Chapter 1 Chemistry The Study of Change

CHAPTER 1 CHEMISTRY: THE STUDY OF CHANGE 1.3 (a) Quantitative. This statement clearly involves a measurable distance. (b) Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic excellence. (c) Qualitative. If the numerical values for the densities of ice and water were given, it would be a quantitative statement. (d) Qualitative. Another value judgment. (e) Qualitative. Even though numbers are involved, they are not the result of measurement. 1.4 (a) hypothesis 1.11 (a) Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are changed. (b) Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter (different composition). (c) Physical property. The measurement of the boiling point of water does not change its identity or composition. (d) Physical property. The measurement of the densities of lead and aluminum does not change their composition. (e) Chemical property. When uranium undergoes nuclear decay, the products are chemically different substances. (a) Physical change. The helium isn't changed in any way by leaking out of the balloon. (b) Chemical change in the battery. (c) Physical change. The orange juice concentrate can be regenerated by evaporation of the water. (d) Chemical change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter. (e) Physical change. The salt can be recovered unchanged by evaporation. 1.12 (b) law (c) theory 1.13 Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum; Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon. 1.14 (a) (f) K Pu 1.15 (a) element 1.16 (a) (d) (g) homogeneous mixture homogeneous mixture heterogeneous mixture (b) (g) Sn S (b) (c) (h) compound (b) (e) Cr Ar (d) (i) (c) B Hg element element heterogeneous mixture (e) (d) (c) (f) Ba compound compound homogeneous mixture 2 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.21 density = 1.22 Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid. Rearrange the density equation, Equation (1.1) of the text, to solve for mass. mass 586 g = = 3.12 g/mL volume 188 mL density = mass volume Solution: mass = density × volume mass of ethanol = 1.23 ? °C = (°F − 32°F) × 0.798 g × 17.4 mL = 13.9 g 1 mL 5°C 9°F 5°C = 35°C 9°F 5°C = − 11°C (12°F − 32°F) × 9 °F 5°C (102°F − 32°F) × = 39°C 9°F 5°C = 1011°C (1852°F − 32°F) × 9°F 9°F ⎞ ⎛ ⎜ °C × 5°C ⎟ + 32°F ⎝ ⎠ (a) ? °C = (95°F − 32°F) × (b) ? °C = (c) ? °C = (d) ? °C = (e) ? °F = 9°F ⎞ ⎛ ? °F = ⎜ −273.15 °C × + 32°F = − 459.67°F °C ⎟⎠ 5 ⎝ 1.24 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the problem into the appropriate equation. (a) Conversion from Fahrenheit to Celsius. ? °C = (°F − 32°F) × 5°C 9°F ? °C = (105°F − 32°F) × (b) 5°C = 41°C 9°F Conversion from Celsius to Fahrenheit. 9°F ⎞ ⎛ ? °F = ⎜ °C × + 32°F 5°C ⎟⎠ ⎝ 9°F ⎞ ⎛ ? °F = ⎜ −11.5 °C × + 32°F = 11.3 °F 5 °C ⎟⎠ ⎝ CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE (c) Conversion from Celsius to Fahrenheit. 9°F ⎞ ⎛ ? °F = ⎜ °C × + 32°F 5 °C ⎟⎠ ⎝ 9°F ⎞ ⎛ ? °F = ⎜ 6.3 × 103 °C × + 32°F = 1.1 × 104 °F 5°C ⎟⎠ ⎝ (d) Conversion from Fahrenheit to Celsius. ? °C = (°F − 32°F) × 5°C 9°F ? °C = (451°F − 32°F) × 1.25 1.26 K = (°C + 273°C) 1K 1°C (a) K = 113°C + 273°C = 386 K (b) K = 37°C + 273°C = 3.10 × 10 K (c) K = 357°C + 273°C = 6.30 × 10 K (a) 2 2 1K 1°C °C = K − 273 = 77 K − 273 = −196°C K = (°C + 273°C) (b) °C = 4.2 K − 273 = −269°C (c) °C = 601 K − 273 = 328°C 1.29 (a) 2.7 × 10 1.30 (a) 10 −2 −8 3.56 × 10 (b) 2 (c) 10 −8 (a) (b) 4 (d) −2 = 0.0152 indicates that the decimal point must be moved 8 places to the left. 7.78 × 10 1.31 4.7764 × 10 indicates that the decimal point must be moved two places to the left. 1.52 × 10 (b) 5°C = 233°C 9°F −8 −1 = 0.0000000778 145.75 + (2.3 × 10 ) = 145.75 + 0.23 = 1.4598 × 10 79500 2.5 × 102 = 7.95 × 104 2.5 × 102 −3 2 = 3.2 × 102 −4 −3 −3 (c) (7.0 × 10 ) − (8.0 × 10 ) = (7.0 × 10 ) − (0.80 × 10 ) = 6.2 × 10 (d) (1.0 × 10 ) × (9.9 × 10 ) = 9.9 × 10 4 6 10 −3 9.6 × 10 −2 3 4 1.32 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE (a) Addition using scientific notation. n Strategy: Let's express scientific notation as N × 10 . When adding numbers using scientific notation, we must write each quantity with the same exponent, n. We can then add the N parts of the numbers, keeping the exponent, n, the same. Solution: Write each quantity with the same exponent, n. 3 n 3 Let’s write 0.0095 in such a way that n = −3. We have decreased 10 by 10 , so we must increase N by 10 . Move the decimal point 3 places to the right. 