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CHAPTER 1

AN INTRODUCTION TO CHEMISTRY SOLUTIONS TO REVIEW QUESTIONS 1. (a)

(b)

A hypothesis is a tentative explanation of certain facts to provide a basis for further experimentation. A theory is an explanation of the general principles of certain phenomena with considerable evidence to support it. A theory is an explanation of the general principles of certain phenomena with considerable evidence to support it. A scientific law is a simple statement of natural phenomena to which no exceptions are known under the given conditions.

2. (a)

hypothesis

(b)

hypothesis

(c)

observation

(d)

theory

(e)

observation

(f)

scientific law

3. (a)

A liquid has a definite volume but not a definite shape.

(b)

A gas has an indefinite volume and high compressibility.

(c)

A solid has a definite shape.

(d)

A liquid has an indefinite shape and slight compressibility.

4. A crystalline solid has a regular, repeating, three-dimensional, geometric pattern. An amorphous solid does not. (a)

A solid that has a regular, repeating pattern is a crystalline solid.

(b)

A plastic solid is amorphous.

(c)

A solid that has no regular repeating pattern is amorphous.

(d)

Glass is an amorphous solid.

(e)

Gold is a crystalline solid.

5. A phase is a homogeneous part of a system separated from other parts by a physical boundary. 6. There are six phases present. 7. Another name for a homogeneous mixture is solution. 8. Alcohol, mercury, and water are the only liquids in the table which are not mixtures. Mercury is an element; alcohol and water are compounds.

– Chapter 1 –

9. Air is the only gas mixture found in the table. The other gases are elements or compounds. 10. Three phases are present within the bottle; solid and liquid are observed visually, while gas is detected by the immediate odor. 11. The system is heterogeneous as three phases are present. 12. A system containing only one substance is not necessarily homogeneous. Two phases may be present. Example: ice in water. 13. A system containing two or more substances is not necessarily heterogeneous. In a solution only one phase is present. Examples: sugar dissolved in water, dilute sulfuric acid. 14. Homogeneous mixtures contain only one phase, while heterogeneous mixtures contain two or more phases. 15. (a) sugar, a compound and (c) gold, an element 16. Using the steps of the scientific method to help determine why your cell phone has stopped working. (a)

Observation:

My cell phone has stopped working.

(b)

Hypothesis:

I think that the battery needs to be recharged.

(c)

Experiment:

Plug in the phone to recharge the battery and allow sufficient time for the battery to fully recharge. Turn the phone back on. The phone now works again.

(d)

Theory:

The battery in the phone has a limited charge time and needs to be recharged on a regular basis in order to keep it in working order.

– Chapter 1 – SOLUTIONS TO EXERCISES 1. Two states are present; solid and gas. 2. Two states are present; solid and liquid. 3. The photo represents a heterogeneous mixture. 4. The maple leaf represents a heterogeneous mixture. 5. (a)

homogeneous

(b)

heterogeneous

(c)

heterogeneous

(d)

heterogeneous

6. (a)

homogeneous

(b)

homogeneous

(c)

heterogeneous

(d)

heterogeneous

7. Typical answers could be Substance chocolate syrup margarine nondairy creamer beef bouillon toothpaste antibacterial soap first aid spray sunblock lotion

Main or Active Ingredient high fructose corn syrup vegetable oil blend corn syrup solids salt sodium fluoride triclocarban lidocaine HCl ethylhexyl p-methoxycinnamate

8. The steps of the scientific method can be used to predict the outcome of the semester in the following way: (a)

Collect the facts or data. They include the number of classes you have enrolled in, the amount of time you will give to each class, the number of hours required for your off campus job, and the amount of time for social activities.

(b)

Formulate a hypotheses. You predict that the amount of time you have allocated each week for class work will be sufficient to result in good grades at the end of the semester.

(c)

Plan and do additional experiments to test the hypothesis. In the first two weeks of the semester keep a record of your performance in each class, including grades on homework, quizzes, and exams.

(d)

Modify the hypothesis. If your grades in any class are not good, your hypothesis was incorrect. It will be necessary to increase the amount of time allocated to that class.

– Chapter 1 – 9. (a)

water, a pure substance

(b)

chicken stock, a mixture

(c)

salt, a pure substance

(d)

mustard flour, a mixture

10. Hypothesis – The soup is not a good source of nutrition because it contains 12 grams of fat per serving. Test – Search for the number of grams of fat per day recommended for an adult. Compare this number to the grams per fat on the label. 11. Observations – Calcite crystals have the ability to reflect light around an object rendering it “unseeable”. “Invisibility cloaks” work only with laser light aimed directly at the crystal. Hypotheses – If larger crystals were used, larger objects could be hidden. These “invisibility cloaks” will improve in the future. 12. (a), (b), Picture (2) best represents a homogeneous mixture. Pictures (1) and (c) (3) show heterogeneous mixtures, and picture (4) does not show a mixture, as only one species is present. Picture (1) likely shows a compound, as one of the components of the mixture is made up of more than one type of “ball”. Picture (2) shows a component with more than one part, but the parts seem identical, and therefore it could be representing a diatomic molecule. 13. (a) two phases, solid and gas (b) two phases, liquid and gas (c)

two phases, solid and liquid

CHAPTER 2

STANDARDS FOR MEASUREMENT SOLUTIONS TO REVIEW QUESTIONS 1.

The exponent will be positive for a large number and negative for a small number.

The exponent will decrease.

The last digit in a measurement is uncertain because if the quantity were to be measured multiple times, the last digit would vary.

It must be written in scientific notation as 6.420 × 105 g.

Zeroes are significant when they are between nonzero digits or at the end of a number that includes a decimal point.

Rule 1. When the first digit after those you want to retain is 4 or less, that digit and all others to its right are dropped. The last digit retained is not changed. Rule 2. When the first digit after those you want to retain is 5 or greater, that digit and all others to the right of it are dropped and the last digit retained is increased by one.

No, the number of significant digits in the calculated value may not be more than the number of significant figures in any of the measurements.

Yes, the number of significant digits depends on the precision of each of the individual measurements and the calculated value may have more or fewer significant figures than the original measurements as long as the precision is no greater than the measurement with the lowest precision. An example of a calculation with an increase in significant figures is in Example 2.9.

100 cm = 1 m

1,000,000,000 nm = 1 m

 1 m   1,000,000,000 nm  (1 cm)    = 10,000,000 nm m   100 cm  

10. 1000 mg = 1 g

1000 g = 1 kg

 1000 g   1000 mg    = 1,000,000 mg g  kg   

(1 kg ) 

11. Weight is a measure of how much attraction the earth’s gravity has for an object (or person). In this case, the farther the astronaut is from the Earth the less gravitational force is pulling on him or her. Less gravitational attraction means the astronaut will weigh less. The mass of the astronaut is the amount of matter that makes up him or her. This does not change as the astronaut moves away from the Earth. 12. They are equivalent units. 13.

