Skip to content
Textbook Guides

Solution Manual For Chemistry, 13th Edition

Gain deeper insight into your textbook problems with Solution Manual For Chemistry, 13th Edition, featuring well-explained solutions.

Preview 16 of 969 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    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

  2. Page 2
  3. Page 3
    Read page 3 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 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 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. 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. = mass density volume Solution: mass = density ´ volume = ´ = 0.7918 g 89.9 mL 1 mL mass of methanol 71.2 g

  4. Page 4
    Read page 4 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 3 1.25   =  -  ´  5 C ? C ( F 32 F) 9 F (a)   = -  ´ =  5 C ? C (95 32) F 9 F 35 C  (b)   = -  ´ = -  5 C ? C (12 32) F 9 F 11 C  (c)   = -  ´ =  5 C ? C (102 32) F 9 F 39 C  (d)   = -  ´ =  5 C ? C (1852 32) F 9 F 1011 C  (e) 9 F ? F C 32 F 5 C æ ö  ÷ ç ÷  =  ´ +  ç ÷ ç ÷ ç  è ø æ ö  ÷ ç ÷  = -  ´ +  = - ç ÷ ç ÷ ç  è ø 9 F ? F 273.15 C 32 F 5 C 459.67 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 temperature values given in the problem into the appropriate equation. (a) Conversion from Fahrenheit to Celsius.   =  -  ´  5 C ? C ( F 32 F) 9 F  = -  ´ =  5 C (105 32) F 9 F ? C 41 C   (b) Conversion from Celsius to Fahrenheit. æ ö  ÷ ç ÷  =  ´ +  ç ÷ ç ÷ ç  è ø 9 F ? F C 32 F 5 C æ ö  ÷ ç ÷ = -  ´ +  = ç ÷ ç ÷ ç  è ø 9 F 11.5 C 32 F 5 C ? F 11.3 F   (c) Conversion from Celsius to Fahrenheit. æ ö  ÷ ç ÷  =  ´ +  ç ÷ ç ÷ ç  è ø 9 F ? F C 32 F 5 C æ ö  ÷ ç ÷ = ´  ´ +  = ç ÷ ç ÷ ç  è ø 3 9 F 6.3 10 C 32 F 5 C 4 ? F 1.1 10 F  ´  (d) Conversion from Fahrenheit to Celsius.   =  -  ´  5 C ? C ( F 32 F) 9 F  = -  ´ =  5 C (451 32) F 9 F ? C 233 C  

  5. Page 5
    Read page 5 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 4 1.27 =  +   1 K K ( C 273 C) 1 C (a) K = 113  C + 273  C = 386 K (b) K = 37  C + 273  C = 3.10 ´ 10 2 K (c) K = 357  C + 273  C = 6.30 ´ 10 2 K 1.28 (a) =  +   1 K K ( C 273 C) 1 C  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 ´ 10 2 (c) 4.7764 ´ 10 4 (d) 9.6 ´ 10 - 2 1.32 (a) 10 - 2 indicates that the decimal point must 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 1.33 (a) 145.75 + (2.3 ´ 10 - 1 ) = 145.75 + 0.23 = 1.4598 ´ 10 2 (b) 4 2 2 7.95 10 79500 2.5 10 2.5 10 ´ = = ´ ´ 2 3.2 × 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 ´ 10 4 ) ´ (9.9 ´ 10 6 ) = 9.9 ´ 10 10 1.34 (a) Addition using scientific notation. Strategy: Let’s express scientific notation as N ´ 10 n . 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 .

  6. Page 6
    Read page 6 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 5 Let’s write 0.0095 in such a way that n = - 3. We have decreased 10 n by 10 3 , so we must increase N by 10 3 . 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 10 n 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 ´ 10 n . When dividing numbers using scientific notation, divide the N parts of the numbers in the usual way. To come up with the correct exponent, n , we subtract the exponents. Solution: Make sure that all numbers are expressed in scientific notation. 653 = 6.53 ´ 10 2 Divide the N parts of the numbers in the usual way. 6.53 ¸ 5.75 = 1.14 Subtract the exponents, n . 1.14 ´ 10 + 2 - ( - 8) = 1.14 ´ 10 + 2 + 8 = 1.14 ´ 10 10 (c) Subtraction using scientific notation. Strategy: Let’s express scientific notation as N ´ 10 n . When subtracting numbers using scientific notation, we must write each quantity with the same exponent, n . We can then subtract the N parts of the numbers, keeping the exponent, n , the same. Solution: Write each quantity with the same exponent, n . Let’s write 850,000 in such a way that n = 5. This means to move the decimal point five places to the left. 850,000 = 8.5 ´ 10 5