0.0095 = 9.5 × 10 −3 Add the N parts of the numbers, keeping the exponent, n, the same. −3 9.5 × 10 −3 + 8.5 × 10 18.0 × 10 −3 The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of n 10 to express N between 1 and 10 (1.8), we must increase 10 by a factor of 10. The exponent, n, is increased by 1 from −3 to −2. 18.0 × 10 (b) −3 = 1.8 × 10 −2 Division using scientific notation. n Strategy: Let's express scientific notation as N × 10 . When dividing numbers using scientific notation, divide the N parts of the numbers in the usual way. To come up with the correct exponent, n, we subtract the exponents. Solution: Make sure that all numbers are expressed in scientific notation. 653 = 6.53 × 10 2 Divide the N parts of the numbers in the usual way. 6.53 ÷ 5.75 = 1.14 Subtract the exponents, n. 1.14 × 10 (c) +2 − (−8) = 1.14 × 10 +2 + 8 = 1.14 × 10 10 Subtraction using scientific notation. n Strategy: Let's express scientific notation as N × 10 . When subtracting numbers using scientific notation, we must write each quantity with the same exponent, n. We can then subtract the N parts of the numbers, keeping the exponent, n, the same. Solution: Write each quantity with the same exponent, n. Let’s write 850,000 in such a way that n = 5. This means to move the decimal point five places to the left. 850,000 = 8.5 × 10 5 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 5 Subtract the N parts of the numbers, keeping the exponent, n, the same. 5 8.5 × 10 5 − 9.0 × 10 −0.5 × 10 5 The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10 n to express N between 1 and 10 (5), we must decrease 10 by a factor of 10. The exponent, n, is decreased by 1 from 5 to 4. 5 −0.5 × 10 = −5 × 10 (d) 4 Multiplication using scientific notation. n Strategy: Let's express scientific notation as N × 10 . When multiplying numbers using scientific notation, multiply the N parts of the numbers in the usual way. To come up with the correct exponent, n, we add the exponents. Solution: Multiply the N parts of the numbers in the usual way. 3.6 × 3.6 = 13 Add the exponents, n. 13 × 10 −4 + (+6) = 13 × 10 2 The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of n 10 to express N between 1 and 10 (1.3), we must increase 10 by a factor of 10. The exponent, n, is increased by 1 from 2 to 3. 2 3 13 × 10 = 1.3 × 10 1.33 (a) (e) four three 1.34 (a) (e) one two or three 1.35 (a) 10.6 m 1.36 (a) Division (b) (f) two one (b) (f) (b) three one 0.79 g (c) (c) (g) five one (c) (g) three one or two (d) (h) two, three, or four two (d) four 2 16.5 cm Strategy: The number of significant figures in the answer is determined by the original number having the smallest number of significant figures. Solution: 7.310 km = 1.283 5.70 km The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits. Therefore, the answer has only three significant digits. The correct answer rounded off to the correct number of significant figures is: 1.28 (Why are there no units?) 6 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE (b) Subtraction Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers. Solution: Writing both numbers in decimal notation, we have 0.00326 mg − 0.0000788 mg 0.0031812 mg The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the decimal point. Therefore, we carry five digits to the right of the decimal point in our answer. The correct answer rounded off to the correct number of significant figures is: 0.00318 mg = 3.18 × 10 (c) −3 mg Addition Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers. Solution: Writing both numbers with exponents = +7, we have 7 7 7 (0.402 × 10 dm) + (7.74 × 10 dm) = 8.14 × 10 dm 7 Since 7.74 × 10 has only two digits to the right of the decimal point, two digits are carried to the right of the decimal point in the final answer. 