atm = 760 torr = 760 mm Hg = kpa = psi. = atm. = atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

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1 CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions, and can be compressed or expanded by changes in those conditions. Although the particles of matter in solids are essentially fixed in position (the solid is rigid), the particles in liquids and gases are free to move. 3. A small amount of water is added to a metal can and then the can is heated so as to boil the water and fill the can with steam (gaseous water). The heat is then removed and the can is sealed off. As the steam in the can cools, it condenses back to a liquid. Since the gas in the can has condensed, the pressure of the atmosphere is much larger than the pressure of gas in the can, and the atmospheric pressure causes the can to collapse. 4. Figure 3. in the text shows a simple mercury barometer: a tube filled with mercury is inverted over a reservoir (containing mercury) that is open to the atmosphere. When the tube is inverted, the mercury falls to a level at which the pressure of the atmosphere is sufficient to support the column of mercury. One standard atmosphere of pressure is taken to be the pressure capable of supporting a column of mercury to a height of mm above the reservoir level. 5. mix 6. ressure units include mm Hg, torr, pascals, and psi. The unit mm Hg is derived from the barometer, since in a traditional mercury barometer, we measure the height of the mercury column (in millimeters) above the reservoir of mercury atm = 760 torr = 760 mm Hg = 0.35 ka = 4.70 psi a. 45. ka b. 755 mm Hg c. 80 torr d..04 atm atm 0.35 ka atm 760 mm Hg = atm = atm atm 0.35 ka = 07 ka 760 torr atm 760 mm Hg atm = 790. mm Hg atm = 760 torr = 760 mm Hg = 0.35 ka = 4.70 psi atm a. 4.9 psi =.0 atm 4.70 psi 67

2 b. 795 torr c. 743 mm Hg atm 760 torr =.05 atm 0.35 ka 760 mm Hg = 99. ka d. 99,436 a ka 000 a = ka atm = 760 torr = 760 mm Hg = 0.35 ka = 4.70 psi atm a. 699 mm Hg = 0.90 atm 760 mm Hg b. 8. psi 760 mm Hg 4.70 psi = 94 mm Hg c. 86 mm Hg = 86 torr d psi 795 mm Hg = 5.4 psi 760 mm Hg atm = 760 torr = 760 mm Hg = 0.35 ka = 4.70 psi 0.35 ka a. 7.3 psi = 9 ka 4.70 psi b..5 atm c. 4.5 atm d. 4 psi 4.70 psi atm 760 mm Hg atm atm 4.70 psi = 6.9 psi = mm Hg = 5. atm..00 atm = 760 torr = 760 mm Hg = 0.35 ka = 4.70 psi a atm a =.5 atm 0,35 a b.. atm c. 97,35 a 0,35 a atm 760 mm Hg 0,35 a = a = mm Hg d..3 ka 000 a ka = a 68

3 ..00 atm = 760 torr = 760 mm Hg = 0.35 ka = 4.70 psi 0.35 ka a. 6.4 atm = 65 ka atm b. 4. atm c. 794 mm Hg d. 7. psi atm 4.70 psi 760 torr atm = torr atm 760 mm Hg =.85 atm =.04 atm 3. The volume of a sample of an ideal gas at constant temperature will decrease if the pressure on the gas is increased. 4. Additional mercury increases the pressure on the gas sample, causing the volume of the gas upon which the pressure is exerted to decrease (Boyle s Law) 5. pressure 6. = k; = 7. a. = 755 mm Hg = 780 mm Hg = 5 ml =? (755 mm Hg)(5 ml) = = = ml (780 mm Hg) b. =.08 atm = 0.95 atm = 3 ml =? (.08 atm)(3 ml) = = = 53 ml (0.95 atm) c. = 03 ka = ka = 3.0 L =? (03 ka)(3.0 L) = = =.57 L ( ka) 8. a. =.5 atm = 775 mm Hg =.00 atm = 375 ml =? (.5 atm)(375 ml) = = = 43 ml (.00 atm) b. =.08 atm = 35 ka =.33 atm = 95 ml =? 69