(3.5 in.) 

2.54 cm   = 8.9 cm  1 in. 

14. Heat is a form of energy, while temperature is a measure of the intensity of heat (how hot the system is). 1

– Chapter 2 –

15. The number of degrees between the freezing and boiling point of water are Fahrenheit

180°F

Celsius

100°C

Kelvin

100 K

16. The three materials would sort out according to their densities with the most dense (mercury) at the bottom and the least dense (glycerin) at the top. In the cylinder, the solid magnesium would sink in the glycerin and float on the liquid mercury. 17. Order of increasing density: ethyl alcohol, vegetable oil, salt, lead 18. The density of ice must be less than 0.91 g/mL and greater than 0.789 g/mL. 19. The density of water is 1.0 g/mL at approximately 4°C. However, when water changes from a liquid to a solid at 0°C there is actually an increase in volume. The density of ice at 0°C is 0.917 g/mL. Therefore, ice floats in water because solid water is less dense than liquid water. 20. If you collect a container of oxygen gas, you should store it with the mouth up. Oxygen gas is denser than air. 21. density of gold = 19.3 g/mL

density of silver = 10.5 g/mL

 1 mL   = 1.3 mL  19.3 g 

( 25 g gold ) 

25 g of silver has the greater volume.  1 mL   = 2.4 mL  10.5 g 

( 25 g silver ) 

– Chapter 2 – SOLUTIONS TO EXERCISES

(a) kilogram = 1000 grams (b) centimetre = 1/100 of a meter (0.01 m) (c) microliter = 1/1,000,000 of a liter (0.000001 L) (d) millimetre = 1/1000 of a meter (0.001 m) (e) decilitre = 1/10 of a liter (0.1 L)

(a) 1000 meters = 1 kilometer

(d)

0.01 meter = 1 centimeter

(b) 0.1 gram = 1 decigram

(e)

0.001 liter = 1 milliliter

(a) gram = g

(d)

micrometer = μm

(b) microgram = μg

(e)

millilitre = mL

(f)

decilitre = dL

(a) milligram = mg

(e)

angstrom= Å

(b) kilogram = kg

(f)

microlitre = μL

(a) 2050

the first zero is significant; the last zero is not significant

(b) 9.00 × 102

zeros are significant

the first two zeros are not significant; the last zero is significant

(d) 0.075

zeros are not significant

(e) 300.

zeros are significant

(f) 285.00

zeros are significant

(a) 0.005

zeros are not significant

(b) 1500

zeros are not significant

zero is significant

(d) 10.000

zeros are significant

(e) 6.070 × 104

zeros are significant

(f) 0.2300

the first zero is not significant; the last two zeros are significant 3

– Chapter 2 –

Significant figures (a) 0.025

(2 sig. fig.)

(3 sig. fig.)

(b) 22.4

(3 sig. fig.)

(d) 5.50 × 103

(3 sig. fig.)

Significant figures (a) 40.0 (b) 0.081

(6 sig. fig.) −3

(2 sig. fig.)

(d) 4.090 × 10

Round each of the following numbers to four significant figures: (a) 47.70

(c)

100.2

(b) 1.026

(d)

16.50

10. Round each of the following numbers to four significant figures: (a) 0.07794

(c)

5.005

(b) 21.63

(d)

74.29

(a) 6.7 × 104

(c)

7.8 × 10−3

(b) 6.54 × 10−2

(d)

4.11 × 105

(a) 4.56 × 10−2

(c)

4.030 × 101

(b) 4.0822 × 103

(d)

1.2 × 107

11. Exponential notation

12. Exponential notation

13. (a)

12.62 1.5 0.25 14.37 = 14.4

(b) (2.25 × 103)(4.80 × 104) = 10.8 × 107 = 1.08 × 108 (c)

( 452 )( 6.2 ) = 195.97 = 2.0×102 14.3

(d) (0.0394) (12.8) = 0.504 (e)

0.4278 = 0.00718 = 7.18×10 −3 59.6

(f) 10.4 + (3.75)(1.5 × 104) = 5.6 × 104

(4 sig. fig.)

– Chapter 2 –

14. (a)

15.2 −2.75 15.67 28.1

(b) (4.68)(12.5) = 58.5 (c)

182.6 = 40. 4.6

1986 (d)

23.84 0.012 2009.852 = 2010. = 2.010 × 103

(e)

29.3 = 2.49 ×10 −4 ( 284 )( 415 )

(f) (2.92 × 10−3)(6.14 × 105) = 1.79 × 103 15. Fractions to decimals (3 significant figures) (a)

5 = 0.833 6

(c)

12 = 0.750 16

(b)

3 = 0.429 7

(d)

9 = 0.500 18

(c)

1.67 = 1

2 5 or 3 3

(d)

0.888 =

8 9

(c)

0.525 = 0.25 x

16. Decimals to fractions

17.

1 4

(a)

0.25 =

(b)

0.625 =

(a)

3.42x = 6.5 6.5 = 1.9 x= 3.42

(b)

5 8

x = 7.05 12.3 x = ( 7.05 ) (12.3 ) = 86.7

0.525 = 0.25x 0.525 x= = 2.1 0.25

– Chapter 2 –

18.

(a)

212 − 32 1.8

72 = 1.8x + 32

(c)

72 − 32 = 1.8x 40. = 1.8x 40. =x 1.8 22 = x

x = 1.0 × 10 2

(b) 8.9

g 40.90 g = mL x

g    8.9  x = 40.90 g mL   40.90 g = 4.6 mL x= g 8.9 mL

19. (a)

( 28.0 cm ) 

1m   = 0.280 m  100 cm 

(b)

(1000. m ) 

(c)

( 9.28 cm ) 

(d)

(10.68 g ) 

(e)

1 g   1 kg  ( 6.8 ×10 mg )  1000   = 6.8×10 kg mg 1000 g

(f)

(8.54 g ) 

(g)

( 25.0 mL ) 

(h)

( 22.4 L ) 

20. (a)

( 4.5 cm ) 

(b)

(12 nm ) 

1 km   = 1.000 km  1000 m  10 mm   = 92.8 mm  1 cm 

 1000 mg  4  = 1.068 × 10 mg  1g  −2







 1 kg   = 0.0854 kg  1000 g 

1L  −2  = 2.50 × 10 L 1000 mL  

 106 μL  7  = 2.24 × 10 μL 1 L   1 m  1 Å  8   −10  = 4.5 × 10 Å  100 cm   10 m 

 10 −9 m   100 cm  −6   = 1.2 × 10 cm  1 nm   1 m  6

– Chapter 2 –

(c)

(8.0 km ) 

1000 m   1000 mm  6   = 8.0 × 10 mm 1 km 1 m   

(d)

(164 mg ) 

(e)

( 0.65 kg ) 

(f)

(5.5 kg ) 

(g)

( 0.468 L ) 

(h)

( 9.0 μL ) 

21. (a)

( 42.2 in.) 

