CHAPTER ONE
CHEMICAL FOUNDATIONS
For Review
1. a. Law versus theory: A law is a concise statement or equation that summarizes observed behavior. A theory is a set of hypotheses that gives an overall explanation of some phenomenon. A law summarizes what happens; a theory (or model) attempts to explain why it happens.
b. Theory versus experiment: A theory is an explanation of why things behave the way they do, while an experiment is the process of observing that behavior. Theories attempt to explain the results of experiments and are, in turn, tested by further experiments.
c. Qualitative versus quantitative: A qualitative observation only describes a quality while a quantitative observation attaches a number to the observation. Examples: Qualitative observations: The water was hot to the touch. Mercury was found in the drinking water. Quantitative observations: The temperature of the water was 62C. The concentration of mercury in the drinking water was 1.5 ppm.
d. Hypothesis versus theory: Both are explanations of experimental observation. A theory is a set of hypotheses that has been tested over time and found to still be valid, with (perhaps) some modifications.
2. No, it is useful whenever a systematic approach of observation and hypothesis testing can be used.
3. a. No. b. Yes c. Yes
Only statements b and c can be determined from experiment.
4. Volume readings are estimated to one decimal place past the markings on the glassware. The assumed uncertainty is ±1 in the estimated digit. For glassware a, the volume would be estimated to the tenths place since the markings are to the ones place. A sample reading would be 4.2 with an uncertainty of ±0.1. This reading has two significant figures. For glassware b, 10.52 ±0.01 would be a sample reading and the uncertainty; this reading has four significant figures. For glassware c, 18 ±1 would be a sample reading and the uncertainty, with the reading having two significant figures.
5. Accuracy: How close a measurement or series of measurements are to an accepted or true value.
1 2 CHAPTER 1 CHEMICAL FOUNDATIONS
Precision: How close a series of measurements of the same thing are to each other. The results, average = 14.91 0.03%, are precise (close to each other) but are not accurate (not close to the true value. 6. In both sets of rules, the lease precise number determines the number of significant figures in the final result. For multiplication/division, the number of significant figures in the result is the same as the number of significant figures in the least precise number used in the calculation. For addition/subtraction, the result has the same number of decimal places as the least precise number used in the calculation (not necessarily the number with the fewest significant figures).
7. Consider gold with a density of 19.32 g/cm3. The two conversion factors are:
19.32 g 1cm3 or 1cm3 19.32 g
Use the first form when converting from the volume of gold in cm3 to the mass of gold and use the second form when converting from mass of gold to volume of gold. When using conversion factors, concentrate on the units crossing off.
8. To convert from Celsius to Kelvin, a constant number of 273 is added to the Celsius temperature. Because of this, T(C) = T(K). When converting from Fahrenheit to Celsius, one conversion that must occur is to multiply the Fahrenheit temperature by a factor less than one (5/9). Therefore, the Fahrenheit scale is more expansive than the Celsius scale, and 1F would correspond to a smaller temperature change than 1C or 1K.
9. Chemical changes involve the making and breaking of chemical bonds. Physical changes do not. The identity of a substance changes after a chemical change, but not after a physical change.
10. Many techniques of chemical analysis require relatively pure samples. Thus, a separation step often is necessary to remove materials that will interfere with the analytical measurement.
Questions
17 A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction or the ideal gas law, PV = nRT. A theory (model) is an attempt to explain why something happens. Dalton’s atomic theory explains why mass is conserved in a chemical reaction. The kinetic molecular theory explains why pressure and volume are inversely related at constant temperature and moles of gas present as well as explaining the other mathematical relationships summarized in PV = nRT.
18. The fundamental steps are: 1. making observations 2. formulating hypotheses 3. performing experiments to test the hypotheses CHAPTER 1 CHEMICAL FOUNDATIONS 3
The key to the scientific method is performing experiments to test hypotheses. If after the test of time, the hypotheses seem to account satisfactorily for some aspect of natural behavior, then the set of tested hypotheses turn into a theory (model). However, scientists continue to perform experiments to refine or replace existing theories. 19. A qualitative observation expresses what makes something what it is; it does not involve a number, e.g., the air we breathe is a mixture of gases, ice is less dense than water, rotten milk stinks.
The SI units are mass in kilograms, length in meters, and volume in the derived units of m3. The assumed uncertainty in a number is 1 in the last significant figure of the number. The precision of an instrument is related to the number of significant figures associated with an experimental reading on that instrument. Different instruments for measuring mass, length, or volume have varying degrees of precision. Some instruments only give a few significant figures for a measurement while others will give more significant figures.
20. Precision: reproducibility; Accuracy: the agreement of a measurement with the true value. a. imprecise and inaccurate data: 12.32 cm, 9.63 cm, 11.98 cm, 13.34 cm b. precise but inaccurate data: 8.76 cm, 8.79 cm, 8.72 cm, 8.75 cm c. precise and accurate data: 10.60 cm, 10.65 cm, 10.63 cm, 10.64 cm
Data can be imprecise if the measuring device is imprecise as well as if the user of the measuring device has poor skills. Data can be inaccurate due to a systematic error in the measuring device or with the user. For example, a balance may read all masses as weighing 0.2500 g too high or the user of a graduated cylinder may read all measurements 0.05 mL too low.