4 (.08 atm)(95 ml) = = = 58 ml (.33 atm) c. = 3 ka = 98.6 mm Hg = 765 mm Hg = 6.75 L =? (98.6 mm Hg)(6.75 L) = = = 8.67 L (765 mm Hg) 9. a. = 0. ka =? ka = 9.3 L = 0.0 L (0. ka)(9.3 L) = = = 97 ka (0.0 L) b. = 755 torr = 755 mm Hg = 76 mm Hg = 5.7 ml =? ml (755 mm Hg)(5.7 ml) = = = 5.5 ml (76 mm Hg) c. =.05 atm =. ka =.07 atm = 5. L =? (.05 atm)(5. L) = = (.07 atm) = 48.6 L 0. a. = 755 mm Hg =? = 5 ml = 37 ml (755 mm Hg)(5 ml) = = = 689 mm Hg (37 ml) b. =.08 atm =? = 33 ml = 99 ml (.08 atm)(33 ml) = = =.0 atm (99 ml) c. = 789 mm Hg = 35 ka = 03 mm Hg = 3.0 L =? (789 mm Hg)(3.0 L) = = (03 mm Hg) =.35 L. =.0 atm =.99 atm = 5 ml =? 70

5 = (.0 atm)(5 ml) = = 76.8 ml (.99 atm). = = =.04 L =? L = ( )(.04 L) = =.04 L = 0.50 L ( ) 3. = 785 mm Hg =? = 9. ml = 5. ml (785 mm Hg)(9. ml) = = (5. ml) =.5 03 mm Hg 4. = (.00 atm)(7. ml) = = 7. atm (.00 ml) 5. Absolute zero is the lowest temperature that can exist. Absolute zero is the temperature at which the volume of an ideal gas sample would be predicted to become zero. Absolute zero is the zeropoint on the Kelvin temperature scale (and corresponds to 73 C). 6. Charles s Law indicates that an ideal gas decreases by /73 of its volume for every degree Celsius its temperature is lowered. This means an ideal gas would approach a volume of zero at 73 C. 7. directly 8. = kt; /T = /T 9. =.5 L =? ml T = 5. C = 98. K T = 78.5 C = 94.5 K T (.5 L)(94.5 K) = T = (98. K) = L 30. = 375 ml =? ml T = 78 C = 35 K T = C = 95 K T = T (375 ml)(95 K) = = 35 ml (35 K) 3. a. =.03 L = 3.0 L T = 4 C = 97 K T =? 7

6 T (3.0 L)(97 K) T = = = 440 K = 67 C (.03 L) b. = 7 ml =? T = 73 K T = 373 K T = T (7 ml)(373 K) = = 74 ml (73 K) c. = 49.7 ml =? T = 34 C= 307 K T = 350 K T = T (49.7 ml)(350 K) = = 56.7 ml (307 K) 3. a. = 73.5 ml =? L T = 0 C = 73 K T = 5 C = 98 K T = T (73.5 ml)(98 K) = = 80. ml (73 K) b. = 5. L = 0.0 L T = 5 C = 98 K T =? C T (0.0 L)(98 K) T = = = 96 K = 77 C (5. L) c. =.75 ml =? ml T =.3 K T = 0 C = 73 K T (.75 ml)(73 K) = T = (.3 K) = 08 ml (. 0 ml) 33. a. = 9.4 L =? T = 4 C = 97 K T = 48 C = 3 K T (9.4 L)(3 K) = T = (97 K) = 9.88 L b. = 4.9 ml = 49.9 ml T = C = 6 K T =? T (49.9 ml)(6 K) T = = = 53 K = 50. C (4.9 ml) c. = 95 ml =? T = 5 K T = 73 K 7

7 T (95 ml)(73 K) = T = (5 K) =.0 04 ml 34. a. =.0 0 L = 5.00 L T = 50 C = 43 K T =? C T (5.00 L)(43 K) T = = = 35.4 K = 38 C (0 L) b. = 44. ml =? ml T = 98 K T = 0 T = T (44. ml)(0 K) = = 0 ml (0 K is absolute zero) (98 K) c. = 44. ml =? ml T = 98 K T = 0 C = 73 K T = T (44. ml)(73 K) = = 40.5 ml (98 K) 35. =.5L =? ml T = 9 K T = 78 K T = T (.5 L)(78 K) = = L = 0.34 L (9 K) T 36. = T (5 ml)(50 K) = = 69.4 ml = 69 ml to two significant figures (450 K) C + 73 = 97 K 7 C + 73 = 345 K T = T (375 ml)(345 K) = = 436 ml (97 K) T 38. = T Temp, C olume, ml directly 40. = an; /n = /n 73

