CHAPTER CHEMISTRY: THE STUDY OF CHANGE Problem Categories Biological: 1.26, 1.50, 1.71, 1.72, 1.80, 1.86, 1.96, 1.97, 1.105, 1.114 Conceptual: 1.3, 1.4, 1.17, 1.18, 1.11, 1.12, 1.56, 1.64, 1.91, 1.94, 1.101, 1.103, 1.117 Environmental: 1.72, 1.89, 1.91, 1.98, 1.109, 1.112 Industrial: 1.53, 1.57, 1.83 Difficulty Level Easy: 1.3, 1.9, 1.10, 1.11, 1.12, 1.17, 1.23, 1.24, 1.25, 1.26, 1.27, 1.28, 1.31, 1.32, 1.33, 1.34, 1.35, 1.36, 1.56, 1.57, 1.66, 1.79, 1.82, 1.86, 1.91 Medium: 1.4, 1.18, 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, 1.54, 1.55, 1.58, 1.59, 1.60, 1.61, 1.62, 1.63, 1.64, 1.65, 1.72, 1.73, 1.74, 1.75, 1.76, 1.77, 1.78, 1.80, 1.81, 1.83, 1.84, 1.85, 1.87, 1.93, 1.96, 1.97, 1.98 Difficult: 1.67, 1.68, 1.69, 1.70, 1.71, 1.88, 1.89, 1.90, 1.92, 1.94, 1.95, 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 involved, they are not the result of measurement 1.4 (a) hypothesis 1.9 Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum; (b) law (c) theory Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon 1.10 1.11 (a) Cs (b) Ge (c) Ga (d) Sr (e) U (f) Se (g) Ne (h) Cd (a) element (b) compound (c) element (d) compound CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 1.12 (a) homogeneous mixture (b) element (c) compound (d) homogeneous mixture (e) heterogeneous mixture (f) heterogeneous mixture (g) element (a) Chemical property Oxygen gas is consumed in a combustion reaction; its composition and identity are 1.17 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 1.18 (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 586 g mass = = 3.12 g/mL volume 188 mL 1.23 density = 1.24 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 density = mass volume Solution: mass = density ´ volume mass of methanol = 0.7918 g ´ 89.9 mL = 71.2 g mL Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang Full file at https://TestbankDirect.eu/Solution-Manual-for-Chemistry-13th-Edition-By-Chang 5C 1.25 ? C = (F - 32F) ´ (a) (b) (c) (d) (e) 9F 5C = 35C 9F 5C ? C = (12 - 32)F ´ = - 11C 9F 5C ? C = (102 - 32)F ´ = 39C 9F 5C ? C = (1852 - 32)F ´ = 1011C 9F ỉ 9F ư÷ ữ + 32F ? F = ỗỗC 5C ứữữ ốỗ ? C = (95 - 32)F ổ 9F ö÷ ÷ + 32F = - 459.67F ? F = çç-273.15 C ´ 5C ÷÷ø èç 1.26 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 - 32F) ´ 5C 9F ? C = (105 - 32)F ´ (b) 5C = 41C 9F Conversion from Celsius to Fahrenheit ỉ 9F ư÷ ÷ + 32F ? F = ỗỗC ỗố 5C ứữữ ổ 9F ÷ö ÷ + 32F = 11.3 F ? F = çç-11.5 C ´ çè 5C ÷÷ø (c) Conversion from Celsius to Fahrenheit ỉ 9F ư÷ ÷ + 32F ? F = ỗỗC 5C ứữữ ốỗ ổ 9F ữử ữữ + 32F = 1.1 ´ 10 F ? F = ỗỗ6.3 103 C ỗố 5C ữứ (d) Conversion from Fahrenheit to Celsius ? C = (F - 32F) ´ 5C 9F ? C = (451 - 32)F ´ 5C = 233C 9F Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 1.27 K = (C + 273C) 1.28 1K 1C (a) K = 113C + 273C = 386 K (b) K = 37C + 273C = 3.10 ´ 10 K (c) K = 357C + 273C = 6.30 ´ 10 K (a) K = (C + 273C) 2 1K 1C C = K - 273 = 77 K - 273 = -196C (b) C = 4.2 K - 273 = -269C (c) C = 601 K - 273 = 328C 1.31 (a) 2.7 ´ 10 1.32 (a) 10 -2 -8 3.56 ´ 10 (b) (c) -2 -8 10 -8 (a) (b) 1.34 = 0.0152 indicates that the decimal point must be moved places to the left 7.78 ´ 10 1.33 -2 9.6 ´ 10 (d) indicates that the decimal point must be moved two places to the left 1.52 ´ 10 (b) 4.7764 ´ 10 = 0.0000000778 -1 145.75 + (2.3 ´ 10 ) = 145.75 + 0.23 = 1.4598 ´ 10 79500 2.5 ´ 10 = 7.95 ´ 10 2.5 ´ 10 -3 = 3.2 × 10 -4 -3 -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 (a) Addition using scientific notation 10 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 Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE n 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 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 -3 18.0 ´ 10 The usual practice is to express N as a number between and 10 Since we must decrease N by a factor of 10 