A set of measurements which are imprecise implies that all the numbers are not close to each other. If the numbers aren’t reproducible, then all of the numbers can’t be very close to the true value. Some say that if the average of imprecise data gives the true value, then it is accurate data; a better description is that the data takers are extremely lucky.
21. Significant figures are the digits we associate with a number. They contain all of the certain digits and the first uncertain digit (the first estimated digit). What follows is one thousand indicated to varying numbers of significant figures: 1000 or 1 × 10 3 (1 S.F.); 1.0 × 103 (2 S.F.); 1.00 × 103 (3 S.F.); 1000. or 1.000 × 103 (4 S.F.).
To perform the calculation, the addition/subtraction significant figure rule is applied to 1.5 1.0. The result of this is the one significant figure answer of 0.5. Next, the multiplication/division rule is applied to 0.5/0.50. A one significant figure number divided by a two significant figure number yields an answer with one significant figure (answer = 1).
22. The volume per mass is the reciprocal of the density (1/density). The volume per mass conversion factor has units of cm3/g and is useful when converting from the mass of an object to its volume in cm3.
23. Straight line equation: y = mx + b where m is the slope of the line and b is the y-intercept. For
the TF vs. TC plot: 4 CHAPTER 1 CHEMICAL FOUNDATIONS
TF = (9/5)TC + 32 y = m x + b
The slope of the plot is 1.8 (= 9/5) and the y-intercept is 32F.
For the TC vs. TK plot:
TC = TK 273 y = m x + b The slope of the plot is one and the y-intercept is 273C.
24. a. coffee; saltwater; the air we breathe (N2 + O2 + others); brass (Cu + Zn) b. book; human being; tree; desk
c. sodium chloride (NaCl); water (H2O); glucose (C6H12O6); carbon dioxide (CO2)
d. nitrogen (N2); oxygen (O2); copper (Cu); zinc (Zn) e. boiling water; freezing water; melting a popsicle; dry ice subliming f. Elecrolysis of molten sodium chloride to produce sodium and chlorine gas; the explosive reaction between oxygen and hydrogen to produce water; photosynthesis which converts
H2O and CO2 into C6H12O6 and O2; the combustion of gasoline in our car to produce CO2 and H2O
Exercises
Significant Figures and Unit Conversions
25. a. exact b. inexact
c. exact d. inexact (π has an infinite number of decimal places.)
26. a. one significant figure (S.F.). The implied uncertainty is 1000 pages. More significant figures should be added if a more precise number is known.
b. two S.F. c. four S.F.
d. two S.F. e. infinite number of S.F. (exact number) f. one S.F.
27. a. 6.07 × 1015 ; 3 S.F. b. 0.003840; 4 S.F. c. 17.00; 4 S.F.
d. 8 × 108; 1 S.F. e. 463.8052; 7 S.F. f. 300; 1 S.F.
g. 301; 3 S.F. h. 300.; 3 S.F.
28. a. 100; 1 S.F. b. 1.0 × 102; 2 S.F. c. 1.00 × 103; 3 S.F.
d. 100.; 3 S.F. e. 0.0048; 2 S.F. f. 0.00480; 3 S.F.
g. 4.80 × 103 ; 3 S.F. h. 4.800 × 103 ; 4 S.F. CHAPTER 1 CHEMICAL FOUNDATIONS 5
29. When rounding, the last significant figure stays the same if the number after this significant figure is less than 5 and increases by one if the number is greater than or equal to 5.
a. 3.42 × 104 b. 1.034 × 104 c. 1.7992 × 101 d. 3.37 × 105 30. a. 5 × 102 b. 4.8 × 102 c. 4.80 × 102 d. 4.800 × 102 31. For addition and/or subtraction, the result has the same number of decimal places as the number in the calculation with the fewest decimal places. When the result is rounded to the correct number of significant figures, the last significant figure stays the same if the number after this significant figure is less than 5 and increases by one if the number is greater than or equal to 5. The underline shows the last significant figure in the intermediate answers.
a. 212.2 + 26.7 + 402.09 = 640.99 = 641.0 b. 1.0028 + 0.221 + 0.10337 = 1.32717 = 1.327 c. 52.331 + 26.01 0.9981 = 77.3429 = 77.34 d. 2.01 × 102 + 3.014 × 103 = 2.01 × 102 + 30.14 × 102 = 32.15 × 102 = 3215
When the exponents are different, it is easiest to apply the addition/subtraction rule when all numbers are based on the same power of 10.
e. 7.255 6.8350 = 0.42 = 0.420 (first uncertain digit is in the third decimal place).
32. For multiplication and/or division, the result has the same number of significant figures as the number in the calculation with the fewest significant figures.