1.37 1.38 1 dm = 226 dm 0.1 m (a) ? dm = 22.6 m × (b) ? kg = 25.4 mg × (c) ? L = 556 mL × (d) ? g cm 3 = 10.6 kg 1 m3 0.001 g 1 kg × = 2.54 × 10−5 kg 1 mg 1000 g 1 × 10−3 L = 0.556 L 1 mL 3 × 1000 g ⎛ 1 × 10−2 m ⎞ ×⎜ ⎟ = 0.0106 g/cm 3 ⎜ 1 cm ⎟ 1 kg ⎝ ⎠ (a) Strategy: The problem may be stated as ? mg = 242 lb A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This relationship will allow conversion from pounds to grams. A metric conversion is then needed to convert −3 grams to milligrams (1 mg = 1 × 10 g). Arrange the appropriate conversion factors so that pounds and grams cancel, and the unit milligrams is obtained in your answer. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 7 Solution: The sequence of conversions is lb → grams → mg Using the following conversion factors, 1 mg 453.6 g 1 lb 1 × 10−3 g we obtain the answer in one step: ? mg = 242 lb × 453.6 g 1 mg × = 1.10 × 108 mg 1 lb 1 × 10−3 g Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many mg are in 1 lb? There are 453,600 mg in 1 lb. (b) Strategy: The problem may be stated as 3 3 ? m = 68.3 cm Recall that 1 cm = 1 × 10 −2 3 3 m. We need to set up a conversion factor to convert from cm to m . Solution: We need the following conversion factor so that centimeters cancel and we end up with meters. 1 × 10−2 m 1 cm Since this conversion factor deals with length and we want volume, it must therefore be cubed to give ⎛ 1 × 10−2 m ⎞ 1 × 10−2 m 1 × 10−2 m 1 × 10−2 m × × = ⎜ ⎟ ⎜ 1 cm ⎟ 1 cm 1 cm 1 cm ⎝ ⎠ 3 We can write 3 ⎛ 1 × 10−2 m ⎞ ? m = 68.3 cm × ⎜ ⎟ = 6.83 × 10−5 m 3 ⎜ 1 cm ⎟ ⎝ ⎠ 3 3 3 Check: We know that 1 cm = 1 × 10 −6 −5 1 × 10 gives 6.83 × 10 . −6 3 1 3 m . We started with 6.83 × 10 cm . Multiplying this quantity by (c) Strategy: The problem may be stated as 3 ? L = 7.2 m 3 3 In Chapter 1 of the text, a conversion is given between liters and cm (1 L = 1000 cm ). If we can convert 3 3 −2 m to cm , we can then convert to liters. Recall that 1 cm = 1 × 10 m. We need to set up two conversion 3 3 3 factors to convert from m to L. Arrange the appropriate conversion factors so that m and cm cancel, and the unit liters is obtained in your answer. 8 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE Solution: The sequence of conversions is 3 3 m → cm → L Using the following conversion factors, ⎛ 1 cm ⎞ ⎜ ⎟ ⎜ 1 × 10−2 m ⎟ ⎝ ⎠ 3 1L 1000 cm3 the answer is obtained in one step: 3 ⎛ 1 cm ⎞ 1L = 7.2 × 103 L ? L = 7.2 m3 × ⎜ ⎟ × ⎜ 1 × 10−2 m ⎟ 1000 cm3 ⎝ ⎠ 3 3 3 Check: From the above conversion factors you can show that 1 m = 1 × 10 L. Therefore, 7 m would 3 equal 7 × 10 L, which is close to the answer. (d) Strategy: The problem may be stated as ? lb = 28.3 μg A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This relationship will allow conversion from grams to pounds. If we can convert from μg to grams, we can then −6 convert from grams to pounds. Recall that 1 μg = 1 × 10 g. Arrange the appropriate conversion factors so that μg and grams cancel, and the unit pounds is obtained in your answer. Solution: The sequence of conversions is μg → g → lb Using the following conversion factors, 1 × 10−6 g 1 μg 1 lb 453.6 g we can write ? lb = 28.3 μg × 1 × 10−6 g 1 lb × = 6.24 × 10−8 lb 1 μg 453.6 g Check: Does the answer seem reasonable? What number does the prefix μ represent? Should 28.3 μg be a very small mass? 1.39 1255 m 1 mi 3600 s × × = 2808 mi/h 1s 1609 m 1h 1.40 Strategy: The problem may be stated as ? s = 365.24 days You should know conversion factors that will allow you to convert between days and hours, between hours and minutes, and between minutes and seconds. Make sure to arrange the conversion factors so that days, hours, and minutes cancel, leaving units of seconds for the answer. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE Solution: The sequence of conversions is days → hours → minutes → seconds Using the following conversion factors, 24 h 1 day 60 min 1h 60 s 1 min we can write ? s = 365.24 day × 24 h 60 min 60 s × × = 3.1557 × 107 s 1 day 1h 1 min Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year? 1.609 km 1000 m 1s 1 min × × × = 8.3 min 8 1 mi 1 km 60 s 3.00 × 10 m 1.41 (93 × 106 mi) × 1.42 (a) ? in/s = (b) ? m/min = (c) ? km/h = 1.43 6.0 ft × 168 lb × 1 mi 5280 ft 12 in 1 min × × × = 81 in/s 13 min 1 mi 1 ft 60 s 1 mi 1609 m × = 1.2 × 102 m/min 13 min 1 mi 1 mi 1609 m 1 km 60 min × × × = 7.4 km/h 13 min 1 mi 1000 m 1h 1m = 1.8 m 3.28 ft 453.6 g 1 kg × = 76.2 kg 1 lb 1000 g 55 mi 1.609 km × = 88 km/h 1h 1 mi 1.44 ? km/h = 1.45 62 m 1 mi 3600 s × × = 1.4 × 102 mph 1s 1609 m 1h 1.46 0.62 ppm Pb = 0.62 g Pb 1 × 106 g blood 0.62 g Pb 6.0 × 103 g of blood × = 3.7 × 10−3 g Pb 6 1 × 10 g blood 9 10 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.47 (a) 1.42 yr × 365 day 24 h 3600 s 3.00 × 108 m 1 mi × × × × = 8.35 × 1012 mi 1 yr 1 day 1h 1s 1609 m (b) 32.4 yd × 36 in 2.54 cm × = 2.96 × 103 cm 1 yd 1 in (c) 3.0 × 1010 cm 1 in 1 ft × × = 9.8 × 108 ft/s 1s 2.54 cm 12 in (a) ? m = 185 nm × (b) ? s = (4.5 × 109 yr) × (c) ⎛ 0.01 m ⎞ −5 3 ? m 3 = 71.2 cm3 × ⎜ ⎟ = 7.12 × 10 m 1 cm ⎝ ⎠ (d) ⎛ 1 cm ⎞ 1L ? L = 88.6 m × ⎜ = 8.86 × 104 L ⎟ × ⎜ 1 × 10−2 m ⎟ 1000 cm3 ⎝ ⎠ 1.48 1 × 10−9 m = 1.85 × 10−7 m 1 nm 365 day 24 h 3600 s × × = 1.4 × 1017 s 1 yr 1 day 1h 3 3 3 2.70 g 3 density = 1.50 density = 1.51 Substance (a) water (b) carbon (c) iron (d) hydrogen gas (e) sucrose (f) table salt (g) mercury (h) gold (i) air 1.52 See Section 1.6 of your text for a discussion of these terms. 1 cm3 × ⎛ 1 cm ⎞ 1 kg 3 3 ×⎜ ⎟ = 2.70 × 10 kg/m 1000 g ⎝ 0.01 m ⎠ 1.49 0.625 g 1L 1 mL × × = 6.25 × 10−4 g/cm 3 1L 1000 mL 1 cm3 Qualitative Statement colorless liquid black solid (graphite) rusts easily colorless gas tastes sweet tastes salty liquid at room temperature a precious metal a mixture of gases Quantitative Statement freezes at 0°C 3 density = 2.26 g/cm 3 density = 7.86 g/cm melts at −255.3°C at 0°C, 179 g of sucrose dissolves in 100 g of H2O melts at 801°C boils at 357°C 3 density = 19.3 g/cm contains 20% oxygen by volume (a) Chemical property. Iron has changed its composition and identity by chemically combining with oxygen and water. (b) Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids, thus changing the composition and identity of the water. (c) Physical property. The color of the hemoglobin can be observed and measured without changing its composition or identity. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE (d) Physical property. The evaporation of water does not change its chemical properties. Evaporation is a change in matter from the liquid state to the gaseous state. (e) Chemical property. The carbon dioxide is chemically converted into other molecules. 1 ton 1.53 (95.0 × 109 lb of sulfuric acid) × 1.54 Volume of rectangular bar = length × width × height density = 1.55 11 3 2.0 × 10 lb = 4.75 × 107 tons of sulfuric acid m 52.7064 g = = 2.6 g/cm 3 V (8.53 cm)(2.4 cm)(1.0 cm) mass = density × volume 3 4 3 4 π(10.0 cm) ] = 8.08 × 10 g 3 (a) mass = (19.3 g/cm ) × [ (b) ⎛ 1 cm ⎞ −6 mass = (21.4 g/cm3 ) × ⎜ 0.040 mm × ⎟ = 1.4 × 10 g 10 mm ⎝ ⎠ (c) mass = (0.798 g/mL)(50.0 mL) = 39.9 g 3 1.56 You are asked to solve for the inner diameter of the tube. If you can calculate the volume that the mercury 2 occupies, you can calculate the radius of the cylinder, Vcylinder = πr h (r is the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2r. volume of Hg filling cylinder = volume of Hg filling cylinder = mass of Hg density of Hg 105.5 g 13.6 g/cm 3 = 7.757 cm3 Next, solve for the radius of the cylinder. 2 Volume of cylinder = πr h r = volume π×h r = 7.757 cm3 = 0.4409 cm π × 12.7 cm The cylinder diameter equals 2r. Cylinder diameter = 2r = 2(0.4409 cm) = 0.882 cm 1.57 From the mass of the water and its density, we can calculate the volume that the water occupies. The volume that the water occupies is equal to the volume of the flask. volume = mass density 12 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE Mass of water = 87.39 g − 56.12 g = 31.27 g Volume of the flask = mass 31.27 g = = 31.35 cm 3 density 0.9976 g/cm3 1.58 343 m 1 mi 3600 s × × = 767 mph 1s 1609 m 1h 1.59 The volume of silver is equal to the volume of water it displaces. 