n to express N between and 10 (1.8), we must increase 10 by a factor of 10 The exponent, n, is increased by 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 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) + - (- 8) +2+8 = 1.14 ´ 10 = 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 = This means to move the decimal point five places to the left 850,000 = 8.5 ´ 10 Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Subtract the N parts of the numbers, keeping the exponent, n, the same 8.5 ´ 10 - 9.0 ´ 10 -0.5 ´ 10 The usual practice is to express N as a number between and 10 Since we must increase N by a factor of 10 n to express N between and 10 (5), we must decrease 10 by a factor of 10 The exponent, n, is decreased by from to -0.5 ´ 10 = -5 ´ 10 (d) 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 - + (+ 6) = 13 ´ 10 The usual practice is to express N as a number between and 10 Since we must decrease N by a factor of 10 n to express N between and 10 (1.3), we must increase 10 by a factor of 10 The exponent, n, is increased by from to 13 ´ 10 = 1.3 ´ 10 (a) four (b) two (c) five (d) two, three, or four (e) three (f) one (g) one (h) two (a) one (b) three (c) three (e) two or three (f) one (g) one or two 1.37 (a) 10.6 m 1.38 (a) Division 1.35 1.36 (b) 0.79 g (c) 16.5 cm (d) (d) four × 10 g/cm Strategy: The number of significant figures in the answer is determined by the original number having the smallest number of significant figures Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Solution: 7.310 km = 1.283 5.70 km The (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 (b) (Why are there no units?) 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: -3 0.00318 mg = 3.18 ´ 10 (c) 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 (0.402 ´ 10 dm) + (7.74 ´ 10 dm) = 8.14 ´ 10 dm 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 (d) Subtraction, addition, and division Strategy: For subtraction and addition, the number of significant 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 Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 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) (1.15 s + 0.82 s) 1.39 = 7.5 m 1.97 s = 3.8 m/s Calculating the mean for each set of data, 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 precision in the measurements made by both students B and C are fairly high, while the measurements made by student A are less precise In summary: Student A: neither accurate nor precise Student B: both accurate and precise Student C: precise, but not accurate 1.40 Calculating the mean for each set of data, 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 seam measurements made by Tailor Z were the most accurate of the three tailors The precision in the measurements made by both 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 1.41 dm (a) ? dm = 22.6 m ´ (b) ? kg = 25.4 mg ´ (c) ? L = 556 mL ´ 0.1 m = 226 dm 0.001g 1mg ´ ´ 10-3 L mL 1kg 1000 g = 2.54 ´ 10 kg = 0.556 L Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE (d) 1.42 ? g cm = æ1 10-2 m ữử3 ỗ ữữữ = 0.0106 g / cm ỗỗỗ kg ỗỗ cm ÷÷÷ è ø 10.6 kg 1000 g m3 (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 = ´ 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 mg lb ´ 10-3 g we obtain the answer in one step: ? mg = 242 lb ´ mg 453.6 g ´ = 1.10 ´ 108 mg -3 lb ´ 10 g Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many mg are in lb? There are 453,600 mg in lb (b) Strategy: The problem may be stated as 3 ? m = 68.3 cm -2 Recall that cm = ´ 10 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 ´ 10-2 m cm Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 10 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Since this conversion factor deals with length and we want volume, it must therefore be cubed to