(Solution Manual) Chemistry

Chang, Goldsby, 13

Chemistry

Solution Manual

13th Edition

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Solution Manual for Chemistry, The downloadable file consists of 975 pages, and the total size of the file is 6.61 MB. Purchase now only for highly discounted price.

Author(s): Chang, Goldsby

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CHAPTER 1 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 (b) law (c) theory 1.9 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. 1.10 (a) Cs (b) Ge (c) Ga (d) Sr (e) U (f) Se (g) Ne (h) Cd 1.11 (a) element (b) compound (c) element (d) compound (Chemistry, 13e Raymond Chang, Kenneth Goldsby) (Solution Manual all Chapter) CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is proprietary material so lely for authorized instructor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 2 1.12 (a) homogeneous mixture (b) element (c) compound (d) homogeneous mixture (e) heterogeneous mixture (f) heterogeneous mixture (g) element 1.17 (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 boilin g 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 uran ium undergoes nuclear decay, the pr oducts 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. 1.23 586 g mass volume 188 mL=== density 3.12 g/mL 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. =massdensityvolume Solution: mass = density ´ volume =´ =0.7918 g 89.9 mL 1 mLmass of methanol 71.2 g CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is proprietary material so lely for authorized instructor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 3 1.25 =-´5C?C ( F 3 2 F )9F (a) =- ´=5C? C (95 32) F9F35 C (b) =- ´= -5C?C ( 1 2 3 2 ) F9F11 C (c) =- ´=5C?C ( 1 0 2 3 2 ) F9F39 C (d) =- ´=5C? C (1852 32) F9F1011 C (e) 9F? F C 32 F5Cæö÷ ç ÷  =  ´ +  ç ÷ ç ÷ ç èø æö ÷ ç ÷ =- ´+ =- ç ÷ ç ÷ ç  èø9F? F 273.15 C 32 F5C459.67 F 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 te mperature values given in the problem into the ap propriate equation. (a) Conversion from Fahrenheit to Celsius. =-´5C?C ( F 3 2 F )9F =- ´=5C(105 32) F9F?C 4 1 C (b) Conversion from Celsius to Fahrenheit. æö÷ ç ÷ =´+  ç ÷ ç ÷ ç èø9F? F C 32 F5C æö ÷ ç ÷ =- ´+ = ç ÷ ç ÷ ç  èø9F11.5 C 32 F5C?F 1 1 . 3 F (c) Conversion from Celsius to Fahrenheit. æö÷ ç ÷ =´+  ç ÷ ç ÷ ç èø9F? F C 32 F5C æö ÷ ç ÷ =´ ´+ = ç ÷ ç ÷ ç  èø3 9F6.3 10 C 32 F5C4?F 1 . 1 1 0 F´  (d) Conversion from Fahrenheit to Celsius. =-´5C?C ( F 3 2 F )9F =- ´=5C(451 32) F9F?C 2 3 3 C CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is proprietary material so lely for authorized instructor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 4 1.27 =+1K K( 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 1.28 (a) =+1K K( 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 1.31 (a) 2.7 ´ 10-8 (b) 3.56 ´ 102 (c) 4.7764 ´ 104 (d) 9.6 ´ 10-2 1.32 (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 mu st be moved 8 places to the left. 7.78 ´ 10-8 = 0.0000000778 1.33 (a) 145.75 + (2.3 ´ 10-1) = 145.75 + 0.23 = 1.4598 ´ 102 (b) 4 227.95 10 79500 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 1.34 (a) Addition using scie ntific notation. Strategy: Let’s express scien tific notation as N ´ 10n. When adding numbers usin g 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. CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 5 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 scien tific 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 ex pressed 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 scien tific 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. 