(b)

( 0.64 mi ) 

(c)

cm  2 ( 2.00 in.2 )  2.54  = 12.9 cm 1 in. 

(d)

( 42.8 kg ) 

(e)

(3.5 qt ) 

(f)

( 20.0 L ) 



1g   = 0.164 g  1000 mg   1000 g   1000 mg  5   = 6.5 × 10 mg  1 kg   1 g 

 1000 g  3  = 5.5 ×10 g  1 kg  1000 mL   = 468 mL  1L 

 1 L   1000 mL  −3  = 9.0 ×10 mL  6 μ 10 L 1 L    

2.54 cm   = 107 cm  1 in.  5280 ft   12 in.  4   = 4.1×10 in. 1 mi 1 ft    2

 2.205 lb   = 94.4 lb  kg 

 946 mL  3  = 3.3 × 10 mL  1 qt  1 qt   1 gal   = 5.29 gal   0.946 L   4 qt 

22. (a) The conversation is: m → cm → in. → ft

(35.6 m ) 

100 cm   1 in.   1 ft     = 117 ft  1 m   2.54 cm   12 in. 

(b)

(16.5 km ) 

1 mi   = 10.3 mi  1.609 km  3

 2.54 cm   10 mm  4 3 (c) ( 4.5 in. )     = 7.4 ×10 mm  1 in.   1 cm  3

(d)

( 95 lb ) 

453.6 g  4  = 4.3 ×10 g 1 lb  

– Chapter 2 –

 4 qt   0.946 L  (e) (20.0 gal)    = 75.7 L  1 gal   1 qt  (f)

The conversation is: ft 3 → in.3 → cm 3 → m3 3

cm   1 m  3 3 ( 4.5 ×10 ft )  121 ftin.   2.54    = 1.3 ×10 m 1 in.   100 cm  4

23. First convert 6 ft 6 in. into inches and then convert to centimeters and meters.

( 6 ft ) 

12 in.   + 6 in. = 78 in.  ft 

( 78 in.) 

2.54 cm  2  = ( 2.0 × 10 cm ) in.  

( 2.0 × 10 cm )  1001 mcm  = 2.0 m ( 2 significant figures) 2

24. We must convert ounces to mL.

( 46.0 oz ) 

29.6 mL  3  = 1.36 ×10 mL ( 3 significant figures )  1 oz 

25. The conversion is: L → mL → mg → g

1000 mL  500. mg   1 g   = 5.00 g    1 L  100. mL   1000 mg 

(1 L ) 

26. The conversion is: tablet → g → mg → grains

0.500 g   1000 mg  1 grain    = 8.33 grains   1 tablet   1 g  60 mg 

(1 tablet ) 

27. The conversion is: hr → min → s → m → km → mi

(5 hours ) 

60 min   60 s   0.11 m   1 km   1 mi       = 1.2 miles  1 hr   1 min   1 s   1000 m   1.61 km 

28. The conversion is: cm → in. → ft → yr → day

1 in.  1 ft  1 yr   365 day   = 3.54 days     2.54 cm  12 in.  3.38 ft   1 yr 

(1 cm ) 

29. The conversion is: lb → g → mg → ant

454 g ant   1000 mg ant  1 ant  4   = 3.6 ×10 ant   1 lb ant   1 g ant  85 mg ant 

( 6.7 lb ant ) 

– Chapter 2 –

30. The conversion is: lb → g → egg

 454 g egg  1 lg egg    = 5.8 lg eggs  1 lb egg  58.5 g egg 

( 0.75 lb egg ) 

31. The conversion is: lb body → lb fat → kg fat

 11.2 lb fat   1 kg    = 11.4 kg fat  100 lb body   2.205 lb 

( 225 lb body ) 

32. The conversion is: carats → mg → g → lb

200 mg   1 g  1 lb  −3   = 2.54 ×10 lb   1 carat   1000 mg  453.6 g 

(5.75 carats ) 

33. The conversion is:

yd mi km m m → → → → s s s s min

 100. yd   1 mi   1.609 km  1000 m  60 s  2 m       = 1.1×10 min  52 s   1760 yd   1 mi  1 km  1 min  34. The conversion is:

mi km m cm cm cm → → → → → hr hr hr hr min s

 133 mi   1.609 km   1000 m   100 cm   1 hr   1 min  3 cm        = 5.94 × 10 s  1 hr   1 mi   1 km   1 m   60 min   60 s 

35.

(a)

29,035 ft − 21,002 ft = 8033 ( total feet climbed )

(16 hr ) 

60 min  3  = 960 min 960 min + 42 min = 1.0 ×10 min  hr 

 8003 ft  1 mi  −3 mi    = 1.5 ×10 3 min  1.0 × 10 min  5280 ft 

36.

(b)

 8003 ft   1 mi  1.609 km  1000 m  1 min  −2 m       = 4.1×10 3 mi  km  60 s  s  1.0×10 min   5280 ft 

(a)

The conversion is:

m cm in. ft ft → → → → hr hr hr hr min

ft  4500 m   100 cm   1 in.   1 ft   1 hr        = 50 min  5 hr   m   2.54 cm   12 in.   60 min 

(b)

The conversion is:

m km km km → → → hr hr min s

 4500 m   1 km   1 hr   1 min  −4 km      = 3 × 10 s  5 hr   1000 m   60 min   60 s 

– Chapter 2 –

37. The conversion is:

mi ft in. cm cm cm → → → → → hr hr hr hr min s

 155 mi   5280 ft   12 in   2.54 cm   1 hr   1 min  6930 cm       = sec  hr   1 mi   1 ft   1 in   60 min   60 sec 

38. The conversion is:

mi ft in cm cm cm → → → → → hr hr hr hr min s

 125 mi   5280 ft   12 in   2.54 cm   1 hr   1 min  5590 cm       = sec  hr   1 mi   1 ft   1 in   60 min   60 sec 

39. The conversion is: gal → qt → L → mL → cup

 4 qt  1 L   1000 mL  1 cup  4     = 1.60 ×10 cups coffee  1 gal  1.06 qt   1 L  473 mL 

( 2010 gal ) 

40. The conversion is: lb → g → number of tilapia

g   1 tilapia  ( 475 ×10 lb )  454  = 4.03 ×10 tilapia  1 lb  535 g 6





41. The conversion is: gal → qt → mL → drops

 4 qt  946 mL   20. drops  4    = 7.6 ×10 drops gal qt mL    

(1.0 gal ) 