give ´ 10-2 m ´ cm ´ 10-2 m cm ´ ´ 10-2 m cm æ1 ´ 10-2 m ửữ3 ỗ ữữ = ỗỗỗ ữữ cm ççè ø÷÷ We can write ?m ỉ1 ´ 10-2 m ửữ3 ỗ ữữ -5 = 68.3 cm ỗỗỗ ữữ = 6.83 10 m ỗỗ cm ố ø÷÷ 3 Check: We know that cm = ´ 10 -6 ´ 10 -6 3 m We started with 6.83 ´ 10 cm Multiplying this quantity by -5 gives 6.83 ´ 10 (c) Strategy: The problem may be stated as ? L = 7.2 m 3 In Chapter of the text, a conversion is given between liters and cm (1 L = 1000 cm ) If we can convert m to cm , we can then convert to liters Recall that cm = ´ 10 -2 m We need to set up two conversion 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 Solution: The sequence of conversions is 3 m cm L Using the following conversion factors, æ cm ửữ3 ỗỗ ữữ ỗỗ ữ ỗỗ1 10-2 m ÷÷÷ è ø 1L 1000 cm the answer is obtained in one step: ổ ỗ cm ? L = 7.2 m3 ỗỗỗ ỗỗ1 10-2 m è ư÷3 1L ÷÷÷ ´ = 7.2 ´ 10 L ÷÷ 1000 cm ø÷ 3 Check: From the above conversion factors you can show that m = ´ 10 L Therefore, m would equal ´ 10 L, which is close to the answer Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 1.76 19 Volume = surface area ´ depth Recall that L = dm Let’s convert the surface area to units of dm and the depth to units of dm ( surface area = 1.8 ´ 10 km ( ) depth = 3.9 ´ 103 m ´ ổ1000 m ữử2 ổ dm ữử2 ỗỗ ỗ ữ ữữ = 1.8 1016 dm ỗ ữ ỗỗ ỗỗ km ữữ ỗỗ 0.1 m ữữ è ø è ø ) dm 0.1 m = 3.9 ´ 10 dm 16 20 20 Volume = surface area ´ depth = (1.8 ´ 10 dm ) (3.9 ´ 10 dm) = 7.0 ´ 10 dm = 7.0 ´ 10 L 1.77 31.103 g Au (a) 2.41 troy oz Au ´ (b) troy oz = 31.103 g = 75.0 g Au troy oz Au ? g in oz = oz ´ lb 16 oz ´ 453.6 g lb = 28.35 g A troy ounce is heavier than an ounce 1.78 Volume of sphere = πr 3 ổỗ15 cm ửữữ 3 ỗ Volume = p ỗ ữ = 1.77 10 cm ỗố ø÷÷ ( ) mass = volume ´ density = 1.77 ´ 103 cm ´ 4.0 ´ 101 kg Os ´ 1.79 (a) (b) 1.80 2.205 lb kg 0.864 g - 0.837 g 0.864 g cm ´ kg 1000 g = 4.0 ´ 101 kg Os = 88 lb Os 0.798 g/mL - 0.802 g/mL 0.798 g/mL 22.57 g Os ´ 100% = 0.5% ´ 100% = 3.1% 62 kg = 6.2 ´ 10 g O: 4 (6.2 ´ 10 g)(0.65) = 4.0 ´ 10 g O N: (6.2 ´ 10 g)(0.03) = ´ 10 g N Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 20 1.81 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 4 C: (6.2 ´ 10 g)(0.18) = 1.1 ´ 10 g C H: (6.2 ´ 10 g)(0.10) = 6.2 ´ 10 g H 4 Ca: (6.2 ´ 10 g)(0.016) = 9.9 ´ 10 g Ca P: (6.2 ´ 10 g)(0.012) = 7.4 ´ 10 g P minutes 43.13 seconds = 223.13 seconds Time to run 1500 meters is: 1500 m ´ 1.82 mi 1609 m ´ 223.13 s mi = 208.01 s = 28.01 s ? C = (7.3 ´ 10 - 273) K = 4.6 ´ 10 C ổ 9o F ửữữ ỗ o ? F = çç(4.6 ´ 102 o C) ´ ÷÷ + 32 F = 8.6 10 F ỗỗố 5o C ứữ 1.83 ? g Cu = (5.11 ´ 103 kg ore) ´ 1.84 (8.0 ´ 104 tons Au) ´ 1.85 ? g Au = 34.63% Cu 100% ore 2000 lb Au ton Au 4.0 ´ 10-12 g Au mL seawater ´ ´ 1000 g lb Au 0.001 L = 1.77 ´ 106 g Cu kg 16 oz Au mL value of gold = (6.0 ´ 1012 g Au) ´ ´ ´ $948 = $2.4 ´ 1012 or 2.4 trillion dollars oz Au ´ (1.5 ´ 1021 L seawater) = 6.0 ´ 1012 g Au lb 453.6 g ´ 16 oz lb ´ $948 = $2.0 ´ 1014 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 1.86 ? Fe atoms = 4.9 g Fe 1.87 mass of Earth's crust = (5.9 ´ 1021 tons) ´ 1.0 g Fe = 5.4 ´ 10 22 Fe atoms 0.50% crust 100% Earth mass of silicon in crust = (2.95 ´ 1019 tons crust) ´ = 2.95 ´ 1019 tons 27.2% Si 100% crust ´ 2000 lb ton ´ kg 2.205 lb = 7.3 ´ 1021 kg Si Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 1.88 21 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 equal to the diameter of a Cu atom, which is (2)(1.3 ´ 10- m) Let n be the number of times we can cut the 10 Cu wire in half We can write: ổ ửữn ỗỗ ữ 0.1 m = 2.6 1010 m ốỗ ứữữ ổ ữửn ỗỗ ữ = 2.6 10-9 m ỗố ÷÷ø Taking the log of both sides of the equation: ổ1ử nlog ỗỗ ữữữ = log(2.6 10-9 ) çè ø÷ n = 29 times 5000 mi 1.89 (250 ´ 106 cars) ´ 1.90 Volume = area ´ thickness car ´ gal gas 20 mi ´ 9.5 kg CO2 = 5.9 ´ 1011 kg CO gal gas From the density, we can calculate the volume of