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 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 6 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 scien tific notation as N ´ 10n. When multiplying numb ers 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 1.35 (a) four (b) two (c) five (d) two, three, or four (e) three (f) one (g) one (h) two 1.36 (a) one (b) three (c) three (d) four (e) two or three (f) one (g) one or two 1.37 (a) 10.6 m (b) 0.79 g (c) 16.5 cm2 (d) 1 × 106 g/cm3 1.38 (a) Division Strategy: The number of signifi cant figures in the answer is determined by the original number having the smallest number of significant figures. CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 7 Solution: 7.310 km 5.70 km = 1.28 3 The 3 (bolded) is a nonsignificant digit because the ori ginal 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?) (b) Subtraction Strategy: The number of significant fi gures 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 fi gures 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 significant figures to the right of the decimal point in that part of the calculation is determin ed by the lowest number of digits to the right of the decimal point in CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8 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 1.39 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 volu me 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 meas urements made by both tailors X and Z are fairly high, while the measurements made by tailo r Y are less precise. In summary: Tailor X: most precise Tailor Y: least accurate and least precise Tailor Z: most accurate 1.41 (a) =´ =1d m 22.6 m0.1 m?d m 2 2 6d m (b) 0.001g 1kg 25.4 mg1mg 1000 g=´ ´ =5?k g 2 . 5 4 1 0 k g-´ (c) 311 0 L 556 mL1m L-´ =´ = ? L 0.556 L CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 9 (d) 32 310.6 kg 1000 g 1 10 m 1k g 1c m 1m- æö÷ ´ç ÷ ç ÷ ç =´ ´ = ÷ ç ÷ ç ÷÷ çèø3 3g?0 . 0 1 0 6 g c m cm/ 1.42 (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: -=´ ´ = ´31m g 453.6 g 242 lb 1 lb 11 0 g8? mg 1.10 10 mg ´ 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 ? 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 that centimeters cancel and we end up with meters. -´211 0m 1c m CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 10 Since this conversion factor deals with length and we want volume, it must therefore be cubed to give --- - æö÷ ´´´ ´ ç ÷ ç ÷ ç ´´ = ÷ ç ÷ ç ÷÷ çèø3222 211 0m 11 0m 11 0m 11 0m 1c m 1c m 1c m 1c m We can write - æö÷ ´ç ÷ ç ÷ ç =´ = ÷ ç ÷ ç ÷÷ çèø32 311 0 m 68.3 cm 1 cm 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 conver sion is given be tween 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 appropriate conversion factors so that m3 and cm3 cancel, and the unit liters is obtain ed in your answer. Solution: The sequence of conversions is m3  cm3  L Using the following conversion factors, -æö÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ´ ÷ çèø3 21c m 11 0 m 31L 1000 cm the answer is obta ined in one step: -æö÷ ç ÷ ç ÷ ç =´ ´ = ÷ ç ÷ ç ÷ ´ ÷ çèø3 3 231 cm 1 L 7.2 m 1 10 m 1000 cm 3? L 7.2 10 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 Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 11 (d) Strategy: The problem may be stated as ? lb = 28.3 mg 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 mg to grams, we can then convert from grams to pounds. Recall that 1 mg = 1 ´ 10-6 g. Arrange the appropri ate conversion factors so that mg and grams cancel, and the unit pounds is obtained in your answer. Solution: The sequence of conversions is mg  g  lb Using the following conversion factors, -´ m611 0 g 1g 1l b 453.6 g we can write -´ =m ´ ´ =m611 0 g 1 lb 28.3 g 1 g 453.6 g 8? lb 6.24 10 lb-´ Check: Does the answer seem reasonabl e? What number does the prefix m represent? Should 28.3 mg be a very small mass? 1.43 1255 m 1 mi 3600 s 1 s 1609 m 1 h´´ = 2808 mi h/ 1.44 Strategy: The problem may be stated as ? s = 365.24 days You should know conversion factors that will allow you to convert betw een 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 CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 12 we can write 24 h 60 min 60 s 365.24 days 1 day 1 h 1 min =´ ´ ´ = ´7? s 3.1557 10 s Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year? 1.45 ´´ ´ ´ ´ = ´6 81.609 km 1000 m 1 s 1 min (93 10 mi)1m i 1k m 6 0s 3.00 10 m8.3 min 1.46 (a) ? in/s = ´´ ´1 mi 5280 ft 12 in 1 min 8.92 min 1 mi 1 ft 60 s = 118 in/s (b) ? m/min = ´1 mi 1609 m 8.92 min 1 mi= 1.80 ´210 m/min (c) ? km/h = ´´´1 mi 1609 m 1 km 60 min 8.92 min 1 mi 1000 m 1 h = 10.8 km/h 1.47 ´=1 m6.0 ft3.28 ft1.8 m ´´ =453.6 g 1 kg 168 lb1l b 1 0 0 0g76.2 kg 1.48 ? mph = ´286 km 1 mi 1 h 1.609 km= 178 mph 1.49 ´´ =62 m 1 mi 3600 s 1 s 1609 m 1 h21.4 10 mph´ 1.50 = ´60.62 g Pb 0.62 ppm Pb 1 10 g blood ´´ = ´ ´3 60.62 g Pb6.0 10 g of blood 1 10 g blood33.7 10 g Pb- 1.51 (a) 8365 days 24 h 3600 s 3.00 10 m 1 mi 1.42 yr1y r 1d a y 1h 1s 1 6 0 9m´ ´´ ´ ´ ´ =128.35 10 mi´ (b) ´´36 in 2.54 cm 32.4 yd1y d 1i n32.96 10 cm=´ CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 13 (c) 103.0 10 cm 1 in 1 ft 1 s 2.54 cm 12 in´ ´´ =89.8 10 ft/s´ 1.52 (a) =´ =1l b 70 kg0.4536 kg2?l b s 1 . 5 1 0 l b s ´ (b) =´ ´ ´ ´ =9365 d 24 h 3600 s 14 10 yr1y r 1d 1h17?s 4 . 4 1 0 s ´ (c) -´=´ ´2 2.54 cm 1 10 m90 in1i n 1c m? m 2.3 m = (d) -æö÷ ç ÷ =´ ´ ç ÷ ç ÷ çèø´3 3 231 cm 1 L88.6 m 1 10 m 1000 cm4? L 8.86 10 L =´ 1.53 3 32.70 g 1 kg 1 cm 1000 g 0.01 m 1c mæö÷ ç ÷ ç =´´ = ÷ ç ÷ ç ÷÷ çèø33density 2.70 10 kg m ´/ 1.54 30.625 g 1 L 1 mL 1 L 1000 mL 1 cm-=´ ´ =43density 6.25 10 g cm ´/ 1.55 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 H2O (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 1.56 See Section 1.6 of your text for a discussion of these terms. (a) Chemical property. Iron has changed its compos ition and identity by che mically 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. CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 14 (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 liquid state to the gaseous state. (e) Chemical property. The carbon dioxide is chemically converted into other molecules. 