(Test Bank) Chemistry: The Study Of Change

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CHAPTER 1 CHEMISTRY: THE STUDY OF CHANGE Problem Categories Biological : 1.24, 1.48, 1.69, 1.70, 1.78, 1.84, 1.93, 1.95, 1.96, 1.97, 1.105. Conceptual : 1.3, 1.4, 1.11, 1.12, 1.15, 1.16, 1.54, 1.62, 1.89, 1.101, 1.103. Environmental : 1.70, 1.87, 1.89, 1.92, 1.98. Industrial : 1.51, 1.55, 1.72, 1.81, 1.91. Difficulty Level Easy : 1.3, 1.11, 1.13, 1.14, 1.15, 1.21, 1.22, 1.23, 1.24, 1.25, 1.26, 1.29, 1.30, 1.31, 1.32, 1.33, 1.34, 1.54, 1.55, 1.63, 1.64, 1.77, 1.80, 1.84, 1.89, 1.91. Medium : 1.4, 1.12, 1.16, 1.35, 1.36, 1.37, 1.38, 1.39, 1.40, 1.41, 1.42, 1.43, 1.44, 1.45, 1.46, 1.47, 1.48, 1.49, 1.50, 1.51, 1.52, 1.53, 1.56, 1.57, 1.59, 1.60, 1.61, 1.62, 1.70, 1.71, 1.72, 1.73, 1.74, 1.75, 1.76, 1.78, 1.79, 1.81, 1.82, 1.83, 1.85, 1.94, 1.95, 1.96, 1.97, 1.98. Difficult : 1.58, 1.65, 1.66, 1.67, 1.68, 1.69, 1.86, 1.87, 1.88, 1.90, 1.92, 1.93, 1.99, 1.100, 1.101, 1.102, 1.103, 1.104, 1.105, 1.106. 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 involve d, they are not the result of measurement. 1.4 (a) hypothesis (b) law (c) theory 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 nucl ear decay, the products are chemically different substances. 1.12 (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. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 2 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) K (b) Sn (c) Cr (d) B (e) Ba (f) Pu (g) S (h) Ar (i) Hg 1.15 (a) element (b) compound (c) element (d) compound 1.16 (a) homogeneous mixture (b) element (c) compound (d) homogeneous mixture (e) heterogeneous mixture (f) homogeneous mixture (g) heterogeneous mixture 1.21 mass 586 g volume 188 mL=== density 3.12 g/mL 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. massdensityvolume= Solution: mass = density × volume 0.798 g17.4 mL1m L=×= mass of ethanol 13.9 g 1.23 5C?C = ( F 3 2 F )9F°°° − ° ×° (a) 5C?C = ( 9 5 3 2 ) F9F°°− ° × =°35 C° (b) 5C? C=( 1 2 3 2 )F9F°°− ° × =°11 C−° (c) 5C? C = (102 32) F9F°°− ° × =°39 C° (d) 5C? C = (1852 32) F9F°°− ° × =°1011 C° (e) 9F?F C 3 2 F5C°⎛⎞°=°× +°⎜⎟°⎝⎠ 9F? F 273.15 C 32 F5C° ⎛⎞°=− °× + °=⎜⎟° ⎝⎠459.67 F− ° 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 te xt. Substitute the temperature values given in the problem into the appropriate equation. (a) Conversion from Fahrenheit to Celsius. 5C?C = ( F 3 2 F )9F°°° − ° ×° CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 3 5C=( 1 0 5 3 2 )F9F°−° × =°?C 4 1 C °° (b) Conversion from Celsius to Fahrenheit. 9F?F C 3 2 F5C°⎛⎞°=°× +°⎜⎟°⎝⎠ 9F11.5 C 32 F5C° ⎛⎞=− °× + °=⎜⎟° ⎝⎠? F 11.3 F °° (c) Conversion from Celsius to Fahrenheit. 9F?F C 3 2 F5C°⎛⎞°=°× +°⎜⎟°⎝⎠ 3 9F6.3 10 C 32 F5C° ⎛⎞=× ° × + ° =⎜⎟° ⎝⎠4?F 1 . 1 1 0 F °× ° (d) Conversion from Fahrenheit to Celsius. 