the Al foil Volume = 3.636 g mass = = 1.3472 cm density 2.699 g / cm 2 Convert the unit of area from ft to cm ỉ12 in ÷ư2 ỉ 2.54 cm ÷ư2 çç ç ÷ ÷÷ = 929.03 cm 1.000 ft ỗ ữ ỗỗ ỗỗ ft ữữ ỗỗ in ÷÷ è ø è ø thickness = 1.91 1.3472 cm3 volume = = 1.450 × 10-3 cm = 1.450 ´ 10-2 mm area 929.03 cm (a) homogeneous (b) heterogeneous The air will contain particulate matter, clouds, etc This mixture is not homogeneous Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 22 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 1.92 First, let’s calculate the mass (in g) of water in the pool We perform this conversion because we know there is g of chlorine needed per million grams of water (2.0 ´ 104 gollons H2O) ´ gallon ´ 0.001 L ´ mL = 7.58 ´ 107 g H2O 3.79 L mL 1g Next, let’s calculate the mass of chlorine that needs to be added to the pool (7.58 ´ 107 g H2 O) ´ ´ 106 g H O g chlorine = 75.8 g chlorine The chlorine solution is only percent chlorine by mass We can now calculate the volume of chlorine solution that must be added to the pool 75.8 g chlorine ´ 1.93 100% soln 6% chlorine ´ mL soln g soln = 1.3 ´ 10 mL of chlorine solution The volume of the cylinder is: 2 V = πr h = π(0.25 cm) (10 cm) = 2.0 cm The number of Al atoms in the cylinder is: 2.0 cm ´ 2.70 g cm 1.94 (a) ´ Al atom -23 4.48 ´ 10 = 1.2 ´ 10 23 Al atoms g The volume of the pycnometer can be calculated by determining the mass of water that the pycnometer holds and then using the density to convert to volume (43.1195 - 32.0764) g ´ (b) 0.99820 g = 11.063 mL Using the volume of the pycnometer from part (a), we can calculate the density of ethanol (40.8051 - 32.0764) g 11.063 mL (c) mL = 0.78900 g/ mL From the volume of water added and the volume of the pycnometer, we can calculate the volume of the zinc granules by difference Then, we can calculate the density of zinc volume of water = (62.7728 - 32.0764 - 22.8476) g ´ mL 0.99820 g = 7.8630 mL volume of zinc granules = 11.063 mL - 7.8630 mL = 3.200 mL Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang density of zinc = 1.95 22.8476 g 3.200 mL 23 = 7.140 g/mL Let the fraction of gold = x, and the fraction of sand = (1 – x) We set up an equation to solve for x 3 (x)(19.3 g/cm ) + (1 – x) (2.95 g/cm ) = 4.17 g/cm 19.3x – 2.95x + 2.95 = 4.17 x = 0.0746 Converting to a percentage, the mixture contains 7.46% gold 1.96 First, convert 10 mm to units of cm 10 mm ´ ´ 10-4 cm mm = 1.0 ´ 10-3 cm Now, substitute into the given equation to solve for time t = (1.0 ´ 10-3 cm)2 x2 = = 0.88 s 2D 2(5.7 ´ 10-7 cm /s) It takes 0.88 seconds for a glucose molecule to diffuse 10 mm 1.97 11 The mass of a human brain is about kg (1000 g) and contains about 10 cells The mass of a brain cell is: 1000 g 11 ´ 10 cells = ´ 10-8 g/cell Assuming that each cell is completely filled with water (density = g/mL), we can calculate the volume of each cell Then, assuming the cell to be cubic, we can solve for the length of one side of such a cell ´ 10-8 g cell Vcube = a a = (V) ´ mL 1g ´ cm mL = ´ 10-8 cm /cell 1/3 = (1 × 10 -8 1/3 cm ) = 0.002 cm 11 Next, the height of a single cell is a, 0.002 cm If 10 cells are spread out in a thin layer a single cell thick, 11 the surface area can be calculated from the volume of 10 cells and the height of a single cell V = surface area × height Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 24 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 11 The volume of 10 brain cells is: 1000 g ´ mL 1g ´ cm mL = 1000 cm The surface area is: æ1 ´ 10-2 m ÷ư2 1000 cm3 V ÷÷ çç Surface area = = = ´ 10 cm ỗỗ ữ = 10 m height 0.002 cm ỗỗ cm ữữữ ố ứ 1.98 (a) A concentration of CO of 800 ppm in air would mean that there are 800 parts by volume of CO per million parts by volume of air Using a volume unit of liters, 800 ppm CO means that there are 800 L of CO