0.102 0.0821 273 a. 2.2635 2.26 1.01
b. 0.14 × 6.022 × 1023 = 8.431 × 1022 = 8.4 × 1022; Since 0.14 only has two significant figures, the result should only have two significant figures.
c. 4.0 × 104 × 5.021 × 103 × 7.34993 × 102 = 1.476 × 105 = 1.5 × 105
2.00 106 d. 6.6667 1012 6.671012 3.00 107 33. a. Here, apply the multiplication/division rule first; then apply the addition/subtraction rule to arrive at the one decimal place answer. We will generally round off at intermediate steps in order to show the correct number of significant figures. However, you should round off at the end of all the mathematical operations in order to avoid round-off error. The best way to do calculations is to keep track of the correct number of significant figures during intermediate steps, but round off at the end. For this problem, we underlined the last significant figure in the intermediate steps.
2.526 0.470 80.705 = 0.8148 + 0.7544 + 186.558 = 188.1 3.1 0.623 0.4326 6 CHAPTER 1 CHEMICAL FOUNDATIONS
b. Here, the mathematical operation requires that we apply the addition/subtraction rule first, then apply the multiplication/division rule.
6.404 2.91 6.404 2.91 = 12 18.7 17.1 1.6 c. 6.071 × 105 8.2 × 106 0.521 × 104 = 60.71 × 106 8.2 × 106 52.1 × 106 = 0.41 × 106 = 4 × 107
3.8 1012 4.0 1013 38 1013 4.0 1013 42 1013 d. 6.3 1026 4 1012 6.3 1013 4 1012 63 1012 67 1012
9.5 4.1 2.8 3.175 19.575 e. = 4.89 = 4.9 4 4
Uncertainty appears in the first decimal place. The average of several numbers can only be as precise as the least precise number. Averages can be exceptions to the significant figure rules.
8.925 8.905 0.020 f. 100 × 100 = 0.22 8.925 8.925
34. a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025
6.6262 1034 2.998 108 b. 7.82 1017 2.54 109
c. 1.285 × 102 + 1.24 × 103 + 1.879 × 10 1 = 0.1285 × 101 + 0.0124 × 10 1 + 1.879 × 101 = 2.020 × 101
When the exponents are different, it is easiest to apply the addition/subtraction rule when all numbers are based on the same power of 10.
d. 1.285 102 1.24 103 = 1.285 102 0.124 102 = 1.161 10 2
(1.00866 1.00728) 0.00138 27 e. 2.2910 6.022051023 6.02205 1023
9.875102 9.795 102 0.080 102 f. 100 100 8.1 101 9.875 102 9.875 102 9.42 102 8.234 102 1.625103 0.942 103 0.824 103 1.625 103 g. 3 3 = 1.130 × 103 CHAPTER 1 CHEMICAL FOUNDATIONS 7
1m 1000 mm 1m 35. a. 8.43 cm × 84.3 mm b. 2.41 × 102 cm × = 2.41 m 100 cm m 100 cm
1m 100 cm 5 c. 294.5 nm 2.94510 cm 1109 nm m
1 km 1000 mm d. 1.445 ×104 m × = 14.45 km e. 235.3 m × = 2.353 × 105 mm 1000 m m
1 m 1 106 μm f. 903.3 nm 0.9033 μm 1 109 nm m
11012 g 1 kg 36. a. 1 Tg 1 109 kg Tg 1000 g
11012 m 1109 nm b. 6.50 × 102 Tm 6.50 1023 nm Tm m
1g 1 kg 18 17 c. 25 fg 25 10 kg 2.5 10 kg 11015 fg 1000 g
1L d. 8.0 dm3 × = 8.0 L (1 L = 1 dm3 = 1000 cm3 = 1000 mL) dm 3
1L 1 106 μL e. 1 mL 1 103 μL 1000 mL L
1g 1 1012 pg f. 1 μg 1 106 pg 1 106 μg g
37. a. Appropriate conversion factors are found in Appendix 6. In general, the number of significant figures we use in the conversion factors will be one more than the number of significant figures from the numbers given in the problem. This is usually sufficient to avoid round-off error.
1lb 16 oz 3.91 kg × = 8.62 lb; 0.62 lb × = 9.9 oz 0.4536 kg lb
Baby’s weight = 8 lb and 9.9 oz or to the nearest ounce, 8 lb and 10. oz.
1in 51.4 cm × = 20.2 in ≈ 20 1/4 in = baby’s height 2.54 cm 8 CHAPTER 1 CHEMICAL FOUNDATIONS
1.61km 1000 m b. 25,000 mi × = 4.0 × 104 km; 4.0 × 104 km × = 4.0 × 107 m mi km
1m 1 m 2 3 c. V = 1 × w × h = 1.0 m × 5.6 cm 2.1dm = 1.2 × 10 m 100 cm 10 dm 3 10 dm 1L 1.2 × 102 m3 × = 12 L m dm3
3 3 3 1000 cm 1in 3 3 1ft 3 12 L × = 730 in ; 730 in × = 0.42 ft L 2.54 cm 12 in
1lb 0.4536 kg 38. a. 908 oz × = 25.7 kg 16 oz lb 1qt 1gal b. 12.8 L × = 3.38 gal 0.9463 L 4 qt 1 L 1qt c. 125 mL × = 0.132 qt 1000 mL 0.9463 L 4 qt 1 L 1000 mL d. 2.89 gal × = 1.09 × 104 mL 1gal 1.057 qt 1 L 453.6 g e. 4.48 lb × = 2.03 × 103 g 1lb 1 L 1.06 qt f. 550 mL × = 0.58 qt 1000 mL L
8 furlongs 40 rods 39. a. 1.25 mi × = 10.0 furlongs; 10.0 furlongs × = 4.00 × 102 rods mi furlong 5.5 yd 36 in 2.54 cm 1 m 4.00 × 102 rods × = 2.01 × 103 m rod yd in 100 cm 1km 2.01 × 103 m × = 2.01 km 1000 m
b. Let's assume we know this distance to ± 1 yard. First convert 26 miles to yards.