5C?C = ( F 3 2 F )9F°°° − ° ×° 5C= (451 32) F9F°−° × =°? C 233 C °° 1.25 1KK( C 2 7 3 C )1C=°+ °° (a) K = 113°C + 273° C = 386 K (b) K = 37°C + 273°C = 3.10 × 102 K (c) K = 357°C + 273° C = 6.30 × 102 K 1.26 (a) 1KK( C 2 7 3 C )1C=°+ °° °C = K − 273 = 77 K − 273 = −196° C (b) °C = 4.2 K − 273 = −269° C (c) °C = 601 K − 273 = 328° C 1.29 (a) 2.7 × 10−8 (b) 3.56 × 102 (c) 4.7764 × 104 (d) 9.6 × 10−2 1.30 (a) 10−2 indicates that the decimal point mu st be moved two places to the left. 1.52 × 10−2 = 0.0152 (b) 10−8 indicates that the decimal point must be moved 8 places to the left. 7.78 × 10−8 = 0.0000000778 CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 4 1.31 (a) 145.75 + (2.3 × 10−1) = 145.75 + 0.23 = 1.4598 × 102 (b) 4 2279500 7.95 10= 2.5 10 2.5 10×= ××23.2 10× (c) (7.0 × 10−3) − (8.0 × 10−4) = (7.0 × 10−3) − (0.80 × 10−3) = 6.2 × 10−3 (d) (1.0 × 104) × (9.9 × 106) = 9.9 × 1010 1.32 (a) Addition using scientific notation. Strategy: Let's express scientific notation as N × 10n. When adding numbers using scientific notation, we must write each quantity w ith 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. Let’s write 0.0095 in such a way that n = −3. We have decreased 10n by 103, so we must increase N by 103. 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. 9.5 × 10−3 + 8.5 × 10−3 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 10 to express N between 1 and 10 (1.8), we must increase 10n by a factor of 10. The exponent, n, is increased by 1 from −3 to − 2. 18.0 × 10−3 = 1.8 × 10−2 (b) Division using scientific notation. Strategy: Let's express scientific notation as N × 10n. 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 × 102 Divide the N parts of the numbers in the usual way. 6.53 ÷ 5.75 = 1.14 Subtract the exponents, n. 1.14 × 10+2 − (−8) = 1.14 × 10+2 + 8 = 1.14 × 1010 (c) Subtraction using scientific notation. Strategy: Let's express scientific notation as N × 10n. 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. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 5 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 × 105 Subtract the N parts of the numbers, keeping the exponent, n, the same. 8.5 × 105 − 9.0 × 105 −0.5 × 105 The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10 to express N between 1 and 10 (5), we must decrease 10n by a factor of 10. The exponent, n, is decreased by 1 from 5 to 4. −0.5 × 105 = −5 × 104 (d) Multiplication using scientific notation. Strategy: Let's express scientific notation as N × 10n. 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 × 102 The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10 to express N between 1 and 10 (1.3), we must increase 10n by a factor of 10. The exponent, n, is increased by 1 from 2 to 3. 13 × 102 = 1.3 × 103 1.33 (a) four (b) two (c) five (d) two, three, or four (e) three (f) one (g) one (h) two 1.34 (a) one (b) three (c) three (d) four (e) two or three (f) one (g) one or two 1.35 (a) 10.6 m (b) 0.79 g (c) 16.5 cm2 (d) 1 × 106 g/cm3 1.36 (a) Division 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 km1.285.70 km= 3 The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits. Therefore, the answer has on ly three significant digits. The correct answer rounded off to the correct number of significant figures is: 1.28 (Why are there no units?) CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 6 (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 0.00318 12 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 × 10−3 mg (c) 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 (0.402 × 107 dm) + (7.74 × 107 