1.57 ´´ = ´9 31t o n (95.0 10 lb of sulfuric acid) 2.0 10 lb74.75 10 tons of sulfuric acid´ 1.58 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/ 1.59 mass = density ´ volume (a) mass = (19.3 g/cm3) ´ [4 3p(10.0 cm)3] = 8.08 ´ 104 g (b) -æö÷ ç ÷ ç =´ ´ = ÷ ç ÷ ç ÷÷ çèø3 31c m (21.4 g/cm ) 0.040 mm10 mm6mass 1.4 10 g ´ (c) mass = (0.798 g/mL) (50.0 mL) = 39.9 g 1.60 You are asked to solve for the inner diameter of the bottle. If we can cal culate the volume that the cooking oil occupies, we can calculate the radius of the cylinder. The volume of the cylinder is, Vcylinder = pr2h (r is the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2 r. =mass of oilvolume of oil filling bottle density of oil == ´ = ´33 3 1360 gvolume of oil filling bottle 1.43 10 mL 1.43 10 cm0.953 g/mL Next, solve for the radius of the cylinder. Volume of cylinder = pr2h volume =p´rh ´ ==p´331.43 10 cm 4.60 cm 21.5 cmr The inner diameter of the bottle equals 2 r. Bottle diameter = 2r = 2(4.60 cm) = 9.20 cm CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 15 1.61 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 ==331.27 g mass=density 0.9976 g/cm3Volume of the flask 31.35 cm 1.62 ´´ =343 m 1 mi 3600 s 1 s 1609 m 1 h767 mph 1.63 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 / 1.64 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. 1.65 The volume of a sphere is: æö÷ ç ÷ ç =p=p = ´ ÷ ç ÷ ç ÷ çèø3 34 348.6 cm 446.01 10 cm33 2Vr ´ == = ´5 436.852 10 g mass volume 6.01 10 cmdensity 11.4 g/cm3 1.66 =massVolumedensity ´ ==3 31.20 10 g 0.53 g / cm33Volume occupied by Li 2.3 10 cm ´ CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 16 1.67 For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C.  ´= 5C? C = 0.1 F 0.056 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 measurement Percent error 100%value of the measurement For the Fahrenheit thermometer, =´ =0.056 C100%38.9 Cpercent error 0.1%  For the Celsius thermometer, =´ =0.1 C100%38.9 Cpercent error 0.3% Which thermometer is more accurate? 1.68 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 dimensiona l analysis method described in Section 1.9 of the text. First, converting from cubic feet to liters: ()-´ æö æ ö÷÷ çç ÷÷ ´´ ´ ´ ´ = ´ çç ÷÷ çç ÷÷ ççèø è ø3 33 73 9 311 0 L 12 in 2.54 cm 1 mL5.0 10 ft 1.42 10 L1 ft 1 in 1 mL 1 cm The mass of vanillin (in g) is: ()- -´ ´´ = ´11 922.0 10 g vanillin 1.42 10 L 2.84 10 g vanillin1 L The cost is: ()-´´ =2 $1122.84 10 g vanillin 50 g vanillin$0.064 6.4 c= 1.69 æö÷ ç ÷ =´+  ç ÷ ç ÷ ç èø9F?F C 3 2 F5C Let temperature = t 9=+ 3 2 F5tt - 9=3 2F5tt - 4=3 2F5t t = - 40F = - 40C CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 17 1.70 There are 78.3 + 117.3 = 195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor. æö÷ ç ÷ ç ÷ ç ÷ ç èø195.6 C 100 S 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. æö÷ ç ÷  - ç ÷ ç ÷ ç èø195.6 C? C = (? S ) 117.3 C100 S Solving for ? °S gives: æö÷ ç ÷   ç ÷ ç ÷ ç èø100 S? S = (? C + 117.3 C)195.6 C For 25°C we have: 100 S (25 + 117.3) C195.6 Cæö÷ ç ÷ =  = ç ÷ ç ÷ ç èø? S 73 S 1.71 The key to solving this problem is to realize that a ll 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, 1m i n 6000 mL of inhaled air 1 min 12 breaths=´ =2volume of air/breath 5 10 mL breath ´/ 1.72 (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 day / 1.73 Twenty-five grams of the least dense metal (solid