is greater than the kinetic energy of the tackle as shown in the calculations below. The tackle will be pushed back because he has less kinetic energy.

KE =

1 mv 2 2 2

 1 m 3 KE halfback = ( 81.8 kg )  8.0  = 2.6 x 10 2 s   2

1  m 2 KE tackle = ( 118.2 kg )  3.0  = 5.3 x 10 2 s  

6.2

N 2 O kinetic energy = 3.14 x 10 KE =

kg  m 2 s2

g cm 2 10 s2

1 44 u  4 cm  mv 2 =  3.78 x 10  2 2  s 

Sevoflurane kinetic energy = 3.13 x 10 KE =

kg  m 2

g cm 2 10 s2

1 200.1 u 44 u  4 cm  mv 2 =  1.77 x 10  2 2 s  

6.3

A gas law is a mathematical relationship that describes the behavior of gases as they are mixed, as they are subjected to pressure or temperature changes, or allowed to diﬀuse.

6.4

atm

 1atm  28.6 in. Hg   = 0.957 atm  29.9 in. Hg 

torr

 1atm   760 torr  28.6 in. Hg    = 727 torr 29.9 in. Hg    1 atm 

psi

 1atm   14.7 psi  28.6 in. Hg    = 14.1 psi  29.9 in. Hg   1 atm 

bars

 1.01bars  28.6 in. Hg   = 0.966 bars  29.9 in. Hg 

Chapter 6

6.5

6.6

6.7

atm

 1atm  210. psi   = 14.3 atm  14.7 psi 

bars

 1atm   1.01 bars  210. psi    = 14.4 bars  14.7 psi   1 atm 

mm Hg

 1atm  760 mm Hg  4 210. psi    = 1.09 x 10 mm Hg   14.7 psi   1 atm

in. Hg

 1atm  29.9 in. Hg  210. psi    = 427 in. Hg  14.7 psi   1 atm 

The kinetic energy of the hydrogen will be greater than the kinetic energy of the helium as shown in the calculations below.

1 mv 2 2 1 KEH2 = (2.02 u)(2v)2 = 4.04 v 2 u 2 1 KEHe = (4.00 u)(v)2 = 2.00 v 2 u 2 KE =

6.8

6.9

K = 273 +  C = 273 + (-268.9  C) = 4.1 K



K = 273 +  C - 273 + 63.7  C = 336.7 K

The melting point of gold, 1337.4K, to degrees in Celsius

1337.4 K – 273 = 1064°C

The melting point of tungsten, 3410°C to kelvins.

3410°C + 273 = 3683 K

The melting point of tin, 505 K, to degrees Celsius.

505 K -273 = 232°C

C = K - 273 = 14.1 K - 273 = 258.9  C

Chapter 6

SECTION 6.2 PRESSURE, TEMPERATURE, AND VOLUME RELATIONSHIPS 6.10

Gas A:

(1.50 atm)(2.00 L) (Pf )(3.00 L) = ; 300. K 450. K

Gas B:

(2.35 atm)(1.97 L) (1.09 atm)(Pf ) = 293. K 310. K

Pf = 1.50 atm Gas C:

Vf = 4.49 L

(9.86 atm)(11.7 L) (5.14 atm)(9.90 L) = 500. K Tf

Tf = 221 K

6.11

(690. torr)(200. mL) (760 torr)(Vf ) ⇒ Vf = 166 mL = 26.0 C + 273 0.0 C + 273

6.12

(610. torr)(200. mL) (760 torr)(Vf ) = ⇒ Vf = 138 mL 45.0 C + 273 0.0 C + 273

6.13

(Pf )(0.50 L) (1.00 atm)(3.00 L) = ⇒ Pf = 7.1 atm  0.0 C + 273 30.0 C + 273

6.14

(Pf )(0.75 L) (1.00 atm)(2.50 L) = ⇒ Pf = 3.70 atm  0.0 C + 273 30.0 C + 273

6.15

(14.7 psi) (1.00 atm) = Vi = 1 atm

(65.0 psi)(1.00 L) ⇒ Vi = 4.42 L

6.16

(14.7 psi) (1.00 atm) = Vi = 1 atm

(32.0 psi)(14.5 L) ⇒ Vi = 31.6 L

6.17

Constant Temp →

V2 =

6.18

760 torr (1000. mL) 620 torr

Constant Temp →

V2 =

P1 V1

P1 V1 T1

2000 psi (5.0 L) 14.0 psi

P2 V2 T2

→ 760 torr (1000. mL) = 620 torr (V2 )

= 1.23 x 10 3 mL

P2 V2 T2

→ 2000 psi (5.0 L) = 14.0 psi (V2 )

= 714 L

Chapter 6

6.19

Constant Temp →

T1 = 4 C = 277 K 

P1 V1 T1

P2 V2 T2

T2 = 37 C = 310 K 

300 mL x mL 300 mL (310 K) = → = 336 mL 277 K 310 K 277 K

6.20

Vf 3.8 L = ⇒ Vf 4.6 L   20. C + 273 85 C + 273

6.21

If the gas pressure remains constant, the equation

P1 V1 P2 V2 V V = reduces to 1 = 2 . T1 T2 T1 T2

Remember the T must be in Kelvins. K = °C = 273 K = 16°C + 273 = 289 K for T1 K = 37°C + 273 = 310 K for T2

V1 = 1.00 L; V2 = ?