  7. Page 7
    Read page 7 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 6 Subtract the N parts of the numbers, keeping the exponent, n , the same. 8.5 ´ 10 5 - 9.0 ´ 10 5 - 0.5 ´ 10 5 The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10 to express N between 1 and 10 (5), we must decrease 10 n by a factor of 10. The exponent, n , is decreased by 1 from 5 to 4. - 0.5 ´ 10 5 = - 5 ´ 10 4 (d) Multiplication using scientific notation. Strategy: Let’s express scientific notation as N ´ 10 n . When multiplying numbers using scientific notation, multiply the N parts of the numbers in the usual way. To come up with the correct exponent, n , we add the exponents. Solution: Multiply the N parts of the numbers in the usual way. 3.6 ´ 3.6 = 13 Add the exponents, n . 13 ´ 10 - 4 + ( + 6) = 13 ´ 10 2 The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10 to express N between 1 and 10 (1.3), we must increase 10 n by a factor of 10. The exponent, n , is increased by 1 from 2 to 3. 13 ´ 10 2 = 1.3 ´ 10 3 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 cm 2 (d) 1 × 10 6 g/cm 3 1.38 (a) Division Strategy: The number of significant figures in the answer is determined by the original number having the smallest number of significant figures.

  8. Page 8
    Read page 8 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 7 Solution: 7.310 km 5.70 km = 1.28 3 The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits. Therefore, the answer has only three significant digits. The correct answer rounded off to the correct number of significant figures is: 1.28 (Why are there no units?) (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 ´ 10 7 dm) + (7.74 ´ 10 7 dm) = 8.14 ´ 10 7 dm Since 7.74 ´ 10 7 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

  9. Page 9
    Read page 9 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 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 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 (a) = ´ = 1 dm 22.6 m 0.1 m ? dm 226 dm (b) 0.001g 1kg 25.4 mg 1mg 1000 g = ´ ´ = 5 ? kg 2.54 10 kg - ´ (c) 3 1 10 L 556 mL 1 mL - ´ = ´ = ? L 0.556 L

  10. Page 10
    Read page 10 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 9 (d) 3 2 3 10.6 kg 1000 g 1 10 m 1 kg 1 cm 1 m - æ ö ÷ ´ ç ÷ ç ÷ ç = ´ ´ = ÷ ç ÷ ç ÷ ÷ ç è ø 3 3 g ? 0.0106 g cm 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 1 lb 3 1 mg 1 10 g - ´ we obtain the answer in one step: - = ´ ´ = ´ 3 1 mg 453.6 g 242 lb 1 lb 1 10 g 8 ? 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 ? m 3 = 68.3 cm 3 Recall that 1 cm = 1 ´ 10 - 2 m. We need to set up a conversion factor to convert from cm 3 to m 3 . Solution: We need the following conversion factor so that centimeters cancel and we end up with meters. - ´ 2 1 10 m 1 cm

  11. Page 11
    Read page 11 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 10 Since this conversion factor deals with length and we want volume, it must therefore be cubed to give - - - - æ ö ÷ ´ ´ ´ ´ ç ÷ ç ÷ ç ´ ´ = ÷ ç ÷ ç ÷ ÷ ç è ø 3 2 2 2 2 1 10 m 1 10 m 1 10 m 1 10 m 1 cm 1 cm 1 cm 1 cm We can write - æ ö ÷ ´ ç ÷ ç ÷ ç = ´ = ÷ ç ÷ ç ÷ ÷ ç è ø 3 2 3 1 10 m 68.3 cm 1 cm 3 5 3 ? m 6.83 10 m - ´ Check: We know that 1 cm 3 = 1 ´ 10 - 6 m 3 . We started with 6.83 ´ 10 1 cm 3 . Multiplying this quantity by 1 ´ 10 - 6 gives 6.83 ´ 10 - 5 . (c) Strategy: The problem may be stated as ? L = 7.2 m 3 In Chapter 1 of the text, a conversion is given between liters and cm 3 (1 L = 1000 cm 3 ). If we can convert m 3 to cm 3 , 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 m 3 to L. Arrange the appropriate conversion factors so that m 3 and cm 3 cancel, and the unit liters is obtained in your answer. Solution: The sequence of conversions is m 3  cm 3  L Using the following conversion factors, - æ ö ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ´ ÷ ç è ø 3 2 1 cm 1 10 m 3 1 L 1000 cm the answer is obtained in one step: - æ ö ÷ ç ÷ ç ÷ ç = ´ ´ = ÷ ç ÷ ç ÷ ´ ÷ ç è ø 3 3 2 3 1 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 m 3 = 1 ´ 10 3 L. Therefore, 7 m 3 would equal 7 ´ 10 3 L, which is close to the answer.