A) will occupy the great est volume of the three metals, and 25.0 g of the most dense metal (so lid B) will occupy the least volume. We can calculate the volume occupied by each metal an d then add the volume of water (20.0 mL) to find the total volume occupied by the metal and water. CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 18 Solid A: ´= =+ =1m L 25.0 g A 8.6 mL2.9 g A Total volume 8.6 mL 20.0 mL 28.6 mL Solid B: ´= =+ =1m L 25.0 g B 3.0 mL8.3 g B Total volume 3.0 mL 20.0 mL 23.0 mL Solid C: ´= =+ =1m L 25.0 g C 7.6 mL3.3 g C Total volume 7.6 mL 20.0 mL 27.6 mL Therefore, we have: (a) solid C , (b) solid B , and (c) solid A . 1.74 The diameter of the basketball can be calculated from its circumference. We can then use the diameter of a ball as a conversion factor to determine the number of basketballs needed to circle the equator. Circumference = 2π r == = =pp29.6 in circumference29 . 4 2 i n dr The circumference of the Earth is 2π r = (2 × π × 6400 km) = 40212 km -´´ ´ ´ = ´21000 m 1 cm 1 in 1 ball40212 km 168,000,000.balls1km 2.54 cm 9.42 in 11 0 / m We round up to an integer number of basketballs with 3 significant figures. 1.75 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 g Volume of crucible 21.45 g/cm340.10 cm - ==3(860.2 820.2)g Volume of water displaced 0.9986 g/cm340.1 cm The volumes are the same (within experimental error), so the crucible is made of platinum. CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 19 1.76 Volume = surface area ´ depth Recall that 1 L = 1 dm3. Let’s convert the su rface area to units of dm2 and the depth to units of dm. ()æö æ ö÷÷ çç ÷÷ çç =´ ´ ´ =´ ÷÷ çç ÷÷ çç ÷÷÷÷ ççèø è ø22 82 1 621000 m 1 dm surface area 1.8 10 km 1.8 10 dm 1k m 0 . 1m ()=´ ´ = ´341d m depth 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 1.77 (a) ´=31.103 g Au 2.41 troy oz Au1 troy oz Au75.0 g Au (b) 1 troy oz = 31.103 g =´ ´ =1 lb 453.6 g ? g in 1 oz 1 oz16 oz 1 lb28.35 g A troy ounce is heavier than an ounce. 1.78 34Volume of sphere π3= r æö÷ ç ÷ ç =p = ´ ÷ ç ÷ ç ÷ çèø3 3315 cm4Volume 1.77 10 cm32 () =´= ´ ´ ´ =33 322.57 g Os 1 kg mass volume density 1.77 10 cm 1000 g 1c m14.0 10 kg Os´ 12.205 lb 4.0 10 kg Os 1k g´ ´ = 88 lb Os 1.79 (a) - ´=0.798 g/mL 0.802 g/mL 100%0.798 g/mL0.5% (b) - ´=0.864 g 0.837 g 100%0.864 g3.1% 1.80 62 kg = 6.2 ´ 104 g O: (6.2 ´ 104 g)(0.65) = 4.0 ´ 104 g O N: (6.2 ´ 104 g)(0.03) = 2 ´ 103 g N CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 20 C: (6.2 ´ 104 g)(0.18) = 1.1 ´ 104 g C Ca: (6.2 ´ 104 g)(0.016) = 9.9 ´ 102 g Ca H: (6.2 ´ 104 g)(0.10) = 6.2 ´ 103 g H P : (6.2 ´ 104 g)(0.012) = 7.4 ´ 102 g P 1.81 3 minutes 43.13 seconds = 223.13 seconds Time to run 1500 meters is: ´´ =1 mi 223.13 s 1500 m1609 m 1 mi208.01 s 3 min 28.01 s = 1.82 ? C = (7.3 ´ 102 - 273) K = 4.6 ´ 102C 2 9F(4.6 10 C) 32 F 5Cooo oæö÷ ç ÷ ç=´ ´ + = ÷ ç ÷ ç ÷÷ çèø2? F 8.6 10 F´  1.83 =´ ´ ´ =334.63% Cu 1000 g (5.11 10 kg ore)100% ore 1 kg6? g Cu 1.77 10 g Cu ´ 1.84 ()´´´ ´ =42 0 0 0l bA u 1 6o zA u $9488.0 10 tons Au or 1t o nA u 1l bA u 1o zA u12$2.4 10 2.4 trillion dollars´ 1.85 -´ =´ ´ ´ = ´12 214.0 10 g Au 1 mL (1.5 10 L seawater)1 mL seawater 0.001 L12? g Au 6.0 10 g Au =´ ´ ´ ´ = ´121l b 1 6o z $948(6.0 10 g Au)453.6 g 1 lb 1 oz14value of gold $2.0 10 No one has become rich mining gold from the ocean, b ecause the cost of recovering the gold would outweigh the price of the gold. 1.86 ´ == ´221.1 10 Fe atoms 4.9 g Fe1.0 g Fe22?F ea t o m s 5 . 