42. The conversion is: gal → qt → L

 4 qt  0.946 L    = 160 L  gal  qt 

( 42 gal ) 

43. The conversion is: ft3 → in.3 → cm3 → mL 3

cm   1 mL  = 2.83 × 10 4 mL (1.00 ft3 )  121 ftin.   2.54   3  1 in.   1 cm 

44. V = A × h A = area h = height V = volume cm 3 cm 3 → → m2 nm m 3 V  200 cm 3   1 nm  1 m  5 2 A = =  −9   = 4 ×10 m h  0.5 nm   10 m  100 cm 

The conversion is:

45. (a) ( 27 cm) ( 21 cm)( 4.4 cm) = 2.5 ×10 cm 3

 1L  3 3 3 (b) 2.5 ×10 cm is 2.5 ×10 mL   = 2.5 L  1000 mL  3

(c)

1 in.  2 3 ( 2.5 ×103 cm3 )  2.54  = 1.5×10 in. cm 

– Chapter 2 –

2.54 cm   1 mL   1 L   1 qt   1 gal  46. (16 in.)( 8 in.)(10 in.)   = 6 gal     3   1 in.   1 cm   1000 mL   0.946 L   4 qt  3

 73.14 gC   = 183 g C  100 g ketone 

47.

( 250. g ketone ) 

48.

( 475. g nectaryl ) 



10.98 g H   = 52.2 g H  100 g nectaryl 

  283.4 g potassium dichromate  mass part  49. (a)  ×100  ×100 =   mass whole   283.4 g potassium dichromate+ 650.0 g water 

 283.4 g potassium dichromate  =  ×100 = 30.36% potassium dichromate 933.4 g soln    30.36 g potassium dichromate   = 83.5 g potassium dichromate 100 g solution  

(b)

( 275. g solution ) 

(c)

(15.0 g potassium dichromate ) 



 100 g soln  = 49.4 g soln  30.36 g potassium dichromate 

  95.4 g sodium nitrate  mass part  50. (a)   ×100 ×100 =   mass whole   95.4 g sodium nitrate + 753 g water   95.4 g sodium nitrate  =  ×100 = 11.3% sodium nitrate 848 g soln    11.3 g sodium nitrate   = 39.6 g sodium nitrate 100 g solution  

(b)

(350. g solution ) 

(c)

(50.0 g sodium nitrate ) 



 100 g soln  = 442 g soln  11.3 g sodium nitrate 

51. (a) °F=1.8°C + 32 = (1.8) (38.8) + 32 = 101.8°F (b) Yes, the child has a fever since 101.8°F > 98.6°F 52. °F = 1.8°C +32 53. (a)

(1.8) ( 45) +32 =113°F Summer!

Remember to express the answer to the same 162 −32 = 72.2°C precision as the original measurement. 1.8

(b) °C + 273 = K

0.0 − 32 + 273 = 255.2 K 1.8

−8.6 − 32 = −22.6°C 1.8

– Chapter 2 –

273 + 273 = 546 K

(c)

°C = 100 − 273 = −173°C

(d)

( −173)(1.8) + 32 = −279°F = −300°F (1 significant figure in 100 K )

55. °C =

( °F − 32 ) = ( 425°F − 32°F ) = 218°C 1.8

1.8°F/°C

K = °C + 273 = 218 + 273 = 491 K

56. °C =

( °F − 32 ) = ( 247°F − 32°F ) = 119°C 1.8

1.8°F/°C

K = °C + 273 = 119 + 273 = 392 K

°F = °C

57.

°F = 1.8 ( °C ) + 32 °F = 1.8 ( °F ) + 32

substitute °F for °C

−32 = 0.8 ( °F ) −32 = °F 0.8 −40 = °F −40°F = −40°C

°F = −°C

58.

°F = 1.8 ( °C ) + 32

substitute − °C for °F

−°C = 1.8 ( °C ) + 32

2.8 ( °C ) = −32

−32 2.8 °C = −11.4 −11.4°C = 11.4°F °C =

59. °C =

−60° − 32 −92° = = −51°C 1.8 1.8

60. °F = (1.8 × 801°) + 32 =1470°F 61. d =

m 59.82 g g = = 0.920 v 65.0 mL mL

62. d =

m 20.41 g g = = 0.810 v 25.2 mL mL

63. 50.92 g − 25.23 g = 25.69 g ( mass of liquid ) d=

m 25.69 g g = = 1.03 v 25.0 mL mL

– Chapter 2 –

64. 54.6 mL − 50.0 mL = 4.6 mL ( volume of zinc ) d=

m 32.95 g g = = 7.2 v 4.6 mL mL

65. The conversion is g → mL

 1 mL   = 16 mL  0.929 g 

(15 g ) 

66. The conversion is g → mL

 1 mL   = 63 mL  1.20 g 

( 75 g ) 

67. The conversion is kg cayenne → kg salsa → lb salsa

 100 kg salsa  2.20 lb salsa    = 85.3 lb  8.41 kg cayenne  1 kg salsa 

(3.26 kg cayenne ) 

68. The conversion is kg pecan → NCB ( nutty candy bars ) → lb NCB

 100 kg NCB  2.20 lb NCB    = 70.lb NCB  22.0 kg pecan  1 kg NCB 

( 7.0 kg pecan )  69. (a)

91.67 g Cu 8.33 g Ni and 100 g quarter 100 g quarter

(b) The conversion is ton Cu →g Cu →g quarters →quarters 1 quarters       1 ton Cu  1 lb Cu   91.67 g Cu  5.670 g quarters 

( 3.0 ton Cu ) 

2000 lb Cu  454 g Cu   100 g quarters 

= 5.2 × 105 quarters

70. (a)

92.5 g Ag 7.50 g Cu and 100 g sterling 100 g sterling

(b) The conversion is lb Ag → g Ag → g sterling → silver hearts

 454 g Ag  100 g sterling  1 silver heart     = 701 silver hearts  1 lb Ag  9.5 g Ag  35.0 g sterling 

(50.0 lb Ag )  71. a) 1055 kg 1)

1055 is the numerical value

kg is the unit

4 significant figures

105 is known, 5 is estimated

1.1 × 103 kg

– Chapter 2 –

b) 1,650,763 wavelengths 1) 1,650,763 is the numerical value 2) wavelengths is the unit 3) 7 significant figures 4) 165076 is known, 3 is estimated 5) 1.7 × 106 wavelengths c)