5280 ft 1 yd 26 mi × = 45,760. yd mi 3 ft
26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards CHAPTER 1 CHEMICAL FOUNDATIONS 9
1 rod 1furlong 46,145 yard × = 8390.0 rods; 8390.0 rods × = 209.75 furlongs 5.5 yd 40 rods
36 in 2.54 cm 1m 1km 46,145 yard × = 42,195 m; 42,195 m × = 42.195 km yd in 100 cm 1000 m
2 2 10,000 m 1km 2 2 40. a. 1 ha × = 1 10 km ha 1000 m 2 2 160 rod 5.5 yd 36 in 2.54 cm 1m 4 2 b. 5.5 acre × = 2.2 × 10 m acre rod yd in 100 cm
1 ha 2 4 2 4 2 1km 2 2.2 × 10 m × 4 2 = 2.2 ha; 2.2 × 10 m × = 0.022 km 110 m 1000 m
c. Area of lot = 120 ft × 75 ft = 9.0 × 103 ft2 2 1 yd 1 rod 1acre $6,500 $31,000 9.0 × 103 ft2 × = 0.21 acre; 2 3 ft 5.5 yd 160 rod 0.21acre acre
We can use our result from (b) to get the conversion factor between acres and ha (5.5 acre = 2.2 ha.). Thus, 1 ha = 2.5 acre.
1 ha $6,500 $77,000 0.21 acre × = 0.084 ha; The price is: 2.5 acre 0.084 ha ha
12 troy oz 20 pw 24 grains 0.0648 g 1kg 41. a. 1 troy lb × = 0.373 kg troy lb troy oz pw grain 1000 g
2.205 lb 1 troy lb = 0.373 kg × = 0.822 lb kg
20 pw 24 grains 0.0648 g b. 1 troy oz × = 31.1 g troy oz pw grain
1carat 1 troy oz = 31.1 g × = 156 carats 0.200 g
1000 g 1cm3 c. 1 troy lb = 0.373 kg; 0.373 kg × = 19.3 cm3 kg 19.3 g
1scruple 1dram ap 3.888 g 42. a. 1 grain ap × = 0.06480 g 20 grain ap 3 scruples dram ap 10 CHAPTER 1 CHEMICAL FOUNDATIONS
From the previous question, we are given that 1 grain troy = 0.0648 g = 1 grain ap. So, the two are the same.
8 dram ap 3.888 g 1oz troy * b. 1 oz ap × = 1.00 oz troy *See Exercise 41b. oz ap dram ap 31.1g
1g 1dram ap 3 scruples c. 5.00 × 102 mg × = 0.386 scruple 1000 mg 3.888 g dram ap 20 grains ap 0.386 scruple × = 7.72 grains ap scruple
1dram ap 3.888 g d. 1 scruple × = 1.296 g 3 scruples dram ap
3.00 108 m 1.094 yd 60 s 60 min 1 knot 43. warp 1.71 = 5.00 s m min hr 2000 yd/ hr
= 2.95 × 109 knots 3.00 108 m 1 km 1mi 60 s 60 min 9 5.00 = 3.36 × 10 mi/hr s 1000 m 1.609 km min hr
100. m 100. m 1km 60 s 60 min 44. = 10.2 m/s; = 36.8 km/hr 9.77 s 9.77 s 1000 m min hr
100. m 1.0936 yd 3 ft 33.6 ft 1mi 60 s 60 min = 33.6 ft/s; = 22.9 mi/hr 9.77 s m yd s 5280 ft min hr
1 m 9.77 s 1.00 × 102 yd × = 8.93 s 1.0936 yd 100. m
65 km 0.6214 mi 45. = 40.4 = 40. mi/hr hr km
To the correct number of significant figures, 65 km/hr does not violate a 40. mi/hr speed limit.
0.6214 mi 1 hr 46. 112 km × = 1.1 hr = 1 hr and 6 min km 65 mi
0.6214 mi 1gal 3.785 L 112 km × = 9.4 L of gasoline km 28 mi gal CHAPTER 1 CHEMICAL FOUNDATIONS 11
$17.25 U.S. $1.00 Canadian 1gal 47. = $0.68 Canadian/L 8.21gal $0.82 U.S. 3.7854 L
80. mg acet 48. 1.5 teaspoons × = 240 mg acetaminophen 0.50 teaspoon
240 mg acet 1lb = 22 mg acetaminophen/kg 24 lb 0.454 kg
240 mg acet 1lb = 15 mg acetaminophen/kg 35 lb 0.454 kg The range is from 15 mg to 22 mg acetaminophen per kg of body weight.