dm) = 8.14 × 107 dm Since 7.74 × 107 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. (d) Subtraction, addition, and division Strategy: For subtraction and addition, the number of significa nt figures to the right of the decimal point in that part of the calculation is determined by the lowest number of digits to the right of the decimal point in any of the original numbers. For the division part of the calculation, the number of significant figures in the answer is determined by the number having the smallest number of significant figures. First, perform the subtraction and addition parts to the correct number of significant figures, and then perform the division. Solution: (7.8 m 0.34 m) 7.5 m/(1.15 s 0.82 s) 1.97 s−==+3.8 m s 1.37 Calculating the mean for each set of date, we find: Student A: 87.6 mL Student B: 87.1 mL Student C: 87.8 mL From these calculations, we can conclude that the volume measurements made by Student B were the most accurate of the three students. The pr ecision in the measurements made by both students B and C are fairly high, while the measurements made by st udent A are less precise. In summary: Student A: neither accurate nor precise Student B: both accurate and precise Student C: precise, but not accurate CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 7 1.38 Calculating the mean for each set of date, we find: Tailor X: 31.5 in Tailor Y: 32.6 in Tailor Z: 32.1 in From these calculations, we can conclude that the s eam measurements made by Tailor Z were the most accurate of the three tailors. The preci sion in the measurements made by bo th tailors X and Z are fairly high, while the measurements made by tailor Y are less precise. In summary: Tailor X: most precise Tailor Y: least accurate and least precise Tailor Z: most accurate 1.39 (a) 1d m22.6 m0.1 m=× = ? dm 226 dm (b) 0.001 g 1 kg25.4 mg1 mg 1000 g=×× =5?k g 2 . 5 4 1 0 k g−× (c) 311 0 L556 mL1m L−×=× = ? L 0.556 L (d) 32 310.6 kg 1000 g 1 10 m 1k g 1c m 1m−⎛⎞×=× × = ⎜⎟⎜⎟⎝⎠3 3g?0 . 0 1 0 6 g / cm cm 1.40 (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 grams to milligrams (1 mg = 1 × 10−3 g). Arrange the appropriate conversion factors so that pounds and grams cancel, and the unit milligrams is obtained in your answer. Solution: The sequence of conversions is lb → grams → mg Using the following conversion factors, 453.6 g 1l b 31m g 11 0 g−× we obtain the answer in one step: 3453.6 g 1 mg242 lb1l b 11 0 g−=× × = ×8?m g 1 . 1 0 1 0 m 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. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 8 (b) Strategy: The problem may be stated as ? m3 = 68.3 cm3 Recall that 1 cm = 1 × 10−2 m. We need to set up a conversion factor to convert from cm3 to m3. Solution: We need the following conversion factor so th at centimeters cancel and we end up with meters. 