1.077 g/cm3 1) 1.077 is the numerical value 2) g/cm3 is the unit 3) 4 significant figures 4) 1.07 is known, 7 is estimated 5) 1.1 × 100 g/cm3

d) 0.00845 L 1) 0.00845 is the numerical value 2) L is the unit 3) 3 significant figures 4) 84 is known, 5 is estimated 5) 8.5 × 10−3L e)

776,000 g 1) 776,000 is the numerical value 2) g is the unit 3) 3 significant figures 4) 77 is known, 6 is estimated 5) 7.8 × 105 g

72. (a) report 10.01 grams (b) report 10.012 grams (c) report 10.0124 grams 73. (a)

(175 Skittles ) 

1.134 g Skittle   =198 g Skittles  1 Skittle 

 5.3 mL Skittle   1 L Skittles  (b) (175 Skittles )    = 0.15 L Skittles  6 Skittle   1000 mL Skittles  (c)



1 Skittles   = 286.6 Skittles  1.134 g Skittles 

(325.0 Skittles ) 

 1000 mL Skittle   6 Skittles  (d) ( 0.550 L Skittles )    = 620 Skittles  1 L Skittle   5.3 mL Skittles  14

– Chapter 2 –

(e) The mass measurement is more precise. It is also more accurate. Using calculations similar to those above there should be 309 Skittles in 350 grams and the average value is 310 Skittles. There should be 367 Skittles in 0.325 L of Skittles and the average value is 384 Skittles. 74. A graduated cylinder would be the best choice for adding 100 mL of solvent to a reaction. While the volumetric flask is also labeled 100 mL, volumetric flasks are typically used for doing dilutions. The other three pieces of glassware could also be used, but they hold smaller volumes so it would take a longer time to measure out 100 mL. Also, because you would have to repeat the measurement many times using the other glassware, there is a greater chance for error. 75. The conversion is: g → mL

 1 mL   = 14.5 mL  1.484 g 

( 21.5 g ) 

76. The conversion is: g → mL

 1 mL   = 26 mL  0.97 g 

( 25.27 g ) 

77. The conversion is: day → cups → mg → g → lb

 4.00 ×108 cups  160 mg  1 g  1 lb  5     = 1×10 lb day cup 1000 mg 453.6 g     

(1 day ) 

78. Mass of gear carried by paladin after drinking strength potion = 115 lb + 50.0 lb = 165 lb Amount of mass potion paladin can carry = 165 lb – 92 lb = 73 lb

The conversion is: lb → g → mL → vials potion 454 g   1 ml   1 vial potion     = 3.43 vials potion  1 lb   93 g   50.0 mL 

( 73 lb ) 

You would only be able to collect 3 vials, because 4 would put you over your mass limit. 79. The conversion is : sequins → cm3 → g → kg

 0.0241 cm 3   41.6 g sequins   1 kg  4560 sequins ( )   = 4.57 kg sequins  1 cm3   1000 g   1 sequins   The conversion is: kg → lb  2.20 lb   = 10.1 lb sequins  1 kg 

( 4.57 kg sequins ) 

– Chapter 2 –

80. V = side3 = ( 0.50 m ) = 0.13 m 3 3

( 0.13 m )  100.mcm   10001 Lcm  = 130 L 3

( volume of the cube ) ( volume of cube )

Yes, the cube will hold the solution. 130 L − 8.5 L = 120 L additional solution is necessary to fill the container. 81. The conversion is:

μg m

→

μg L

→

μg day

 180 μ g   1 m   4 L  = 4000 μ g ingested/day    2 × 10 3  1 m 1000 L day     3

(1 significant figure )

Yes, the nurse is at risk. This is well over the toxic limit. 82. (a)

The conversion is ac ft → L → kL

 1.233×106 L   1 kL  6   = 1.70 ×10 kL 1 ac ft 1000 L   

(1380 ac ft )  (b)

The conversion is day → kL → L → qt → gal  1.70×106 kL   1000 L   1.06 qt   1 gal  10 (31 day )    = 1.40 ×10 gal    1 day   1 kL   1 L   4 qt 

83. (a)

Convert 20.27 K to °C K − 273.15 = °C 20.27 K − 273.15 = −252.88°C

(b)

Convert 20.27 K to °F °F = (1.8× °C ) + 32

°F = (1.8× −252.88 ) + 32 = −455.18 + 32

°F = −423.18°F 84. °F = 1.8°C + 32 Convert 36°C to °F

Convert 38°C to °F

(1.8)(36 ) + 32 = 97°F (1.8)( 38) + 32 = 100°F

Sauropods have a body temperature close to the body temperature of other warmblooded mammals such as dogs, humans, and cows. 85.

The conversion is : L → dL → mg → g 10 dL   130 mg   1 g   = 6.1 g    1 L   1 dL   1000 mg 

( 4.7 L ) 

86. A sample of gold will sink to the bottom of the mercury and a sample of iron pyrite will float.

– Chapter 2 –

87.

The conversion is : El → ton → lb

( 5.3 El ) 

6.00 ton   2000 lb  4   = 6.4 × 10 lb  1 El   1 ton 

88. The conversion is: hands → in. → cm → m

(14.2 hands ) 

4 in.   2.54 cm  1 m     = 1.44 m  1 hand   1 in.  100 cm 

89. The conversion is: days → hr → gal → qt → L

 24 hr   22.5 gal   4 qt  0.946 L  3    = 5.1×10 L   1 day   12 hr   1 gal  1 qt 

(30 days )  90. d =

m The cube with the largest volume has the lowest density. Use Table 2.6. V

Cube A - lowest density Cube B

1.74 g/mL - magnesium 2.7 g/mL - aluminium

Cube C - highest density 10.5 g/mL - silver

91. The conversion is: ft → in. → cm → m → nm → nanotubes 9  12 in.  2.54 cm  1 m   10 nm   1 nanotube  9 40.0 ft ( )      = 9.4 ×10 nanotubes  1 ft  1 in.  100 cm   1 m   1.3 nm  92.

 0.25 c peanuts   = 1.4 c peanuts  67 mg Mg 

(380 mg Mg ) 

2 2 2 93. The conversion is: acres → ft → mi → km

 43560 ft 2   1 mi   1.609 km  2     = 0.506 km 1 acre 5280 ft 1 mi     

(125 acres ) 

94. The volume of the aluminum cube is m 500. g V= = = 185 mL d 2.70 g mL

Density of Al is 2.70 g/mL. This is the same volume as the gold cube thus

m = dV = (185 mL)(19.3 g/mL) = 3.57 ×103 g of gold Density of Au is 19.3 g/mL. m 24.12 g 0.965 g = = V 25.0 mL mL 96. 150.50 g − 88.25 g = 62.25 g

95. d =

( density of water at 90°C )

m m 62.25 g thus V = = = 49.8 mL V d 1.25 g mL The container must hold at least 50 mL. d=

( mass of liquid ) ( volume of liquid )

– Chapter 2 –

97. H2 O

50 g = 50 mL g 1.0 mL alcohol 50 g = 60 mL g 0.789 mL

Ethyl alcohol has the greater volume due to its lower density. 98. Volume of sulfuric acid

 1 mL    (100. g ) = 54.3 mL  1.84 g  99.