per million liters of air The volume in liters occupied by CO in the room is: ổ cm ữử3 1L ỗ ữữ 17.6 m ´ 8.80 m ´ 2.64 m = 409 m3 ỗỗỗ = 4.09 105 L air ữữ ỗỗ1 10 m ữữ 1000 cm è ø 4.09 ´ 10 L air ´ (b) m3 ´ ´ 10-3 g mg -3 mg = × 10 120 mg dL 1.99 ´ 106 L air = 327 L CO mg = × 10- g and L = 1000 cm We convert mg/m to g/L: 0.050 mg (c) 8.00 ´ 102 L CO ´ 3 ổ1 10-2 m ữử3 1000 cm3 ỗ ữữ ỗỗỗ = 5.0 10-8 g/L ữữ L ỗỗ cm ữ ữứ ố -2 mg and mL = × 10 mg ´ 10-3 mg ´ ´ 10-2 dL mL dL We convert mg/dL to mg/mL: = 1.20 ´ 10 μg /mL This problem is similar in concept to a limiting reagent problem We need sets of coins with quarters, nickel, and 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) ´ quarter 5.645 g nickel 4.967 g dime 2.316 g = 6000 quarters = 2100 nickels = 3450 dimes Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 25 Next, we need to find which coin limits the number of sets that can be assembled For each set of coins, we need dimes for every nickel 2100 nickels ´ dimes nickel = 4200 dimes We not have enough dimes For each set of coins, we need dimes for every quarters 6000 quarters ´ dimes quarters = 4000 dimes Again, we not have enough dimes, and therefore the number of dimes is our “limiting reagent” If we need dimes per set, the number of sets that can be assembled is: 3450 dimes ´ set dimes = 1725 sets The mass of each set is: ổ 5.645 g ữử ổỗ 4.967 g ữử ổỗ 2.316 g ữử ỗỗ ữữ + ỗ1 nickel ữữ + ỗ2 dimes ữữ = 26.534 g/set quarters ỗỗ ỗ ỗ quarter ữữứ ỗỗố nickel ữữứ ỗỗố dime ữữứ ỗố Finally, the total mass of 1725 sets of coins is: 1725 sets ´ 1.100 26.534 g set = 4.577 ´ 104 g 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 40.00 mL = 0.8620 g/mL From the second experiment: Mass of oil = 50.952 g - 18.713 g = 32.239 g Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 26 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition By Chang Volume of oil = 32.239 g ´ mL 0.8620 g = 37.40 mL The volume of the ball bearing is obtained by difference 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 2.60 cm = 7.20 g/cm Using the formula for the volume of a sphere, we can solve for the radius of the ball bearing V = pr 2.60 cm3 = pr 3 r = 0.621 cm r = 0.853 cm 1.101 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) 1.102 We want to calculate the mass of the cylinder, which can be calculated from its volume and density The volume of a cylinder is pr 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: V = pr l V = p(6.44 cm) (44.37 cm) V = 5781 cm The density of the cylinder is: 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 ì volume Copyright â McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 3 27 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 1.103 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 1.104 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 1.105 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 Mass of hydrochloric acid = 40.00 mL ´ 1.140 g mL = 45.60 g Mass of solution before reaction = 45.60 g + 1.328 g = 46.928 g 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 CO2 released = 0.229 g ´ 1.106 1L 1.81 g = 0.127 L As water freezes, it expands First, calculate the mass of the water at 20C Then, determine the volume that this mass of water would occupy at -5C Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 28 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition By Chang Mass of water = 242 mL ´ 0.998 g mL Volume of ice at - 5C = 241.5 g ´ = 241.5 g mL 0.916 g = 264 mL The volume occupied by the ice is larger than the volume of the glass bottle The glass bottle would crack! 