4 1 0 F ea t o m s 1.87 =´ ´ = ´21 190.50% crust mass of Earth's crust (5.9 10 tons) 2.95 10 tons100% Earth =´ ´ ´ ´ = ´1927.2% Si 2000 lb 1 kg (2.95 10 tons crust)100% crust 1 ton 2.205 lb21mass of silicon in crust 7.3 10 kg Si CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 21 1.88 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 ´ 10-10 m). Let n be the number of times we can cut the Cu wire in half. We can write: æö÷ç÷´= ´ ç÷ç÷çèø10 10.1 m 2.6 10 m2n -æö÷ç÷=´ ç÷ç÷çèø9 12.6 10 m2n Taking the log of both sides of the equation: -æö÷ç÷=´ç÷ç÷çèø9 1log log(2.6 10 )2n n = 29 times 1.89 ´´´ ´ = ´2 65000 mi 1 gal gas 9.5 kg CO (250 10 cars)1 car 20 mi 1 gal gas11 25.9 10 kg CO 1.90 Volume = area ´ thickness. From the density, we can calculate the volume of the Al foil. == =3 33.636 g massVolume 1.3472 cmdensity 2.699 g / cm Convert the unit of area from ft2 to cm2. æö æ ö÷÷ çç ÷÷ çç´´ = ÷÷ çç ÷÷ çç ÷÷÷÷ ççèø è ø22 2212 in 2.54 cm 1.000 ft 929.03 cm1f t 1i n -== = =3 3 21.3472 cm volume1.450 × 10 cmarea 929.03 cm2thickness 1.450 10 mm-´ 1.91 (a) homogeneous (b) heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous. CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 22 1.92 First, let’s calculate the mass (in g) of water in the p ool. We perform this conversion because we know there is 1 g of chlorine needed per million grams of water. ()´´ ´ ´ = ´47 223.79 L 1 mL 1 g 2.0 10 gollons H O 7.58 10 g H O1 gallon 0.001 L 1 mL Next, let’s calculate the mass of chlorine that needs to be added to the pool. () ´´ ´7 26 21gc h l o r i n e 7.58 10 g H O 75.8 g chlorine 11 0 g H O= 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. ´´ =100% soln 1 mL soln 75.8 g chlorine 6% chlorine 1 g soln31.3 10 mL of chlorine solution´ 1.93 The volume of the cylinder is: V = πr2h = π(0.25 cm)2(10 cm) = 2.0 cm3 The number of Al atoms in the cylinder is: -´´ = ´3 32 32.70 g 1 Al atom 2.0 cm 1c m 4 . 4 8 1 0 g231.2 10 Al atoms´ 1.94 (a) The volume of the pycnom eter can be calculated by determining the mass of water that the pycnometer holds and then using the density to convert to volume. -´ =1m L (43.1195 32.0764) g0.99820 g11.063 mL (b) Using the volume of the pycnometer from part (a), we can calculate the density of ethanol. (40.8051 32.0764)g 11.063 mL- =0.78900 g mL / (c) 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. =- - ´ =1m L volume of water (62.7728 32.0764 22.8476)g0.99820 g7.8630 mL =- = volume of zinc granules 11.063 mL 7.8630 mL 3.200 mL CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 23 ==22.8476 g 3.200 mLdensity of zinc 7.140 g/mL 1.95 Let the fraction of gold = x, and the fraction of sand = (1 – x). We set up an equation to solve for x. ( x)(19.3 g/cm3) + (1 – x) (2.95 g/cm3) = 4.17 g/cm3 19.3 x – 2.95 x + 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. - -´ m´ = ´m4 311 0c m 10 m 1.0 10 cm1m Now, substitute into the given equation to solve for time. - -´ == = ´322 72(1.0 10 cm) 2 2(5.7 10 cm /s)0.88 sx Dt It takes 0.88 seconds for a glucose molecule to diffuse 10 mm. 1.97 The mass of a human brain is about 1 kg (1000 g) and contains about 1011 cells. The mass of a brain cell is: -=´ ´8 