The conversion is: cup → oz → qt → L → mg 











31.4 mg NMP oz coffee 1 qt coffee 1 L coffee     ( 2.00 cup coffee )  10 1 cup coffee   32 oz coffee   1.057 qt coffee   1 L coffee  = 18.6 mg NMP

100.

m , as the volume increases, the density decreases. As solids are heated V the density decreases due to an increase in the volume of the solid.

Since d =

1 mL  101. V = ( 2.00 cm )(15.0 cm )( 6.00 cm )  = 180.mL ( volume of bar ) 3   1 cm  m d = = 3300 g/180.mL = 18.3 g/mL V

The density of pure gold is 19.3 g/mL (from Table 2.5), therefore, the gold bar is not pure gold, since its density is only 18.3 g/mL, or it is hollow inside. 102. m = dV = ( 0.789 g/mL ) ( 35.0 mL ) = 27.6 g ethyl alcohol 27.6 g + 49.28 g = 76.9 g ( mass of cylinder and alcohol )

103. The conversion is g → oz → gr → scruples

 12 oz   480 gr  1 scruple    = 537 scruples   373 g   oz   20 gr 

( 695 g ) 

– Chapter 2 –

104. The conversion is: days → teaspoons → mL

 2 teaspoons× 4   5 mL    = 400 mL day    teaspoon 

(10 days ) 

Since you will need a total of 400 mL for the 10 days and the bottle contains 500 mL, you have purchased enough. 105. (a)

The conversion is kg → g → mL → L

 1000 g brine  1 mL brine   1 L brine     = 6.36 L brine  1 kg brine  1.28 g brine   1000 mL brine 

(8.14 kg brine )  (b)

The conversion is g NaCl → g brine  100 g brine   = 1990 g brine  7.55 g NaCl 

(150.0 g NaCl ) 

106. The density of lead is 11.34 g/mL. The density of aluminum is 2.70 g/mL. The density of silver is 10.5 g/mL. The density of the unknown piece of metal can be calculated from the mass (20.25 g) and the volume (57.5 mL − 50 mL = 7.5 mL) of the metal. Density of the unknown metal = 20.25 g/7.5 mL = 2.7 g/mL. The metal must be aluminum. 107.

Volume of slug

30.7 mL − 25.0 mL = 5.7 mL

Density of slug

Mass of liquid, cylinder, and slug

m 15.454 g = = 2.7 g/mL V 5.7 mL 125.934 g

Mass of slug ( subtract )

Mass of cylinder ( subtract )

−89.450 g

Mass of the liquid

21.030 g

Density of liquid d =

108.

−15.454 g

m 21.030 g = = 0.841 g/mL V 25.0 mL

7.8 g bronze needed   100 g bronze melted     1 mL bronze needed   90.0 g bronze needed 

( 225 mL bronze needed ) 

= 2.0 ×103 g bronze melted 109. The conversion is: km → m → cm → in. → ft → mi → hr → min → s → ns 











m 100 cm 1 in. 1 ft 1 mi 1 hr 60 min 60 s         (730.0 km )  1000 1 km 1 m  2.54 cm  12 in.   5280 ft  1.86 ×108 mi  1 hr  1 min  109 ns    = 8.78 ×106 ns  1s   

8,780,000 ns is good to only 3 significant figures. You would need at least 6 significant digits to detect the difference between 8,780,000 ns and (8,780,000 + 60) ns. Sometimes experimenters do not see differences due to the precision of their measuring techniques.

– Chapter 2 –

110.

(1.5 m ) 

100 cm   1 in.   1 ft     = 4.9 ft  1 m   2.54 cm   12 in.  100 cm  1 in.  1 ft  ( 4 m )     = 13 ft  1 m  2.54 cm  12 in. 

( 27°C ×1.8) + 32 = 81°F

CHAPTER 3

ELEMENTS AND COMPOUNDS SOLUTIONS TO REVIEW QUESTIONS 1.

Silicon

25.7%

Hydrogen 0.9%

25.7 g Si = 30 g Si/1 g H (1 sig. fig.) 0.9 g H

In 100 g

Si is 28 times heavier than H, thus since 30 > 28, there are more Si atoms than H atoms. 2. 3.

(a)

(c)

(e)

(g)

(b) F

(d)

(f)

(h)

(a)

Iron

(c)

Carbon

(e)

Beryllium

(g)

Argon

(b) Magnesium

(d)

Phosphorus

(f)

Cobalt

(h)

Mercury

The symbol of an element represents the element itself. It may stand for a single atom or a given quantity of the element.

sodium

silver

potassium

tungsten

iron

gold

antimony

mercury

tin

lead

hydrogen

sulfur

boron

potassium

carbon

vanadium

nitrogen

yttrium

oxygen

iodine

fluorine

tungsten

phosphorus

uranium

0 metalloids

5 nonmetals

1 metal

Hydrogen

Chlorine

Cl2

Nitrogen

Bromine

Br2

Oxygen

Iodine

Fluorine

– Chapter 3 –

(a) CO – 1 atom of carbon and 1 atom of oxygen (b) H2 – 1 molecule of hydrogen (made of 2 atoms of hydrogen) (c) S8 – 1 molecule of sulfur (made of 8 atoms of sulfur) (d) CS – 1 atom of carbon and 1 atom of sulfur

Co – 1 atom of cobalt 2 H – 2 atoms of hydrogen 8 S – 8 atoms of sulfur Cs – 1 atom of cesium

10. In an element all atoms are alike, while a compound contains two or more elements (different atoms) which are chemically combined. Compounds can be decomposed into simpler substances while elements cannot. 11. 92 metals

7 metalloids

19 nonmetals (based on 118 elements)

12. 7 metals

1 metalloid

2 nonmetals

13. (a) iodine

(b) bromine

14. A compound is composed of two or more elements which are chemically combined in a definite proportion by mass. Its properties differ from those of its components. A mixture is the physical combining of two or more substances (not necessarily elements). The composition may vary, the substances retain their properties, and they may generally be separated by physical means. 15. Molecular compounds exist as molecules formed from two or more atoms of elements bonded together. Ionic compounds exist as cations and anions held together by electrical attractions. 16. Compounds are distinguished from one another by their characteristic physical and chemical properties. 17. Cations are positively charged, while anions are charged negatively.