1.107 Basic approach: · Estimate the mass of one ant In this problem a reasonable estimate is provided, but in future problems of this type you will need to provide such estimates and look up certain constants See page 24 of your textbook for some general advice on looking up chemical information 23 · Multiply by ´ 10 to obtain the mass of one mole of ants, and convert mg to kg The mass of one mole of ants in kilograms is ´ 1023 ´ mg ´ 1´10 –3 g kg ´ » ´ 1018 kg mg 1000 g It is interesting to compare this mass with the total mass of Earth’s human population The world population is very close to 6.9 billion people Assuming an average body mass of 150 lb, we write 6.9 ´ 109 ´ 150 lb ´ 453.6 g kg ´ » ´ 1011 kg lb 1000 g Thus, one mole of ants outweigh all humans by million times! 1.108 Basic approach: · Assume an average amount of time that a person sleeps in one night · Multiple that value by the number of nights in 80 years We assume that an average person sleeps eight hours a night Although an infant or a young child sleeps more than eight hours a day, as one grows older, one tends to sleep less and less Thus, the time spent sleeping in 80 years is h sleep 365 days day yr ´ ´ 80 yr ´ ´ » 27 yr day yr 24 h 365 days So regardless of an adult’s age, on average, a third of his/her life is spent sleeping! Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 29 Full1.109 file atBasic https://TestbankDirect.eu/Solution-Manual-for-Chemistry-13th-Edition-By-Chang approach: · List the major ways water is used indoors by a typical family · Estimate the volume of water used according to the activity and the number of times that activity occurs in one day · Calculate total volume of water used in one day We can estimate the usage of water in the following categories: Showers Each person takes one shower per day; each shower lasts for minutes; gallons of water used per minute people ´ shower/person ´ minutes/shower ´ gallons/minute = 64 gallons Washing Hands Each person washes hands six times a day; each washing uses 0.5 gallons of water people ´ washings/person ´ 0.5 gallon/washing = 12 gallons Brushing Teeth Each person brushes teeth twice a day; each brushing uses 0.5 gallon of water people ´ brushings/person ´ 0.5 gallon/brushing = gallons Flushing Toilets Each person uses the toilet four times a day; each flushing uses 1.5 gallons of water people ´ flushes/person ´ 1.5 gallons/flush = 24 gallons Dishwasher It is used twice a day; for newer dishwashers, each wash uses gallons of water dishwashing ´ gallons/dishwashing = 12 gallons Laundry Seven loads of laundry per week; each load uses 40 gallons of water week ´ loads/week ´ 40 gallons/load = 40 gallons Other Miscellaneous Uses (washing dishes by hand, watering plants, drinking, etc.) 20 gallons Summing all of the activities gives 176 gallons Obviously this is a very rough estimate, and the amount of water used by a family of four in one day will vary considerably depending on the habits of that family; however, this estimate is almost certainly within a factor of two, and it is noteworthy that the daily water consumption of a typical US family is considerable and rather extravagant by international standards 1.110 Basic approach: · Calculate the volume of a bowling ball in cm Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 30 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition By Chang Full ·file at https://TestbankDirect.eu/Solution-Manual-for-Chemistry-13th-Edition-By-Chang Convert the mass of the ball to grams · Calculate the density of the bowling ball and compare to the density of water (1 g/cm ) The volume of the bowling ball is given by 4 ổ 8.6 in ữử ữữ pr = p ỗỗ 3 ỗố ữứ ổ 2.54 cm ửữ3 ỗỗ ữ = 5.5 103 cm3 ốỗ in ứữữ Starting with an lb bowling ball and assuming two significant figures in the mass, converting pounds to grams gives lb ´ 453.6 g = 3.6 ´ 103 g lb So the density of an lb bowling ball would be 3.6 ´ 103 g mass density = = = 0.65 g/cm3 3 volume 5.5 ´ 10 cm Carrying out analogous calculations