Temperature
5 5 49. a. TC = (TF 32) = (459C 32) = 273C; TK = TC + 273 = 273C + 273 = 0 K 9 9 5 b. TC = (40.F 32) = 40.C; TK = 40.C + 273 = 233 K 9 5 c. TC = (68F 32) = 20.C; TK = 20.C + 273 = 293 K 9
5 7 7 7 7 d. TC = (7 × 10 F 32) = 4 × 10 C; TK = 4 × 10 C + 273 = 4 × 10 K 9 5 5 50. 96.1°F ± 0.2°F; First, convert 96.1°F to °C. TC = (TF - 32) = (96.1 - 32) = 35.6°C 9 9
A change in temperature of 9°F is equal to a change in temperature of 5°C. So the uncertainty is:
5 C ± 0.2°F × = ± 0.1°C. Thus, 96.1 ± 0.2°F = 35.6 ± 0.1°C 9 F 9 9 51. a. TF = × TC + 32 = × 39.2°C + 32 = 102.6°F (Note: 32 is exact.) 5 5
TK = TC + 273.2 = 39.2 + 273.2 = 312.4 K
9 b. TF = × (25) + 32 = 13°F; TK = 25 + 273 = 248 K 5 9 c. TF = × (273) + 32 = -459°F; TK = 273 + 273 = 0 K 5 9 d. TF = × 801 + 32 = 1470°F; TK = 801 + 273 = 1074 K 5 12 CHAPTER 1 CHEMICAL FOUNDATIONS
52. a. TC = TK 273 = 233 273 = -40.°C 9 9 TF = × TC + 32 = × (40.) + 32 = 40.°F 5 5 9 b. T = 4 273 = 269°C; T = × (269) + 32 = 452°F C F 5 9 c. T = 298 273 = 25°C; T = × 25 + 32 = 77°F C F 5 9 d. T = 3680 273 = 3410°C; T = × 3410 + 32 = 6170°F C F 5
Density 3 453.6 g 2.54 cm 53. mass = 350 lb = 1.6 × 105 g; V = 1.2 × 104 in3 = 2.0 × 105 cm3 lb in mass 1 105 g density = 0.80 g / cm3 volume 2.0 105 cm3
Because the material has a density less than water, it will float in water.
4 3 4 3 3 2.0 g 54. V πr 3.14 (0.50 cm) 0.52 cm ; d = 3.8 g/cm3 3 3 0.52 cm3 The ball will sink.
3 4 4 1000 m 100 cm 55. V π r3 3.14 7.0 105 km 1.4 1033 cm3 3 3 km m 1000 g 2 1036 kg mass kg 6 3 6 3 density = = 1.4 × 10 g/cm = 1 × 10 g/cm volume 1.4 1033 cm3
56. V = l × w × h = 2.9 cm × 3.5 cm × 10.0 cm = 1.0 × 102 cm3
615.0 g 6.2 g d = density = 1.0 102 cm3 cm3
0.200 g 1cm3 57. 5.0 carat = 0.28 cm3 carat 3.51g
1cm3 3.51g 1carat 58. 2.8 mL = 49 carats mL cm3 0.200 g 33.42 g 59. V = 21.6 mL 12.7 mL = 8.9 mL; density = = 3.8 g/mL = 3.8 g/cm3 8.9 mL CHAPTER 1 CHEMICAL FOUNDATIONS 13
1cm3 60. 5.25 g = 0.500 cm3 = 0.500 mL 10.5 g The volume in the cylinder will rise to 11.7 mL (11.2 mL + 0.500 mL = 11.7 mL).
61. a. Both have the same mass of 1.0 kg.
b. 1.0 mL of mercury; Mercury has a greater density than water. Note: 1 mL = 1 cm3
13.6 g 0.998 g 1.0 mL = 14 g of mercury; 1.0 mL = 1.0 g of water mL mL
c. Same; Both represent 19.3 g of substance. 0.9982 g 19.32 g 19.3 mL = 19.3 g of water; 1.00 mL = 19.3 g of gold mL mL
d. 1.0 L of benzene (880 g vs 670 g)
8.96 g 1000 mL 0.880 g 75 mL = 670 g of copper; 1.0 L = 880 g of benzene mL L mL
2 5280 ft 62. Volume of lake = 100 mi2 × × 20 ft = 6 × 1010 ft3 mi
3 12 in 2.54 cm 1 mL 0.4 μg 6 × 1010 ft3 × = 7 × 1014 μg mercury ft in cm3 mL 1g 1kg 7 × 1014 μg × = 7 × 105 kg of mercury 106 μg 103 g
63. a. 1.0 kg feather; Feathers are less dense than lead.
b. 100 g water; Water is less dense than gold. c. Same; Both volumes are 1.0 L.