– Chapter 3 –

SOLUTIONS TO EXERCISES 1.

Diatomic molecules: (a) HCl, (b) O2 (h) NO

Diatomic molecules: (a) I2 (d) HF (f) Cl2

(a) Potassium, iodine

(d)

Calcium, bromine

(b) Sodium, carbon, oxygen

(e)

Hydrogen, carbon, oxygen

(a) Magnesium, bromine

(d)

Barium, sulfur, oxygen

(b) Carbon, chlorine

(e)

Aluminum, phosphorus, oxygen

(a) ZnO

(c)

NaOH

(b) KClO3

(d)

C2H6O

(a) AlBr3

(c)

PbCrO4

(b) CaF2

(d)

C6H6

(a) C6H10OS2 (b) C18H27NO3 (c) C10H16

(a) C55H86NO24 (b) C15H14O6 (c) C30H48O3

(a) 2 atoms iron, 3 atoms oxygen (b) 2 atoms calcium, 2 atoms nitrogen, 6 atoms oxygen (c) 1 atom cobalt, 4 atoms carbon, 6 atoms hydrogen, 4 atoms oxygen (d) 3 atoms carbon, 6 atoms hydrogen, 1 atom oxygen (e) 2 atoms potassium, 1 atom carbon, 3 atoms oxygen (f) 3 atoms copper, 2 atoms phosphorus, 8 atoms oxygen (g) 2 atoms carbon, 6 atoms hydrogen, 1 atom oxygen

(h) 2 atoms sodium, 2 atoms chromium, 7 atoms oxygen 10. (a) 4 atoms hydrogen, 2 atoms carbon, 2 atoms oxygen (b) 3 atoms nitrogen, 12 atoms hydrogen, 1 atom phosphorus, 4 atoms oxygen (c) 1 atom magnesium, 2 atoms hydrogen, 2 atoms sulfur, 6 atoms oxygen (d) 1 atom zinc, 2 atoms chlorine (e) 1 atom nickel, 1 atom carbon, 3 atoms oxygen (f) 1 atom potassium, 1 atom manganese, 4 atoms oxygen (g) 4 atoms carbon, 10 atoms hydrogen (h) 1 atom lead, 1 atom chromium, 4 atoms oxygen

– Chapter 3 –

11. (a) 9 atoms

(b) 14 atoms

(d)

45 atoms

12. (a) 9 atoms

(b) 12 atoms

(d)

12 atoms

13. (a) 6 atoms C (b) 3 atoms C

(d)

8 atoms C

14. (a) 6 atoms O (b) 4 atoms O

(d)

21 atoms O

15. (a) mixture

(d)

mixture

(b) mixture

(e)

pure substance

(f)

mixture

16. (a) mixture

(d)

pure substance

(b) mixture

(e)

mixture

(f)

mixture

17. (c) compound

(e)

element

18. (c) element

(d)

compound

19. (a) mixture

(c)

element

(c)

mixture

(b) compound 20. (a) compound (b) compound 21. Yes. The gaseous elements are all found on the extreme right of the periodic table. They are the entire last column and in the upper right corner of the table. Hydrogen is the exception and located at the upper left of the table. 22. No. The only common liquid elements (at room temperature) are mercury and bromine. 23.

18 metals ×100 = 50% metals 36 elements

24.

26 solids ×100 = 72% solids 36 elements

25. The formula for water is H2O. There is one atom of oxygen for every two atoms of hydrogen. The molar mass of oxygen is 16.00 g and the molar mass of hydrogen is 1.008 g. For H2O the mass of two hydrogen atoms is 2.016 g and the mass of one oxygen atom is 16.00 g. The ratio of hydrogen to oxygen is approximately 2:16 or 1:8. Therefore, there is 1 gram of hydrogen for every 8 grams of oxygen. 26. The formula for hydrogen peroxide is H2O2. There are two atoms of oxygen for every two atoms of hydrogen. The molar mass of oxygen is 16.00 g and the molar mass of hydrogen is 1.008 g. For hydrogen peroxide the total mass of hydrogen is 2.016 g and the total mass of oxygen is 32.00 g for a ratio of hydrogen to oxygen of approximately 2:32 or 1:16. Therefore, there is 1 gram of hydrogen for every 16 grams of oxygen.

– Chapter 3 –

27. (a) No (b) Compound; molecular (c) No (d) Compound; molecular (e) Compound; molecular 28. (a) Compound; molecular (b) No (c) Compound; ionic (d) No (e) Compound; molecular 29. To a small sample of the mixture, add water and observe that the salt dissolves but the pepper does not. After the small trial, add water to the entire mixture to dissolve the salt. Separate the undissolved pepper from the salt solution by filtering the mixture. Coffee filters or strong paper towels would work well for this process. 30. The atoms that make up each ionic compound are on opposite ends of the periodic table from one another. An ionic compound is made up of a metal-nonmetal combination. 31. (a) 1 carbon atom and 1 oxygen atom, total number of atoms = 2 (b) 1 boron atom and 3 fluorine atoms, total number of atoms = 4 (c) 1 hydrogen atom, 1 nitrogen atom, 3 oxygen atoms, total number of atoms = 5 (d) 1 potassium atom, 1 manganese atom, 4 oxygen atoms, total number of atoms = 6 (e) 1 calcium atom, 2 nitrogen atoms, 6 oxygen atoms, total number of atoms = 9 (f)

3 iron atoms, 2 phosphorus atoms, 8 oxygen atoms, total number of atoms = 13

32. (a) 181 atoms/module

63 C 88 C 1 Co 14 N 14 O 1P 181 atoms (b)

63 C ×100 = 35% C atoms 181 atoms

(c)

1 Co 1 = metals 181 atoms 181

– Chapter 3 –

33. The conversion is: cm3 → L, → mg, → g, → $ −4 1L   4 × 10 mg   1 g   $19.40  15 3  × 1 10 cm ( )  1000 cm3   L   1000 mg   g  = $8 ×106    

34. Ca(H2PO4)2

(10 formula units ) 

4 atoms H   = 40 atoms H  formula unit 

35. C145H293O168

145 C 293 H 168 O 606 atoms/molecule 36. (a)

magnesium, manganese, molybdenum, mendeleevium, mercury, meitnerium

(b)

carbon, phosphorus, sulfur, selenium, iodine, astatine, boron

(c)

sodium, potassium, iron, silver, tin, antimony

37. Add water to the mixture to dissolve the sugar. Filter the mixture to separate the sugar solution from the insoluble sand. Add another small amount of water to remove last traces of sugar. Filter. Allow the water to evaporate from the sugar solution to obtain crystals of sugar. Sand is the insoluble residue. 38. HNO3 has 5 atoms/molecule 7 dozen = 84

(84 molecules )(5 atoms/molecule ) = 420 atoms  12 molecules   5 atoms  or ( 7 dz )    = 420 atoms dz    molecule  39.