211 0 m 1c m−× Since this conversion factor deals with length and we want volume, it must therefore be cubed to give 3222 211 0 m 11 0 m 11 0 m 11 0 m 1c m 1c m 1c m 1c m−−− − ⎛⎞ ××× ×××= ⎜⎟⎜⎟⎝⎠ We can write 323 11 0 m68.3 cm1c m−⎛⎞×=× = ⎜⎟⎜⎟⎝⎠35 3? m 6.83 10 m−× Check: We know that 1 cm3 = 1 × 10−6 m3. We started with 6.83 × 101 cm3. Multiplying this quantity by 1 × 10−6 gives 6.83 × 10−5. (c) Strategy: The problem may be stated as ? L = 7.2 m3 In Chapter 1 of the text, a conversion is given between liters and cm3 (1 L = 1000 cm3). If we can convert m3 to cm3, we can then convert to liters. Recall that 1 cm = 1 × 10−2 m. We need to set up two conversion factors to convert from m3 to L. Arrange the appropriat e conversion factors so that m3 and cm3 cancel, and the unit liters is obtained in your answer. Solution: The sequence of conversions is m3 → cm3 → L Using the following conversion factors, 3 21c m 11 0 m−⎛⎞ ⎜⎟⎜⎟×⎝⎠ 31L 1000 cm the answer is obtained in one step: 3 3 231c m 1L7.2 m 1 10 m 1000 cm−⎛⎞ =× × = ⎜⎟⎜⎟×⎝⎠3?L 7 . 2 1 0 L× Check: From the above conversion factors you can show that 1 m3 = 1 × 103 L. Therefore, 7 m3 would equal 7 × 103 L, which is close to the answer. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 9 (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 convert from grams to pounds. Recall that 1 μg = 1 × 10−6 g. Arrange the appropri ate 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, 611 0 g 1g−× μ 1l b 453.6 g we can write 611 0 g 1 l b28.3 g1 g 453.6 g−×=μ × × =μ8?l b 6 . 2 4 1 0 l b−× Check: Does the answer seem reasonable? What number does the prefix μ represent? Should 28.3 μg be a very small mass? 1.41 1255 m 1 mi 3600 s 1 s 1609 m 1 h××= 2808 mi/h 1.42 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. Solution: The sequence of conversions is days → hours → minutes → seconds Using the following conversion factors, 24 h 1d a y 60 min 1h 60 s 1m i n we can write 24 h 60 min 60 s= 365.24 day1d a y 1h 1m i n×× × =7? s 3.1557 10 s× Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year? 1.43 6 81.609 km 1000 m 1 s 1 min(93 10 mi)1m i 1k m 6 0s 3.00 10 m×× ×× × = ×8.3 min CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 10 1.44 (a) 1 mi 5280 ft 12 in 1 min 13 min 1 mi 1 ft 60 s=× × × = ?i n / s 8 1i n / s (b) 1 mi 1609 m 13 min 1 mi=×=2? m/min 1.2 10 m/min× (c) 1 mi 1609 m 1 km 60 min 13 min 1 mi 1000 m 1 h=×××= ?k m / h 7 . 4k m / h 1.45 1 m6.0 ft3.28 ft×= 1.8 m 453.6 g 1 kg168 lb1l b 1 0 0 0g××= 76.2 kg 1.46 55 mi 1.609 km 1h 1m i=× = ?k m / h 8 8k m / h 1.47 62 m 1 mi 3600 s 1 s 1609 m 1 h××=21.4 10 mph× 1.48 60.62 g Pb0.62 ppm Pb 1 10 g blood= × 3 60.62 g Pb6.0 10 g of blood 11 0 g b l o o d×× = ×33.7 10 g Pb−× 1.49 (a) 8365 day 24 h 3600 s 3.00 10 m 1 mi1.42 yr1y r 1d a y 1h 1s 1 6 0 9m××× × × ×=128.35 10 mi× (b) 36 in 2.54 cm32.4 yd1y d 1i n×× =32.96 10 cm× (c) 103.0 10 cm 1 in 1 ft 1 s 2.54 cm 12 in××× =89.8 10 ft/s× 1.50 (a) 911 0 m185 nm1n m−×=× =7? m 1.85 10 m−× (b) 9 365 day 24 h 3600 s(4.5 10 yr)1y r 1d a y 1h=× × × × =17?s 1 . 4 1 0 s× (c) 3 30.01 m71.2 cm1c m⎛⎞=× = ⎜⎟ ⎝⎠35 3? m 7.12 10 m−× (d) 3 3 231c m 1L88.6 m 1 10 m 1000 cm−⎛⎞ =× × = ⎜⎟⎜⎟×⎝⎠4? L 8.86 10 L× CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 11 1.51 3 32.70 g 1 kg 1 cm 1000 g 0.01 m 1c m⎛⎞=×× = ⎜⎟ ⎝⎠33density 2.70 10 kg/m× 1.52 30.625 g 1 L 1 mL 1 L 1000 mL 1c m=× × =43density 6.25 10 g/cm−× 1.53 Substance Qualitative Statement Quantitative Statement (a) water colorless liquid freezes at 0 °C (b) carbon black solid (graphite) density = 2.26 g/cm3 (c) iron rusts easily