111000 g 1 10 g/cell 1 10 cells Assuming that each cell is comple tely filled with water (density = 1 g/mL), we can calculate the volume of each cell. Then, assuming the cell to be cubic, we ca n solve for the length of one side of such a cell. 83 8311 0 g 1 m L 1 c m 11 0 c m / c e l l1c e l l 1g 1m L- -´ ´´ = ´ Vcube = a3 a = (V)1/3 = (1 × 10-8 cm3)1/3 = 0.002 cm Next, the height of a single cell is a, 0.002 cm. If 1011 cells are spread out in a thin layer a single cell thick, the surface area can be calcula ted from the volume of 1011 cells and the height of a single cell. V = surface area × height CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 24 The volume of 1011 brain cells is: ´´ =3 31m L 1c m 1000 g 1000 cm1g 1m L The surface area is: - æö÷ ´ç ÷ ç ÷ ç == = ´ ´ = ÷ ç ÷ ç ÷÷ çèø232 521000 cm 1 10 m 51 0 c mheight 0.002 cm 1 cm12Surface area 5 10 mV´ 1.98 (a) A concentration of CO of 800 ppm in air would mean that there are 800 parts by volume of CO per 1 million parts by volume of air. Using a volume un it of liters, 800 ppm CO me ans that there are 800 L of CO per 1 million liters of air. The volume in liters occupied by CO in the room is: -æö÷ ç ÷ ç ÷ ç ´´ = ´ ´ = ´ ÷ ç ÷ ç ÷ ´ ÷ çèø3 35 231c m 1L 17.6 m 8.80 m 2.64 m 409 m 4.09 10 L air 1 10 m 1000 cm ´ ´´ = ´2 5 68.00 10 L CO 4.09 10 L air 11 0 L a i r327 L CO (b) 1 mg = 1 × 10-3 g and 1 L = 1000 cm3. We convert mg/m3 to g/L: 332 3 30.050 mg 1 10 g 1 10 m 1000 cm 1m g 1c m 1L 1m-- æö÷ ´´ ç ÷ ç ÷ ç ´´ ´= ÷ ç ÷ ç ÷÷ çèø85.0 10 g L-´/ (c) 1 mg = 1 × 10-3 mg and 1 mL = 1 × 10-2 dL. We convert mg/dL to mg/mL: 2 31 2 0 m g1 g1 1 0 d L 1d L 1m L 11 0 m g- -m´ ´´ = ´31.20 10 μg mL´/ 1.99 This problem is similar in concept to a limiting reagent pr oblem. 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. =´ ´ =31 quarter Number of quarters (33.871 10 g) 6000 quarters5.645 g =´ ´ =31n i c k e l Number of nickels (10.432 10 g) 2100 nickels4.967 g =´ ´ =31d i m e Number of dimes (7.990 10 g) 3450 dimes2.316 g CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE Copyright © McGraw-Hill Education. This is pr oprietary material solely for authorized in structor use. Not authorized for sale o r distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 25 Next, we need to find which coin limits the number of se ts that can be assembled. For each set of coins, we need 2 dimes for every 1 nickel. ´=2d i m e s 2100 nickels 4200 dimes1n i c k e l We do not have enough dimes. For each set of coins, we need 2 dimes for every 3 quarters. ´=2 dimes 6000 quarters 4000 dimes3 quarters Again, we do not have enough d imes, and therefore th e number of dimes is our “limiting reagent”. If we need 2 dimes per set, the number of sets that can be assembled is: ´=1s e t 3450 dimes2d i m e s1725 sets The mass of each set is: æö æ ö æ ö÷÷÷ çç ç ÷÷÷ çç ç ´+ ´ + ´ = ÷÷÷ çç ç ÷÷÷ çç ç ÷÷÷÷÷÷ çç çèø è ø è ø5.645 g 4.967 g 2.316 g 3 quarters 1 nickel 2 dimes 26.534 g/set1 quarter 1 nickel 1 dime Finally, the total mass of 1725 sets of coins is: ´=26.534 g 1725 sets1s e t44.577 10 g´ 1.100 We wish to calculate the density and radius of the ball be aring. 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 ==34.480 g Density of oil 0.8620 g/mL40.00 mL From the second experiment: Mass of oil = 50.952 g - 18.713 g = 32.239 g