1cm3 64. a. H (g): V = 25.0 g × = 3.0 × 105 cm3 [H (g) = hydrogen gas] 2 0.000084 g 2 1cm3 b. H O(l): V = 25.0 g × = 25.0 cm3 [H O(l) = water] 2 0.9982 g 2 1cm3 c. Fe(s): V = 25.0 g × = 3.18 cm3 [Fe(s) = iron] 7.87 g
Notice the huge volume of the gaseous H2 sample as compared to the liquid and solid samples. The same mass of gas occupies a volume that is over 10,000 times larger than the liquid sample. Gases are indeed mostly empty space. 14 CHAPTER 1 CHEMICAL FOUNDATIONS
1cm3 65. V = 1.00 × 103 g × = 44.3 cm3 22.57 g 44.3 cm3 = 1 × w × h = 4.00 cm × 4.00 cm × h, h = 2.77 cm
1cm3 66. V = 22 g × = 2.5 cm3; V = π r2 × l where l = length of the wire 8.96 g 2 2 3 0.25 mm 1cm 3 2.5 cm = π × × × l, l = 5.1 × 10 cm = 170 ft 2 10 mm
Classification and Separation of Matter
67. A gas has molecules that are very far apart from each other while a solid or liquid has molecules that are very close together. An element has the same type of atom, whereas a compound contains two or more different elements. Picture i represents an element that
exists as two atoms bonded together (like H2 or O2 or N2). Picture iv represents a compound (like CO, NO, or HF). Pictures iii and iv contain representations of elements that exist as individual atoms (like Ar, Ne, or He).
a. Picture iv represents a gaseous compound. Note that pictures ii and iii also contain a gaseous compound, but they also both have a gaseous element present.
b. Picture vi represents a mixture of two gaseous elements.
c. Picture v represents a solid element.
d. Pictures ii and iii both represent a mixture of a gaseous element and a gaseous compound.
68. Solid: rigid; has a fixed volume and shape; slightly compressible
Liquid: definite volume but no specific shape; assumes shape of the container; slightly compressible
Gas: no fixed volume or shape; easily compressible
Pure substance: has constant composition; can be composed of either compounds or ele- ments
Element: substances that cannot be decomposed into simpler substances by chemical or physical means.
Compound: a substance that can be broken down into simpler substances (elements) by chemical processes.
Homogeneous mixture: a mixture of pure substances that has visibly indistinguishable parts. CHAPTER 1 CHEMICAL FOUNDATIONS 15
Heterogeneous mixture: a mixture of pure substances that has visibly distinguishable parts. Solution: a homogeneous mixture; can be a solid, liquid or gas
Chemical change: a given substance becomes a new substance or substances with different properties and different composition.
Physical change: changes the form (g, l, or s) of a substance but does no change the chemical composition of the substance.
69. Homogeneous: Having visibly indistinguishable parts (the same throughout). Heterogeneous: Having visibly distinguishable parts (not uniform throughout).
a. heterogeneous (Due to hinges, handles, locks, etc.)
b. homogeneous (hopefully; If you live in a heavily polluted area, air may be heterogeneous.)
c. homogeneous d. homogeneous (hopefully, if not polluted)
e. heterogeneous f. heterogeneous
70. a. pure b. mixture c. mixture d. pure e. mixture (copper and zinc)
f. pure g. mixture h. mixture i. pure
Iron and uranium are elements. Water and table salt are compounds. Water is H2O and table salt is NaCl. Compounds are composed of two or more elements.
71. A physical change is a change in the state of a substance (solid, liquid and gas are the three states of matter); a physical change does not change the chemical composition of the substance. A chemical change is a change in which a given substance is converted into another substance having a different formula (composition).
a. Vaporization refers to a liquid converting to a gas, so this is a physical change. The formula (composition) of the moth ball does not change.
b. This is a chemical change since hydrofluoric acid (HF) is reacting with glass (SiO 2) to form new compounds which wash away.
c. This is a physical change since all that is happening is the conversion of liquid alcohol to
gaseous alcohol. The alcohol formula (C2H5OH) does not change.
d. This is a chemical change since the acid is reacting with cotton to form new compounds.
72. a. Distillation separates components of a mixture, so the orange liquid is a mixture (has an average color of the yellow liquid and the red solid). Distillation utilizes boiling point differences to separate out the components of a mixture. Distillation is a physical change because the components of the mixture do not become different compounds or elements. 16 CHAPTER 1 CHEMICAL FOUNDATIONS
b. Decomposition is a type of chemical reaction. The crystalline solid is a compound, and decomposition is a chemical change where new substances are formed.
c. Tea is a mixture of tea compounds dissolved in water. The process of mixing sugar into tea is a physical change. Sugar doesn’t react with the tea compounds, it just makes the solution sweeter.
Additional Exercises
1capsule 73. 15.6 g × = 24 capsules 0.65 g
4 qt 1 L 74. 126 gal × = 477 L gal 1.057 qt
100 cm 100 cm 75. Total volume = 200. m 300. m × 4.0 cm = 2.4 × 109 cm3 m m
2 2 12 in 2.54 cm 2.54 cm Vol. of topsoil covered by 1 bag = 10. ft 2 1.0 in ft in in = 2.4 × 104 cm3 1 bag 2.4 × 109 cm3 = 1.0 × 105 bags topsoil 2.4 104 cm3
76. a. No; If the volumes were the same, then the gold idol would have a much greater mass because gold is much more dense than sand.