density = 7.86 g/cm3 (d) hydrogen gas colorless gas melts at −255.3° C (e) sucrose tastes sweet at 0 °C, 179 g of sucrose dissolves in 100 g of H 2O (f) table salt tastes salty melts at 801 °C (g) mercury liquid at room temperature boils at 357 °C (h) gold a precious metal density = 19.3 g/cm3 (i) air a mixture of gases contains 20% oxygen by volume 1.54 See Section 1.6 of your text for a discussion of these terms. (a) Chemical property . Iron has changed its composition and id entity 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. (d) Physical property . The evaporation of water does not change its chemical properties. Evaporation is a change in matter from the liqui d state to the gaseous state. (e) Chemical property . The carbon dioxide is chemically converted into other molecules. 1.55 9 31t o n(95.0 10 lb of sulfuric acid) 2.0 10 lb×× = ×74.75 10 tons of sulfuric acid× 1.56 Volume of rectangular bar = length × width × height 52.7064 g=(8.53 cm)(2.4 cm)(1.0 cm)==3density 2.6 g/cmm V 1.57 mass = density × volume (a) mass = (19.3 g/cm3) × [4 3π(10.0 cm)3] = 8.08 × 104 g (b) 3 3 1c m(21.4 g/cm ) 0.040 mm10 mm⎛⎞=× × = ⎜⎟ ⎝⎠6mass 1.4 10 g−× (c) mass = (0.798 g/mL)(50.0 mL) = 39.9 g CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 12 1.58 You are asked to solve for the inner diameter of the tube. If you can cal culate the volume that the mercury occupies, you can calculate the radius of the cylinder, Vcylinder = πr2h (r is the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2 r. mass of Hgvolume of Hg filling cylinderdensity of Hg= 3 3105.5 gvolume of Hg filling cylinder 7.757 cm 13.6 g/cm== Next, solve for the radius of the cylinder. Volume of cylinder = πr2h volume=π×rh 37.757 cm0.4409 cm12.7 cm==π×r The cylinder diameter equals 2 r. Cylinder diameter = 2r = 2(0.4409 cm) = 0.882 cm 1.59 From the mass of the water and its density, we can calcu late the volume that the water occupies. The volume that the water occupies is equa l to the volume of the flask. massvolumedensity= Mass of water = 87.39 g − 56.12 g = 31.27 g 3mass 31.27 g=density 0.9976 g/cm==3Volume of the flask 31.35 cm 1.60 3 4 3m 1m i 3 6 0 0s 1 s 1609 m 1 h××= 767 mph 1.61 The volume of silver is equal to the volume of water it displaces. Volume of silver = 260.5 mL − 242.0 mL = 18.5 mL = 18.5 cm3 3194.3 g 18.5 cm==3density 10.5 g/cm 1.62 In order to work this problem, you need to understand the physical principles involved in the experiment in Problem 1.61. 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. CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 13 1.63 4 33mass 1.20 10 g volume 1.05 10 cm×== = ×3density 11.4 g/cm 1.64 massVolumedensity= 3 31.20 10 g 0.53 g / cm×==33Volume occupied by Li 2.3 10 cm× 1.65 For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C. 5C?C = 0 . 1 F 0 . 0 5 6 C9F°°° × = °° 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. known error in a measurementPercent error 100%value of the measurement=× For the Fahrenheit thermometer, 0.056 C100%38.9 C°=× =°percent error 0.1% For the Celsiu s thermometer, 0.1 C100%38.9 C°=× =°percent error 0.3% Which thermometer is more accurate? 