3 Volume of silver = 260.5 mL − 242.0 mL = 18.5 mL = 18.5 cm 194.3 g density = 1.60 18.5 cm3 = 10.5 g/cm 3 In order to work this problem, you need to understand the physical principles involved in the experiment in Problem 1.59. The volume of the water displaced must equal the volume of the piece of silver. If the silver did not sink, would you have been able to determine the volume of the piece of silver? The liquid must be less dense than the ice in order for the ice to sink. The temperature of the experiment must be maintained at or below 0°C to prevent the ice from melting. 1.61 density = mass 1.20 × 104 g = = 11.4 g/cm 3 volume 1.05 × 103 cm3 1.62 Volume = mass density Volume occupied by Li = 1.63 1.20 × 103 g 0.53 g / cm3 = 2.3 × 103 cm 3 For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C. 5°C ? °C = 0.1°F × = 0.056°C 9°F The percent error is the amount of uncertainty in a measurement divided by the value of the measurement, converted to percent by multiplication by 100. Percent error = known error in a measurement × 100% value of the measurement For the Fahrenheit thermometer, percent error = 0.056°C × 100% = 0.1% 38.9°C For the Celsius thermometer, percent error = 0.1°C × 100% = 0.3% 38.9°C Which thermometer is more accurate? CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.64 To work this problem, we need to convert from cubic feet to L. Some tables will have a conversion factor of 3 28.3 L = 1 ft , but we can also calculate it using the dimensional analysis method described in Section 1.9 of the text. First, converting from cubic feet to liters: 3 3 ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ 1 mL 1 × 10−3 L (5.0 × 107 ft 3 ) × ⎜ × × × = 1.42 × 109 L ⎟ ⎜ ⎟ 3 1 ft 1 in 1 mL ⎝ ⎠ ⎝ ⎠ 1 cm The mass of vanillin (in g) is: 2.0 × 10−11 g vanillin × (1.42 × 109 L) = 2.84 × 10−2 g vanillin 1L The cost is: (2.84 × 10−2 g vanillin) × 1.65 $112 = $0.064 = 6.4¢ 50 g vanillin 9°F ⎞ ⎛ ? °F = ⎜ °C × + 32°F 5°C ⎟⎠ ⎝ Let temperature = t 9 t + 32°F 5 9 t − t = 32°F 5 4 − t = 32°F 5 t = t = −40°F = −40°C 1.66 There are 78.3 + 117.3 = 195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor. ⎛ 195.6o C ⎞ ⎜ ⎟ ⎜ 100oS ⎟ ⎝ ⎠ Set up the equation like a Celsius to Fahrenheit conversion. We need to subtract 117.3°C, because the zero point on the new scale is 117.3°C lower than the zero point on the Celsius scale. ⎛ 195.6°C ⎞ ? °C = ⎜ ⎟ (? °S ) − 117.3°C ⎝ 100°S ⎠ 1.67 13 Solving for ? °S gives: ⎛ 100°S ⎞ ? °S = (? °C + 117.3°C) ⎜ ⎟ ⎝ 195.6°C ⎠ For 25°C we have: ⎛ 100°S ⎞ ? °S = (25°C + 117.3°C) ⎜ ⎟ = 73°S ⎝ 195.6°C ⎠ The key to solving this problem is to realize that all the oxygen needed must come from the 4% difference (20% - 16%) between inhaled and exhaled air. The 240 mL of pure oxygen/min requirement comes from the 4% of inhaled air that is oxygen. 14 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 240 mL of pure oxygen/min = (0.04)(volume of inhaled air/min) Volume of inhaled air/min = 240 mL of oxygen/min = 6000 mL of inhaled air/min 0.04 Since there are 12 breaths per min, volume of air/breath = 1.68 1.69 6000 mL of inhaled air 1 min × = 5 × 102 mL/breath 1 min 12 breaths (a) 6000 mL of inhaled air 0.001 L 60 min 24 h × × × = 8.6 × 103 L of air/day 1 min 1 mL 1h 1 day (b) 8.6 × 103 L of air 2.1 × 10−6 L CO × = 0.018 L CO/day 1 day 1 L of air The mass of the seawater is: (1.5 × 1021 L) × 1 mL 1.03 g × = 1.55 × 1024 g = 1.55 × 1021 kg seawater 0.001 L 1 mL Seawater is 3.1% NaCl by mass. The total mass of NaCl in kilograms is: mass NaCl (kg) = (1.55 × 1021 kg seawater) × mass NaCl (tons) = (4.8 × 1019 kg) × 1.70 3.1% NaCl = 4.8 × 1019 kg NaCl 100% seawater 2.205 lb 1 ton × = 5.3 × 1016 tons NaCl 1 kg 2000 lb First, calculate the volume of 1 kg of seawater from the density and the mass. We chose 1 kg of seawater, because the problem gives the amount of Mg in every kg of seawater. The density of seawater is given in Problem 1.69. volume = mass density volume of 1 kg of seawater = 1000 g = 970.9 mL = 0.9709 L 1.03 g/mL In other words, there are 1.3 g of Mg in every 0.9709 L of seawater. Next, let’s convert tons of Mg to grams of Mg. (8.0 × 104 tons Mg) × 2000 lb 453.6 g × = 7.26 × 1010 g Mg 1 ton 1 lb 4 Volume of seawater needed to extract 8.0 × 10 ton Mg = (7.26 × 1010 g Mg) × 0.9709 L seawater = 5.4 × 1010 L of seawater 1.3 g Mg CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.71 15 Assume that the crucible is platinum. Let’s calculate the volume of the crucible and then compare that to the volume of water that the crucible displaces. volume = mass density Volume of crucible = 860.2 g = 40.10 cm 3 21.45 g/cm3 Volume of water displaced = (860.2 − 820.2)g 0.9986 g/cm 3 = 40.1 cm 3 The volumes are the same (within experimental error), so the crucible is made of platinum. 