for the higher-weight bowling balls gives ball weight (lb) mass (g) density (g/cm ) 0.75 0.82 0.91 0.98 1.1 4.1 ´ 10 10 4.5 ´ 10 11 5.0 ´ 10 12 5.4 ´ 10 13 5.9 ´ 10 Therefore we would expect bowling balls that are 11 lb or lighter to float since they are less dense than water Bowling balls that are 13 lb or heavier would be expected to sink since they are denser than water The 12 lb bowling ball is borderline, but it would probably float Note that the above calculations were carried out by rounding off the intermediate answers, as discussed in Section 1.8 If we carry an additional digit past the number of significant figures to minimize errors from rounding, the following densities are obtained: ball weight (lb) density (g/cm ) 10 0.83 11 0.91 12 1.01 Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 31 The differences in the densities obtained are slight, but the value obtained for the 12 lb ball now suggests that it might sink This problem illustrates the difference rounding off intermediate answers can make in the final answers for some calculations 1.111 Basic approach: · Approximate the shape of the juncture · Estimate the dimensions in nm and calculate the volume, converting to L The volume could be approximated in several ways One approach would be to model the junction as a cylinder with an internal diameter of 200 nm and a height of 200 nm The volume of the cylinder would be æ 10 cm ữử3 mL ổ 200 nm ữử2 1L ỗ ữữ ữ 200 nm ỗỗ pr h = p ỗỗ = 10-18 L ỗỗố109 nm ữữứ 1000 mL ốỗ ữữứ cm To get a sense of that volume relative to a molecule, consider that the volume of a water molecule in liquid water is roughly ´ 10 ( ´ 10 -26 -26 -18 L Therefore a volume of ´ 10 -18 L could still potentially contain (6 ´ 10 L)/ L) ≈ ´ 10 water molecules! Water molecules are very small, and they pack tightly in the liquid state due to strong intermolecular forces (Chapter 11) For larger molecules such as those involved in biological processes, it might be possible to trap a much smaller number in a nanofiber juncture 1.112 Basic approach: · Look up or estimate the number of cars currently operating in the United States · Estimate the average fuel efficiency (miles per gallon) and average miles driven by a car in one year The information from the web shows there are about 250 million passenger cars in the US Assume on average each car covers 10,000 miles at the gas consumption rate of 20 mpg (miles per gallon) The total number of gallons of gasoline consumed in a year is 250 ´ 106 cars ´ 1.113 10000 mi gal ´ » 1´1011 gal car 20 mi Basic approach: · Look up the percentage of Earth’s surface covered by oceans and the average depth of the ocean · Calculate volume based on the surface area of the oceans and the average depth About 70 percent of Earth is covered with water Useful information from the web: Average depth of ocean is 4000 m; radius of Earth = 6400 km or 6.4 ´ 10 cm Using the formula for the surface area of a sphere of radius r as (4pr ), we calculate the area of ocean water as follows: 18 4pr ´ 0.7 = 4p(6.4 ´ 10 cm) ´ 0.7 = 3.6 ´ 10 cm Volume is area ´ depth so the volume of ocean water is 18 24 21 (3.6 ´ 10 cm ) ´ (4 ´ 10 cm) » ´ 10 cm = ´ 10 L Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 32 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition By Chang Full Basic file at https://TestbankDirect.eu/Solution-Manual-for-Chemistry-13th-Edition-By-Chang approach: 1.114 · Model the shape and approximate the dimensions of an average human body, and calculate volume · Look up the percent water in the body, and estimate the fraction of that water contained in the blood stream One way to proceed is to assume that the human body can be treated as a cylinder of height h and diameter d The volume of the body is then given by pr ´ h where r is the radius