1.66 To work this problem, we need to c onvert from cubic feet to L. Some tabl es will have a conversion factor of 28.3 L = 1 ft3, but we can also calculate it using the dimensi onal analysis method described in Section 1.9 of the text. First, converting from cubic feet to liters: 33 3 73 9 312 in 2.54 cm 1 mL 1 10 L(5.0 10 ft ) 1.42 10 L1f t 1i n 1m L 1c m−⎛⎞ ⎛ ⎞ ××× × × × = × ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ The mass of vanillin (in g) is: 1192 2.0 10 g vanillin(1.42 10 L) 2.84 10 g vanillin1L−− ××× =× The cost is: 2 $112(2.84 10 g vanillin)50 g vanillin−×× = = $0.064 6.4¢ 1.67 9F?F = C + 3 2 F5C°⎛⎞°° × °⎜⎟°⎝⎠ Let temperature = t 9=+ 3 2 F5° tt CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 14 9=3 2F5−°tt 4=3 2F5−°t t = −40°F = −40°C 1.68 There are 78.3 + 117.3 = 195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor. 195.6 C 100 S⎛⎞ ⎜⎟⎜⎟⎝⎠D D Set up the equation like a Celsius to Fahrenhe it 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 C100 S°⎛⎞°° −° ⎜⎟°⎝⎠ Solving for ? °S gives: 100 S? S = (? C + 117.3 C)195.6 C°⎛⎞°° ° ⎜⎟°⎝⎠ For 25°C we have: 100 S(25 + 117.3) C195.6 C°⎛⎞=°= ⎜⎟°⎝⎠?S 7 3 S °° 1.69 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. 240 mL of pure oxygen/min = (0.04)(volume of inhaled air/min) 240 mL of oxygen/minVolume of inhaled air/min 6000 mL of inhaled air/min0.04== Since there are 12 breaths per min, 6000 mL of inhaled air 1 min 1 min 12 breaths=× =2volume of air/breath 5 10 mL/breath× 1.70 (a) 6000 mL of inhaled air 0.001 L 60 min 24 h 1m i n 1m L 1h 1d a y××× =38.6 10 L of air/day× (b) 368.6 10 L of air 2.1 10 L CO 1 day 1 L of air−×××= 0.018 L CO/da y 1.71 The mass of the seawater is: 21 24 21 1m L 1 . 0 3g(1.5 10 L) 1.55 10 g 1.55 10 kg seawater0.001 L 1 mL×× ×= ×= × Seawater is 3.1% NaCl by mass. Th e total mass of NaCl in kilograms is: 21 3.1% NaCl(1.55 10 kg seawater)100% seawater=× × =19mass NaCl (kg) 4.8 10 kg NaCl× CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 15 19 2.205 lb 1 ton(4.8 10 kg)1k g 2 0 0 0l b=× × × =16mass NaCl (tons) 5.3 10 tons NaCl× 1.72 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 1.71. massvolumedensity= 1000 gvolume of 1 kg of seawater 970.9 mL 0.9709 L1.03 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. 41 0 2000 lb 453.6 g(8.0 10 tons Mg) 7.26 10 g Mg1t o n 1l b×× × = × Volume of seawater n eeded to extract 8.0 × 104 ton Mg = 10 0.9709 L seawater(7.26 10 g Mg)1.3 g Mg×× =105.4 10 L of seawater× 1.73 Assume that the crucible is platinum. Let’s calculate the volume of the crucible an d then compare that to the volume of water that the crucible displaces. massvolumedensity= 3860.2 gVolume of crucible 21.45 g/cm==340.10 cm 3(860.2 820.2)gVolume of water displaced 0.9986 g/cm−==340.1 cm The volumes are the same (within experimental error), so the crucible is made of platinum. 1.74 Volume = surface area × depth Recall that 1 L = 1 dm3. Let’s convert the surface area to units of dm2 and the depth to units of dm. 22 82 1 62 1000 m 1 dmsurface area (1.8 10 km ) 1.8 10 dm1k m 0 . 1m⎛⎞ ⎛ ⎞=× × × = × ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ 34 1d mdepth (3.9 10 m) 3.9 10 dm0.1 m=× × = × Volume = surface area × depth = (1.8 × 1016 dm2)(3.9 × 104 dm) = 7.0 × 1020 dm3 = 7.0 × 1020 L