(a) As temperatures decreases, density increases. (b) approximately 1.28 g/L at 5°C approximately 1.18 g/L at 25°C approximately 1.09 g/L at 70°C

– Chapter 3 –

40. Each represents eight units of sulfur. In 8 S the atoms are separate and distinct. In S8 the atoms are joined as a unit (molecule). 41. (a) NaCl

(d)

Fe2S3

(g) C6H12O6

(b) H2SO4

(e)

K3PO4

(h) C2H5OH

(f)

Ca(CN)2

(i)

Cr(NO3)3

42. Let X = grams sea water

 5.0 ×10 −8 %I 2    ( X ) = 1.0 g I 2 100   (1.0 g I2 )(100 ) = 2.0 ×109 g sea water X= (5.0 ×10−8%I2 ) 1 kg  ( 2.0 ×10 g )  1000  = 2.0 × 10 kg sea water g 9





43. (a) 12 carbons, 22 hydrogens, 11 oxygens (b) 7 carbons, 5 hydrogens, 3 oxygens, 1 nitrogen, 1 sulfur (c) 14 carbons, 18 hydrogens, 5 oxygens, 2 nitrogens (d) 4 carbons, 4 hydrogens, 3 oxygens, 1 nitrogen, 1 sulfur, 1 potassium (e) 12 carbons, 19 hydrogens, 8 oxygens, 3 chlorines 44. Cobalt should be written Co as CO is the formula for carbon monoxide. 45. C8N4O2H10 46. (a) Picture (3) because fluorine gas exists as a diatomic molecule. (b) Other elements that exist as diatomic molecules are oxygen, nitrogen, chlorine, hydrogen, bromine, and iodine. (c) Picture (2) could represent SO3 gas. 47. (a) noble gases

(d)

alkaline earth metals

(b) metalloids

(e)

transition metals

(f)

halogens

48. (a) 6

(b) 1

(c)

(d) 2

(e) 6

49. (a) Bromine is the only nonmetal element that is a liquid at room temperature. (b) Mercury is the only metal element that is a liquid at room temperature. (c) Iodine (d) (b) diatomic, (c) noble gas 50. (a) NH4Cl

(d)

FeF2

(b) H2SO4

(e)

Pb3(PO4)2

(f)

Al2O3

– Chapter 3 –

51. Group 1A oxides: Li2O, Na2O, K2O, Rb2O, Cs2O Group 2A oxides: BeO, MgO, CaO, SrO, BaO 52. (a) Arachidic acid Arachidonic acid

20 carbons, 40 hydrogens, and 2 oxygens 20 carbons, 32 hydrogens, and 2 oxygens

(b) Stearic acid

18 carbons, 36 hydrogens, and 2 oxygens

Linoleic acid

18 carbons, 32 hydrogens, and 2 oxygens

40 H 2H = 20 C C 36 H 2 H Stearic acid = 18 C C

32 H 1.6 H = 20 C C 32 H 1.8 H Linoleic acid = 18 C C

Arachidonic acid

(d) Saturated molecules have more H’s per C than unsaturated molecules. Saturated molecules must have more hydrogen atoms. 53. (a) Let x = RDA of iron 60% of x = 11 mg Fe x=

11 mg Fe ×100% = 18 mg Fe 60%

(b) density of iron = 7.86 g/mL

 1 g  1 mL  m = (11 mg Fe )    d  1000 mg  7.86 g 

= 1.4 ×10−3 mL Fe 54. If Alfred inspects the bottles carefully, he should be able to see whether the contents are solid (silver) or liquid (mercury). Alternatively, since mercury is more dense than silver, the bottle of mercury should weigh more than the bottle of silver (the question indicated that both bottles were of similar size and both were full). Density is mass/volume.

CHAPTER 4

PROPERTIES OF MATTER SOLUTIONS TO REVIEW QUESTIONS 1.

Gaseous state: 393 K = 120.0°C; boiling point of acetic acid is 118.0°C.

Liquid state: melting point of chlorine is −101.6°C and boiling point of chlorine is −34.6°C.

(a) 118.0°C+273.15=391.2 K (b) (118.0°C)(1.8)+32=244.4°F

(a) (16.7°C)(1.8)+32=62.1°F (b) 16.7°C+273.15=289.9 K

Small bubbles appear at each electrode, and a gas collects above each electrode. The system now contains water and two different gases.

A new substance is always formed during a chemical change, but never formed during physical changes.

Reading the problem carefully and writing down all of the important information including units is the critical first step in solving any chemistry problem.

No, the last step is always a check of the answer to make sure it makes sense.

Potential energy is the energy of position. By the position of an object, it has the potential of movement to a lower energy state. Kinetic energy is the energy matter possesses due to its motion.

10. A food Calorie is equal to 1000 cal or 1 kcal. 11. Iron requires more energy to heat because it has a higher specific heat. 12. A molecule of octane would produce more carbon dioxide because it contains more carbon atoms. 13. All hydrocarbons are composed only of the elements carbon and hydrogen. 14. Carbon 15. Fossil fuels produce carbon dioxide, a greenhouse gas. The supply of fossil fuels is also decreasing and mining or drilling for these fuels is harmful to the environment. Other sources of energy which may become important are wind and solar.

– Chapter 4 –

SOLUTIONS TO EXERCISES

(a) physical

(e) chemical

(b) chemical

(f)

(c)

physical

(g) chemical

(d) physical

(h) chemical

(a) physical

(e) physical

(b) physical

(f)

(c)

chemical

(g) chemical

(d) physical

(h) chemical

physical

Although the appearance of the platinum wire changed during the heating, the original appearance was restored when the wire cooled. No change in the composition of the platinum could be detected.

A copper wire, like the platinum wire, changes to a glowing red color when heated (physical change). Upon cooling, the original appearance of the copper wire is not restored, but a new substance, black copper(II) oxide appears (chemical change).

Reactants:

copper, oxygen

Product:

copper(II) oxide

Reactant:

water

Product:

hydrogen, oxygen

6. 7.

(a) chemical

(d)

chemical

(b) physical

(e)

chemical

(f)

physical

(a) physical

(d)

physical

(b) chemical

(e)

chemical

(f)

chemical

(a) kinetic energy

(d)

potential energy

(b) kinetic energy

(e)

kinetic energy

(d)

kinetic energy

(e)

kinetic energy

The kinetic energy is converted to thermal energy (heat), chiefly in the brake system, and eventually dissipated into the atmosphere.

12.

The transformation of kinetic energy to thermal energy (heat) is responsible for the fiery re-entry of a space vehicle.