1000 cm3 19.32 g 1 kg b. Mass = 1.0 L = 19.32 kg (= 42.59 lb) L cm3 1000 g
It wouldn't be easy to play catch with the idol because it would have a mass of over 40 pounds.
10.0o F 77. 18.5 cm × = 35.2F increase; T = 98.6 + 35.2 = 133.8F 5.25 cm final
Tc = 5/9 (133.8 – 32) = 56.56C 1cm3 78. mass = 58.80 g 25.00 g = 33.80 g; V = 33.80 g = 38.4 cm3 benzene benzene 0.880 g
25.00 g V = 50.0 cm3 38.4 cm3 = 11.6 cm3; density = = 2.16 g/cm3 solid 11.6 cm3 CHAPTER 1 CHEMICAL FOUNDATIONS 17
79. a. Volume × density = mass; the orange block is more dense. Because mass (orange) > mass (blue) and because volume (orange) < volume (blue), the density of the orange block must be greater to account for the larger mass of the orange block.
b. Which block is more dense cannot be determined. Because mass (orange) > mass (blue) and because volume (orange) > volume (blue), the density of the orange block may or may not be larger than the blue block. If the blue block is more dense, its density cannot be so large that its mass is larger than the orange block’s mass.
c. The blue block is more dense. Because mass (blue ) = mass (orange) and because volume (blue) < volume (orange), the density of the blue block must be larger in order to equate the masses.
d. The blue block is more dense. Because mass (blue) > mass (orange) and because the volumes are equal, the density of the blue block must be larger in order to give the blue block the larger mass.
3 4π r3 4π c c3 80. Circumference = c = 2π r; V = 2 3 3 2π 6π
5.25 oz 5.25 oz 0.427 oz Largest density = (9.00 in)3 = = 12.3 in 3 in 3 6π 2
5.00 oz 5.00 oz 0.73 oz Smallest density = (9.25 in)3 = = 13.4 in 3 in 3 6π2 (0.373 0.427) oz Maximum range is: or 0.40 ± 0.03 oz/in3 (Uncertainty in 2nd decimal in 3 place.)
28.90 g 28.90 g 81. V = V - V ; d = = 8.5 g/cm3 final initial 9.8 cm3 6.4 cm3 3.4 cm3
mass max 3 3 3 dmax = ; We get Vmin from 9.7 cm 6.5 cm = 3.2 cm . Vmin
28.93 g 9.0 g massmin 28.87 g 8.0 g dmax = 3 3 ; dmin = = 3 3 3 3.2 cm cm Vmax 9.9 cm 6.3 cm cm
The density is: 8.5 ± 0.5 g/cm3.
Challenge Problems
82. In general, glassware is estimated to one place past the markings. 18 CHAPTER 1 CHEMICAL FOUNDATIONS
a. 128.7 mL glassware b. 18 mL glassware c. 23.45 mL glassware
24 130 30 129 20 128 23 127 10
read to tenth’s place read to one’s place read to two decimal places
128.7 + 18 + 23.45 = 170.15 = 170. (Due to 18, the sum would be only known to the ones place.)
2.70 2.64 |16.12 16.48| 83. a. × 100 = 2% b. × 100 = 2.2% 2.70 16.12
1.000 0.9981 0.002 c. × 100 = × 100 = 0.2% 1.000 1.000
84. a. At some point in 1982, the composition of the metal used in minting pennies was changed because the mass changed during this year (assuming the volume of the pennies were constant).
b. It should be expressed as 3.08 ± 0.05 g. The uncertainty in the second decimal place will swamp any effect of the next decimal places.
85. Heavy pennies (old): mean mass = 3.08 ± 0.05 g
(2.467 2.545 2.518) Light pennies (new): mean mass = = 2.51 ± 0.04 g 3
Because we are assuming that volume is additive, let’s calculate the volume of 100. g of each type of penny then calculate the density of the alloy. For 100. g of the old pennies, 95 g will be Cu (copper) and 5 g will be Zn (zinc).
1cm3 1cm3 V = 95 g Cu × 5 g Zn = 11.3 cm3 (carrying one extra sig. fig.) 8.96g 7.14g 100. g Density of old pennies = = 8.8 g/cm3 11.3 cm 3
For 100. g of new pennies, 97.6 g will be Zn and 2.4 g will be Cu. CHAPTER 1 CHEMICAL FOUNDATIONS 19
1cm3 1cm3 V = 2.4 g Cu × + 97.6 g Zn × = 13.94 cm3 (carrying one extra sig. fig.) 8.96g 7.14g 100. g Density of new pennies = = 7.17 g/cm3 13.94 cm3 mass d = ; Because the volume of both types of pennies are assumed equal, then: volume
3 d new massnew 7.17 g / cm 3 = 0.81 d old massold 8.8 g / cm mass 2.51g The calculated average mass ratio is: new = 0.815 massold 3.08 g
To the first two decimal places, the ratios are the same. If the assumptions are correct, then we can reasonably conclude that the difference in mass is accounted for by the difference in alloy used.