1.72 Volume = surface area × depth 3 2 Recall that 1 L = 1 dm . Let’s convert the surface area to units of dm and the depth to units of dm. 2 2 ⎛ 1000 m ⎞ ⎛ 1 dm ⎞ 16 2 surface area = (1.8 × 108 km2 ) × ⎜ ⎟ ×⎜ ⎟ = 1.8 × 10 dm 1 km 0.1 m ⎝ ⎠ ⎝ ⎠ depth = (3.9 × 103 m) × 1 dm = 3.9 × 104 dm 0.1 m Volume = surface area × depth = (1.8 × 10 1.73 16 2 4 dm )(3.9 × 10 dm) = 7.0 × 10 20 31.103 g Au = 75.0 g Au 1 troy oz Au (a) 2.41 troy oz Au × (b) 1 troy oz = 31.103 g ? g in 1 oz = 1 oz × 1 lb 453.6 g × = 28.35 g 16 oz 1 lb A troy ounce is heavier than an ounce. 1.74 Volume of sphere = 4 3 πr 3 3 Volume = 4 ⎛ 15 cm ⎞ π = 1.77 × 103 cm3 3 ⎜⎝ 2 ⎟⎠ mass = volume × density = (1.77 × 103 cm3 ) × 4.0 × 101 kg Os × 1.75 (a) 22.57 g Os 2.205 lb = 88 lb Os 1 kg |0.798 g/mL − 0.802 g/mL| × 100% = 0.5% 0.798 g/mL 1 cm 3 × 3 dm = 7.0 × 10 1 kg = 4.0 × 101 kg Os 1000 g 20 L 16 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE (b) 1.76 4 62 kg = 6.2 × 10 g O: C: H: 1.77 |0.864 g − 0.837 g| × 100% = 3.1% 0.864 g 4 4 4 (6.2 × 10 g)(0.65) = 4.0 × 10 g O 4 4 (6.2 × 10 g)(0.18) = 1.1 × 10 g C 4 3 (6.2 × 10 g)(0.10) = 6.2 × 10 g H 3 N: (6.2 × 10 g)(0.03) = 2 × 10 g N 4 2 Ca: (6.2 × 10 g)(0.016) = 9.9 × 10 g Ca 4 2 P: (6.2 × 10 g)(0.012) = 7.4 × 10 g P 3 minutes 44.39 seconds = 224.39 seconds Time to run 1500 meters is: 1500 m × 1.78 1 mi 224.39 s × = 209.19 s = 3 min 29.19 s 1609 m 1 mi 2 2 ? °C = (7.3 × 10 − 273) K = 4.6 × 10 °C 9°F ⎞ ⎛ ? °F = ⎜ (4.6 × 102 °C) × + 32°F = 8.6 × 102 °F 5°C ⎟⎠ ⎝ 1.79 ? g Cu = (5.11 × 103 kg ore) × 1.80 (8.0 × 104 tons Au) × 1.81 ? g Au = 34.63% Cu 1000 g × = 1.77 × 106 g Cu 100% ore 1 kg 2000 lb Au 16 oz Au $350 × × = $9.0 × 1011 or 900 billion dollars 1 ton Au 1 lb Au 1 oz Au 4.0 × 10−12 g Au 1 mL × × (1.5 × 1021 L seawater) = 6.0 × 1012 g Au 1 mL seawater 0.001 L value of gold = (6.0 × 1012 g Au) × 1 lb 16 oz $350 × × = $7.4 × 1013 453.6 g 1 lb 1 oz No one has become rich mining gold from the ocean, because the cost of recovering the gold would outweigh the price of the gold. 1.1 × 1022 Fe atoms = 5.4 × 1022 Fe atoms 1.0 g Fe 1.82 ? Fe atoms = 4.9 g Fe × 1.83 mass of Earth's crust = (5.9 × 1021 tons) × 0.50% crust = 2.95 × 1019 tons 100% Earth mass of silicon in crust = (2.95 × 1019 tons crust) × 27.2% Si 2000 lb 1 kg × × = 7.3 × 1021 kg Si 100% crust 1 ton 2.205 lb CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.84 17 10 cm = 0.1 m. We need to find the number of times the 0.1 m wire must be cut in half until the piece left is −10 1.3 × 10 m long. Let n be the number of times we can cut the Cu wire in half. We can write: n ⎛1⎞ −10 m ⎜ 2 ⎟ × 0.1 m = 1.3 × 10 ⎝ ⎠ n ⎛1⎞ −9 ⎜ 2 ⎟ = 1.3 × 10 m ⎝ ⎠ Taking the log of both sides of the equation: ⎛1⎞ n log ⎜ ⎟ = log(1.3 × 10−9 ) ⎝2⎠ n = 30 times 5000 mi 1 gal gas 9.5 kg CO2 × × = 9.5 × 1010 kg CO 2 1 car 20 mi 1 gal gas 1.85 (40 × 106 cars) × 1.86 Volume = area × thickness. From the density, we can calculate the volume of the Al foil. Volume = mass 3.636 g = = 1.3472 cm3 density 2.699 g / cm3 2 2 Convert the unit of area from ft to cm . 