  12. Page 12
    Read page 12 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 11 (d) Strategy: The problem may be stated as ? lb = 28.3 m g 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 m g to grams, we can then convert from grams to pounds. Recall that 1 m g = 1 ´ 10 - 6 g. Arrange the appropriate conversion factors so that m g and grams cancel, and the unit pounds is obtained in your answer. Solution: The sequence of conversions is m g  g  lb Using the following conversion factors, - ´ m 6 1 10 g 1 g 1 lb 453.6 g we can write - ´ = m ´ ´ = m 6 1 10 g 1 lb 28.3 g 1 g 453.6 g 8 ? lb 6.24 10 lb - ´ Check: Does the answer seem reasonable? What number does the prefix m represent? Should 28.3 m g 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 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 1 day 60 min 1 h 60 s 1 min

  13. Page 13
    Read page 13 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 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 8 1.609 km 1000 m 1 s 1 min (93 10 mi) 1 mi 1 km 60 s 3.00 10 m 8.3 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 ´ 2 10 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 m 6.0 ft 3.28 ft 1.8 m ´ ´ = 453.6 g 1 kg 168 lb 1 lb 1000 g 76.2 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 h 2 1.4 10 mph ´ 1.50 = ´ 6 0.62 g Pb 0.62 ppm Pb 1 10 g blood ´ ´ = ´ ´ 3 6 0.62 g Pb 6.0 10 g of blood 1 10 g blood 3 3.7 10 g Pb - 1.51 (a) 8 365 days 24 h 3600 s 3.00 10 m 1 mi 1.42 yr 1 yr 1 day 1 h 1 s 1609 m ´ ´ ´ ´ ´ ´ = 12 8.35 10 mi ´ (b) ´ ´ 36 in 2.54 cm 32.4 yd 1 yd 1 in 3 2.96 10 cm = ´

  14. Page 14
    Read page 14 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 13 (c) 10 3.0 10 cm 1 in 1 ft 1 s 2.54 cm 12 in ´ ´ ´ = 8 9.8 10 ft /s ´ 1.52 (a) = ´ = 1 lb 70 kg 0.4536 kg 2 ? lbs 1.5 10 lbs ´ (b) = ´ ´ ´ ´ = 9 365 d 24 h 3600 s 14 10 yr 1 yr 1 d 1 h 17 ? s 4.4 10 s ´ (c) - ´ = ´ ´ 2 2.54 cm 1 10 m 90 in 1 in 1 cm ? m 2.3 m = (d) - æ ö ÷ ç ÷ = ´ ´ ç ÷ ç ÷ ç è ø ´ 3 3 2 3 1 cm 1 L 88.6 m 1 10 m 1000 cm 4 ? L 8.86 10 L = ´ 1.53 3 3 2.70 g 1 kg 1 cm 1000 g 0.01 m 1 cm æ ö ÷ ç ÷ ç = ´ ´ = ÷ ç ÷ ç ÷ ÷ ç è ø 3 3 density 2.70 10 kg m ´ / 1.54 3 0.625 g 1 L 1 mL 1 L 1000 mL 1 cm - = ´ ´ = 4 3 density 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/cm 3 (c) iron rusts easily density = 7.86 g/cm 3 (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 2 O (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/cm 3 (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 composition and identity by chemically combining with oxygen and water. (b) Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids, thus changing the composition and identity of the water.

  15. Page 15
    Read page 15 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 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 3 1 ton (95.0 10 lb of sulfuric acid) 2.0 10 lb 7 4.75 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) = = 3 density 2.6 g cm m V / 1.59 mass = density ´ volume (a) mass = (19.3 g/cm 3 ) ´ [ 4 3 p (10.0 cm) 3 ] = 8.08 ´ 10 4 g (b) - æ ö ÷ ç ÷ ç = ´ ´ = ÷ ç ÷ ç ÷ ÷ ç è ø 3 3 1 cm (21.4 g/cm ) 0.040 mm 10 mm 6 mass 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 calculate the volume that the cooking oil occupies, we can calculate the radius of the cylinder. The volume of the cylinder is, V cylinder = p r 2 h ( r is the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2 r . = mass of oil volume of oil filling bottle density of oil = = ´ = ´ 3 3 3 1360 g volume of oil filling bottle 1.43 10 mL 1.43 10 cm 0.953 g/mL Next, solve for the radius of the cylinder. Volume of cylinder = p r 2 h volume = p ´ r h ´ = = p ´ 3 3 1.43 10 cm 4.60 cm 21.5 cm r The inner diameter of the bottle equals 2 r . Bottle diameter = 2 r = 2(4.60 cm) = 9.20 cm

  16. Page 16
    Read page 16 as text

    CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 15 1.61 From the mass of the water and its density, we can calculate the volume that the water occupies. The volume that the water occupies is equal to the volume of the flask. = mass volume density Mass of water = 87.39 g - 56.12 g = 31.27 g = = 3 31.27 g mass = density 0.9976 g/cm 3 Volume of the flask 31.35 cm 1.62 ´ ´ = 343 m 1 mi 3600 s 1 s 1609 m 1 h 767 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 cm 3 3 194.3 g 18.5 cm = = 3 density 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: æ ö ÷ ç ÷ ç = p = p = ´ ÷ ç ÷ ç ÷ ç è ø 3 3 4 3 48.6 cm 4 4 6.01 10 cm 3 3 2 V r ´ = = = ´ 5 4 3 6.852 10 g mass volume 6.01 10 cm density 11.4 g/cm 3 1.66 = mass Volume density ´ = = 3 3 1.20 10 g 0.53 g / cm 3 3 Volume occupied by Li 2.3 10 cm ´

953 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Chemistry study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.