86. a. o o A change in temperature of 160°C equals a 100 A 115 C change in temperature of 100°A.
100oA 160oC 160C So, is our unit conversion for a 100A degree change in temperature. o o 0 A -45 C At the freezing point: 0°A = -45°C
Combining the two pieces of information: 100A 5A 8C TA = (TC + 45°C) × = (TC + 45°C) × or TC = TA × 45°C 160C 8C 5A 5 8 5 b. TC = (TF - 32) × ; TC = TA × 45 = (TF - 32) × 9 5 9
9 8 72 72F TF 32 = TA 45 = TA 81, TF = TA × 49F 5 5 25 25A
8 8 3 Tc c. T = T × 45 and T = T ; So, T = T × 45, = 45, T = 75°C = 75°A C A 5 C A C C 5 5 C
8A 72F 2 d. TC = 86°A × 45°C = 93°C; TF = 86°A × 49°F = 199°F = 2.0 × 10 °F 5C 25A 5A e. TA = (45°C + 45°C) × = 56°A 8C
87. Let x = mass of copper and y = mass of silver. 20 CHAPTER 1 CHEMICAL FOUNDATIONS
x y 105.0 g = x + y and 10.12 mL = ; Solving: 8.96 10.5 x 105.0 x 10.12 × 8.96 × 10.5, 952.1 = 10.5 x + 940.8 8.96 x 8.96 10.5 (carrying 1 extra significant figure) 7.3 g 11.3 = 1.54 x, x = 7.3 g; mass %Cu = × 100 = 7.0% Cu 105.0 g
88. a.
2 compounds compound and element (diatomic)
b.
solid element gas element (monoatomic) liquid element atoms/molecules far apart; atoms/molecules close atoms/molecules random order; takes volume together; somewhat close together; of container ordered arrangement; ordered arrangement; takes volume of container has its own volume
89. a. One possibility is that rope B is not attached to anything and rope A and rope C are connected via a pair of pulleys and/or gears.
b. Try to pull rope B out of the box. Measure the distance moved by C for a given movement of A. Hold either A or C firmly while pulling on the other.
90. The bubbles of gas is air in the sand that is escaping; methanol and sand are not reacting. We will assume that the mass of trapped air is insignificant.
mass of dry sand = 37.3488 g 22.8317 g = 14.5171 g CHAPTER 1 CHEMICAL FOUNDATIONS 21
mass of methanol = 45.2613 g 37.3488 g = 7.9125 g
Volume of sand particles (air absent) = volume of sand and methanol volume of methanol
Volume of sand particles (air absent) = 17.6 mL 10.00 mL = 7.6 mL 14.5171g density of dry sand (air present) = = 1.45 g/mL 10.0 mL 7.9125 g density of methanol = = 0.7913 g/mL 10.00 mL 14.5171g density of sand particles (air absent) = = 1.9 g/mL 7.6 mL
Integrative Problems
91. 2.97 108 persons 0.0100 = 2.97 106 persons contributing
$4.75 108 $160. 20 nickels = $160./person; = 3.20 × 103 nickels/person 2.97 106 persons person $1
$160. 1pound sterling = 85.6 pounds sterling/person person $1.869 22610 kg 1000 g 1m3 92. = 22.61 g/cm3 m3 kg 1106 cm3 22.61g volume of block = 10.0 cm 8.0 cm 9.0 cm = 720 cm3; × 720 cm3 = 1.6 × 104 g cm3 5 93. At 200.0 F: TC = (200.0 F 32 F) = 93.33 C; TK = 93.33 + 273.15 = 366.48 K 9 5 At 100.0 F: TC = (100.0 F 32 F) = 73.33 C; TK = 73.33 C + 273.15 9 = 199.82 K
T(C) = [93.33 C (73.33 C)] = 166.66 C; T(K) = [366.48 K 199.82 K] = 166.66 K
The “300 Club” name only works for the Fahrenheit scale; it does not hold true for the Celsius and Kelvin scales.
Marathon Problem 3 2 2 2.54 cm 3 94. a. Vgold = r h = 3.14 × (0.25/2 in) × 1.5 in × = 1.2 cm in 23.1984 g 3 dgold (at 86 F) = = 19 g/cm 1.2 cm3 22 CHAPTER 1 CHEMICAL FOUNDATIONS
b. Calculate the density of the liquid at 86F, then determine the density at 40.F.
massliquid = 79.16 g 73.47 g = 5.69 g
3 3 3 3 volumefinal = 8.5 cm = Vgold + Vliquid, Vliquid = 8.5 cm 1.2 cm = 7.3 cm
5.69 g d (at 86 F) = = 0.78 g/cm3 liquid 7.3 cm3
The density will increase by 1.0 % for every 10. C drop in temperature. The temperature drop is 86 40. = 46 F. Because 1F is equivalent to 5/9C, the temperature drop in C equals 46(5/9) = 26C. Because there is a 1% increase in density for every 10.C drop in temperature, there will be a 2.6% increase in density for the 26C temperature drop.
3 3 densityliquid (at 40.C) = 1.026 × 0.78 g/cm = 0.80 g/cm