2 2 ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ 2 1.000 ft × ⎜ ⎟ ×⎜ ⎟ = 929.03 cm ⎝ 1 ft ⎠ ⎝ 1 in ⎠ 2 thickness = volume 1.3472 cm3 = = 1.450 × 10−3 cm = 1.450 × 10−2 mm 2 area 929.03 cm 1.87 (a) (b) 1.88 First, let’s calculate the mass (in g) of water in the pool. We perform this conversion because we know there is 1 g of chlorine needed per million grams of water. homogeneous heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous. (2.0 × 104 gallons H 2 O) × 3.79 L 1 mL 1g × × = 7.58 × 107 g H 2 O 1 gallon 0.001 L 1 mL Next, let’s calculate the mass of chlorine that needs to be added to the pool. (7.58 × 107 g H 2 O) × 1 g chlorine 1 × 106 g H 2 O = 75.8 g chlorine 18 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE The chlorine solution is only 6 percent chlorine by mass. We can now calculate the volume of chlorine solution that must be added to the pool. 75.8 g chlorine × 100% soln 1 mL soln × = 1.3 × 103 mL of chlorine solution 6% chlorine 1 g soln 1 yr 1.89 (2.0 × 1022 J) × 1.90 We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can calculate the thickness of the oil layer from the volume and surface area. 1.8 × 1020 J = 1.1 × 102 yr 2 ⎛ 1 cm ⎞ 5 2 40 m 2 × ⎜ ⎟ = 4.0 × 10 cm 0.01 m ⎝ ⎠ 3 0.10 mL = 0.10 cm Volume = surface area × thickness thickness = volume 0.10 cm3 = = 2.5 × 10−7 cm 5 2 surface area 4.0 × 10 cm Converting to nm: (2.5 × 10−7 cm) × 1.91 0.01 m 1 nm × = 2.5 nm 1 cm 1 × 10−9 m The mass of water used by 50,000 people in 1 year is: 50, 000 people × 150 gal water 3.79 L 1000 mL 1.0 g H 2 O 365 days × × × × = 1.04 × 1013 g H 2 O/yr 1 person each day 1 gal. 1L 1 mL H 2 O 1 yr A concentration of 1 ppm of fluorine is needed. In other words, 1 g of fluorine is needed per million grams of water. NaF is 45.0% fluorine by mass. The amount of NaF needed per year in kg is: (1.04 × 1013 g H 2 O) × 1g F 6 10 g H 2 O × 100% NaF 1 kg × = 2.3 × 104 kg NaF 45% F 1000 g An average person uses 150 gallons of water per day. This is equal to 569 L of water. If only 6 L of water is used for drinking and cooking, 563 L of water is used for purposes in which NaF is not necessary. Therefore the amount of NaF wasted is: 563 L × 100% = 99% 569 L 1.92 3 3 (a) ⎛ 1 ft ⎞ ⎛ 1 in ⎞ 1 cm3 1 mL ×⎜ × = $3.06 × 10−3 /L ⎟ ×⎜ ⎟ × 3 12 in 2.54 cm 1 mL 0.001 L 15.0 ft ⎝ ⎠ ⎝ ⎠ (b) 2.1 L water × $1.30 0.304 ft 3 gas $1.30 × = $0.055 = 5.5¢ 1 L water 15.0 ft 3 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.93 19 To calculate the density of the pheromone, you need the mass of the pheromone, and the volume that it occupies. The mass is given in the problem. First, let’s calculate the volume of the cylinder. Converting the radius and height to cm gives: 0.50 mi × 40 ft × 1609 m 1 cm × = 8.05 × 104 cm 1 mi 0.01 m 12 in 2.54 cm × = 1.22 × 103 cm 1 ft 1 in 2 volume of a cylinder = area × height = πr × h 4 2 3 volume = π(8.05 × 10 cm) × (1.22 × 10 cm) = 2.48 × 10 13 3 cm Density of gases is usually expressed in g/L. Let’s convert the volume to liters. (2.48 × 1013 cm3 ) × density = 1.94 1 mL 1 cm 3 × 1L = 2.48 × 1010 L 1000 mL mass 1.0 × 10−8 g = = 4.0 × 10−19 g/L 10 volume 2.48 × 10 L This problem is similar in concept to a limiting reagent problem. We need sets of coins with 3 quarters, 1 nickel, and 2 dimes. First, we need to find the total number of each type of coin. Number of quarters = (33.871 × 103 g) × Number of nickels = (10.432 × 103 g) × Number of dimes = (7.990 × 103 g) × 1 quarter = 6000 quarters 5.645 g 1 nickel = 2100 nickels 4.967 g 1 dime = 3450 dimes 2.316 g Next, we need to find which coin limits the number of sets that can be assembled. For each set of coins, we need 2 dimes for every 1 nickel. 2100 nickels × 2 dimes = 4200 dimes 1 nickel We do not have enough dimes. For each set of coins, we need 2 dimes for every 3 quarters. 6000 quarters × 2 dimes = 4000 dimes 3 quarters Again, we do not have enough dimes, and therefore the number of dimes is our “limiting reagent”. If we need 2 dimes per set, the number of sets that can be assembled is: 3450 dimes × 1 set = 1725 sets 2 dimes 20 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE The mass of each set is: ⎛ 5.645 g ⎞ ⎛ 4.967 g ⎞ ⎛ 2.316 g ⎞ ⎜ 3 quarters × ⎟ + ⎜ 1 nickel × ⎟ + ⎜ 2 dimes × ⎟ = 26.534 g/set 1 quarter 1 nickel 1 dime ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ Finally, the total mass of 1725 sets of coins is: 1725 sets × 1.95 26.534 g = 4.577 × 104 g 1 set We wish to calculate the density and radius of the ball bearing. For both calculations, we need the volume of the ball bearing. The data from the first experiment can be used to calculate the density of the mineral oil. In the second experiment, the density of the mineral oil can then be used to determine what part of the 40.00 mL volume is due to the mineral oil and what part is due to the ball bearing. Once the volume of the ball bearing is determined, we can calculate its density and radius. From experiment one: Mass of oil = 159.446 g − 124.966 g = 34.480 g Density of oil = 34.480 g = 0.8620 g/mL 40.00 mL From the second experiment: Mass of oil = 50.952 g − 18.713 g = 32.239 g Volume of oil = 32.239 g × 1 mL = 37.40 mL 0.8620 g The volume of the ball bearing is obtained by difference. 