V1 V2 V2 100 = = = T1 T2 289 K 310 K Cross multiplying gives (1.00 L)(310 K) = (289 K)(V2) Dividing both sides by 289 gives V2 = 1.07 L

6.22

V1 2.0 L = ⇒ V1 = 2.5 L  120. C + 273 40. C + 273

6.23

P1 V1 P2 V2 = T1 T2

T1 = 20 C = 293 K

T2 = 37  C = 310 K

2500 torr (180 mL) 94 torr (x mL) 2500 torr (180 mL)(310 K) = 506 mL = → 293 K 310 K 940 torr (293) K

6.24

P1 V1 P2 V2 = T1 T2

T1 = 37  C = 310 K

T2 = 22 C = 295 K

210 K Pa (x mL) 94.6 K Pa (7.64 mL) = 310 K 295 K 94.6 K Pa (7.64 mL)(310 K) = 3.62 mL 210.0 K Pa (295 K)

Chapter 6

6.25

P1 V1 P2 V2 = T1 T2

T1 = 0.0 C = 283 K

T2 = 37  C = 310 K

3.25 atm (6.25 L) 250 atm (x L) = 283 K 310 K 3.25 atm (6.25 L)(310K) = 89.0 L 0.250 atm + (283 K)

 1 atm  (400. torr)  (V ) 760 torr  f = ⇒ Vf 1.4 x 10 4 ft 3 5 C + 273

6.26

(0.98 atm) (8000. ft 3 ) = 23 C + 273

6.27

(1.50 L) 2(1.50 L) Tf 333 C = ⇒ =  30 C + 273 Tf + 273

6.28

P1 V1 P2 V2 at constant temperature gives P1 V1 = P2 V2 = T1 T2

1.00 atm (4.00 L) = 3.00 atm (x L) 1.00 atm (4.00 L) = 1.33 L 3.00 atm 6.29

 1000 mL  3 (Pi )(250. mL) = (700. torr)(2.00 L)   ⇒ Pi = 5.60 x 10 torr 1 L  

6.30

 760 torr  (760 torr)(2.00 L) = (4.00 atm)   (Vf ) ⇒ Vf = 0.500 L  1 atm 

density =

g 2.50 g mass ⇒ = 5.00 L volume 0.500 L

SECTION 6.3 THE IDEAL GAS LAW 6.31

 1L  L ⋅ atm PV = nRT ⇒ P ⇒ (400. mL)  )((20.0 + 273) K)  = (2.00 mol) (0.0821 ⋅K 1000 mL mol   L ⋅ atm (2.00 mol)(0.0821 )((20.0 + 273) K) ⋅K mol = 120. atm V= (400. mL)

PV = nRT ⇒ (3.00 atm) V= (0.525 mol)(0.0821 V=

L ⋅ atm )((20.0 + 273) K) mol ⋅ K

L ⋅ atm )((15.0 + 273) K) mol ⋅ K = 4.14 L (3.00 atm)

(0.525 mol)(0.0821

Chapter 6

L ⋅ atm  1 atm  PV = nRT ⇒ (300. torr)  )T  (2.25 L) = (0.100 mol)(0.0821 ⋅K 760 torr mol    1 atm  (300. torr)  (2.25 L) torr   760 = T= 108 K - 273 = − 165 C L ⋅ atm (0.100 mol)(0.0821 ) mol ⋅ K

6.32

6.33

PV = nRT ⇒ P(0.750 L) = (0.156 mol)(0.0821 P=

L ⋅ atm )((27 + 273)K) mol ⋅ K

L ⋅ atm )((27 + 273)K) mol ⋅ K = 5.12 atm (0.750 L)

(0.156 mol) (0.0821

6.34

6.35

PV = nRT ⇒ P(0.890 atm) V = (8.75 g)( (8.75 g)( V=

1 mole L ⋅ atm )(0.0821 )((35.0 + 273)K) 32.00 g O 2 mol ⋅ K

1 mole L ⋅ atm )(0.0821 )((35.0 + 273)K) 32.00 g O 2 mol ⋅ K = 7.77 L (0.890 atm)