For a ft tall person with an average width of 15 in, the volume in cubic inches is p(15 in/2) ´ ft ´ 12 in/ft = 1.3 ´ 10 in Expressed in liters, æ 2.54 cm ÷ư3 1L ÷ ´ 1.3 ´ 10 in3 ỗỗ = 210 L ỗố in ữữứ 1000 cm3 About 60 percent of a human body is water, so the volume of water is 0.6 ´ 210 L = 130 L Assuming that one-tenth of the water is blood, the volume of blood would be 13 L This calculation turns out to be a rather rough estimate, where most of our error probably comes from our assumption that 10% of the water in our bodies is contained in the blood stream In reality, the volume of blood in an adult is around L 1.115 Basic approach: · Look up the speed of light and calculate the distance traveled in one nanosecond -9 The speed of light is ´ 10 m/s and a nanosecond is ´ 10 s The distance covered by light in this time period is ´ 108 m/s ´ 100 cm in ft 1´10 –9 s ´ ´ ´ = ft 1m 2.54 cm 12 in ns Our senses tell us that objects are instantly illuminated by light, but this calculation shows that while the speed of light is very fast, it travels finite and measurable distances in very short periods of time 1.116 Basic approach: · Break up period of time over which a professional basketball game is played according to the level of activity of the players · Estimate the time spent engaged in that activity and the average running speed · Calculate the total distance traveled An NBA game lasts 48 minutes We can divide this time period into three parts: (1) dashing, (2) running, and (3) movement during offense and defense under the basket Reasonable estimates for time and speed during these activities are: dashing (3 at 10 m/s); running (5 at m/s); movement under basket (40 at Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1: CHEMISTRY: THE STUDY OF CHANGE Solution Manual for Chemistry 13th Edition ByCHAPTER Chang 33 m/s) (Note that both the 10 m/s and m/s estimates are near world track records.) Summing up, we calculate the average distance covered by the player as (3 ´60 s/min ´10 m/s) + (5 ´60 s/min´7 m/s) + (40 ´60 s/min ´3 m/s) » 1´104 m Expressed in miles, ´ 104 m ´ km mi ´ » mi 1000 m 1.61 km Alternatively, one can make a quicker (and cruder) estimate by assuming that the players traverse the length of the court (94 ft) every 24 seconds based on the shot clock Over 48 minutes, the distance traveled would be 48 ´ 60 s 94 ft mi ´ ´ » mi 24 s 5280 ft Taking the higher estimate, this distance would correspond to quite a bit of running, with an average speed of roughly mile every minutes Keep in mind, however, that hardly any player is on the court for a full 48 minutes of a game, and all players get a break during the timeouts 1.117 Basic approach: · Calculate the thickness of the oil layer from the volume and surface area, and assume that to be the length of one molecule 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 ỉ cm ÷ư2 ÷ = 4.0 ´ 105 cm 40 m ỗỗ ỗố 0.01 m ữữứ Given that 0.10 mL = 0.10 cm , we write volume = surface area ´ thickness and rearranging that equation gives thickness = volume 0.10 cm = = 2.5 ´ 10-7 cm surface area 4.0 ´ 10 cm Converting to nm: 2.5 ´ 10 –7 cm ´ 0.01 m nm ´ = 2.5 nm cm ´ 10 –9 m Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ... Solution Manual for Chemistry 13th Edition ByCHAPTER Chang Full file at https://TestbankDirect.eu /Solution-Manual-for-Chemistry-13th-Edition-By-Chang 5C 1.25 ? C = (F - 32F) ´ (a) (b) (c) (d)... for Chemistry 13th Edition ByCHAPTER Chang 29 Full1.109 file atBasic https://TestbankDirect.eu /Solution-Manual-for-Chemistry-13th-Edition-By-Chang approach: · List the major ways water is used... CHANGE Solution Manual for Chemistry 13th Edition By Chang Full ·file at https://TestbankDirect.eu /Solution-Manual-for-Chemistry-13th-Edition-By-Chang Convert the mass of the ball to grams · Calculate