3 Volume of ball bearing = 40.00 mL − 37.40 mL = 2.60 mL = 2.60 cm Now that we have the volume of the ball bearing, we can calculate its density and radius. Density of ball bearing = 18.713 g 3 = 7.20 g/cm 3 2.60 cm Using the formula for the volume of a sphere, we can solve for the radius of the ball bearing. V = 4 3 πr 3 4 3 πr 3 3 3 r = 0.621 cm 2.60 cm3 = r = 0.853 cm CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 1.96 21 We want to calculate the mass of the cylinder, which can be calculated from its volume and density. The 2 volume of a cylinder is πr l. The density of the alloy can be calculated using the mass percentages of each element and the given densities of each element. The volume of the cylinder is: 2 V = πr l 2 V = π(6.44 cm) (44.37 cm) 3 V = 5781 cm The density of the cylinder is: 3 3 3 density = (0.7942)(8.94 g/cm ) + (0.2058)(7.31 g/cm ) = 8.605 g/cm Now, we can calculate the mass of the cylinder. mass = density × volume 3 3 4 mass = (8.605 g/cm )(5781 cm ) = 4.97 × 10 g The assumption made in the calculation is that the alloy must be homogeneous in composition. 1.97 It would be more difficult to prove that the unknown substance is an element. Most compounds would decompose on heating, making them easy to identify. For example, see Figure 4.13(a) of the text. On heating, the compound HgO decomposes to elemental mercury (Hg) and oxygen gas (O2). 1.98 The density of the mixed solution should be based on the percentage of each liquid and its density. Because the solid object is suspended in the mixed solution, it should have the same density as this solution. The density of the mixed solution is: (0.4137)(2.0514 g/mL) + (0.5863)(2.6678 g/mL) = 2.413 g/mL As discussed, the density of the object should have the same density as the mixed solution (2.413 g/mL). Yes, this procedure can be used in general to determine the densities of solids. This procedure is called the flotation method. It is based on the assumptions that the liquids are totally miscible and that the volumes of the liquids are additive. 1.99 Gently heat the liquid to see if any solid remains after the liquid evaporates. Also, collect the vapor and then compare the densities of the condensed liquid with the original liquid. The composition of a mixed liquid would change with evaporation along with its density. 1.100 When the carbon dioxide gas is released, the mass of the solution will decrease. If we know the starting mass of the solution and the mass of solution after the reaction is complete (given in the problem), we can calculate the mass of carbon dioxide produced. Then, using the density of carbon dioxide, we can calculate the volume of carbon dioxide released. 1.140 g Mass of hydrochloric acid = 40.00 mL × = 45.60 g 1 mL Mass of solution before reaction = 45.60 g + 1.328 g = 46.928 g 22 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE We can now calculate the mass of carbon dioxide by difference. Mass of CO2 released = 46.928 g − 46.699 g = 0.229 g Finally, we use the density of carbon dioxide to convert to liters of CO2 released. Volume of CO 2 released = 0.229 g × 1.101 1L = 0.127 L 1.81 g As water freezes, it expands. First, calculate the mass of the water at 20°C. Then, determine the volume that this mass of water would occupy at −5°C. Mass of water = 242 mL × 0.998 g = 241.5 g 1 mL Volume of ice at − 5°C = 241.5 g × 1 mL = 264 mL 0.916 g The volume occupied by the ice is larger than the volume of the glass bottle. The glass bottle would crack!