8 4. /n = /n = 4 ml n = mol 4 ml = 89 ml =? L n = mol 4. = an; /n = /n Since.08 g of chlorine contains twice the number of moles of gas contained in the.04 g sample, the volume of the.08 g sample will be twice as large = 744 ( ) ml 43. = 00. L =? L n = 3.5 mol n = 4.5 mol 4.5 mol 00. L 3.5 mol = 435 L 44. molar mass of Ar = g.7 g Ar = mol Ar g 4. L.9 mol mol = 80. L 45. Although the definition may seem a little strange, an ideal gas is one which obeys the ideal gas law, =, exactly. That is, if knowledge of three of the properties of a gas (pressure, volume, temperature, and amount) leads to the correct value for the fourth property when using this equation, then the gas under study is an ideal gas. 46. Real gases most closely approach ideal gas behavior under conditions of relatively high temperatures (0 C or higher) and relatively low pressures ( atm or lower). 47. For an ideal gas, = is true under any conditions. Consider a particular sample of gas (so that n remains constant) at a particular fixed temperature (so that T remains constant also). Suppose that at pressure the volume of the gas sample is. Then for this set of conditions, the ideal gas equation would be given by =. If we then change the pressure of the gas sample to a new pressure, the volume of the gas sample changes to a new volume. For this new set of conditions, the ideal gas equation would be given by =. As the right-hand sides of these equations are equal to the same quantity (because we defined n and T to be constant), then the left-hand sides of the equations must also be equal, and we obtain the usual form of Boyle s law. = 74

9 48. For an ideal gas, = is true under any conditions. Consider a particular sample of gas (so that n remains constant) at a particular fixed pressure (so that remains constant also). Suppose that at temperature T the volume of the gas sample is. Then for this set of conditions, the ideal gas equation would be given by =. If we then change the temperature of the gas sample to a new temperature T, the volume of the gas sample changes to a new volume. For this new set of conditions, the ideal gas equation would be given by =. If we make a ratio of these two expressions for the ideal gas equation for this gas sample, and cancel out terms that are constant for this situation (, n, and R) we get = T = T This can be rearranged to the familiar form of Charles s law = T T 49. a. = 78.4 mm Hg =.09 atm; T = 6. C = 99 K = (0.0)( L atm mol K )(99 K) = =.44 L (.09 atm) b. = 7.5 ml = L; T = 6.6 C = 90 K = (0.0078)( L atm mol K )(90 K) = = 6.76 atm (0.075 L) c. = 45. ml = L (.045 atm)(0.045 L) T = = = 57 K nr (0.004)( L atm mol K ) 50. a. = 78 mm Hg =.03 atm; T = 7 C = 300 K = (0.0 mol)( L atm mol K )(300 K) = = 5.0 L (.03 atm) b. = 644 ml = L = (0.09)( L atm mol K )(303 K) = = 3.56 atm (0.644 L) = mm Hg 75

10 c. = 745 mm = atm (0.980 atm)(. L) T = = = 334 K nr (0.40)( L atm mol K ) 5. molar mass Ne = 0.8 g; 5 C = 98 K (.0 atm)(5.00 L) n= = = mol Ne RT ( L atm mol K )(98 K) mol Ne 0.8 g Ne Ne = 4. g Ne 5. molar mass Ar = g; 5 C = 98 K Ar. g Ar = mol Ar g Ar = ( mol)( L atm mol K )(98 K) = = atm (0.0 L) 53. molar mass of He = g; 00 C = 373 K; 785 mm Hg =.033 atm He.04 g He = mol He g He = ( mol)( L atm mol K )(373 K) = = 5. L (.033 atm) 54. molar mass Ar = g; 40.0 g =.00 (.00 atm)(5.00 L) T = = = 304 K = 3 C nr (.00)( L atm mol K ) 55. T = 5 C + 73 = 98 K; molar masses: He, g; O, 3.00 g (55 atm)(00.0 L) n= = = 043 mol = mol RT ( L atm mol K )(98 K) mol of either He or O would be needed g He mol He He = g He g O mol O O = g O 56. molar mass Ne = 0.8 g; 5 C = 98 K; 50 C = 33 K.5 g Ne = mol 0.8 g 76

11 = = ( mol)( L atm mol K )(98 K) = = 0.50 atm (0. L) ( mol)( L atm mol K )(33 K) = = 0.63 atm (0. L) 57. molar mass Ne = 0.8 g; = 500 torr = atm.0 g Ne 0.8 g = mol Ne ( atm)(5.00 L) T = = = K = 8.0 K nr ( mol)( L atm mol K ) 58. molar mass O = 3.00 g; 784 mm Hg =.03 atm O 4.5 g O = 0.38 mol 3.00 g O (.03 atm)(.5 L) T = nr = (0.38 mol)( L atm mol K ) = 38 K = 35 C kg = g; molar mass Ne = 0.8 g Ne g Ne = 47.8 mol Ne 0.8 g Ne = (47.8 mol)( L atm mol K )(300. K) = = 30.5 atm = 3 atm (00. L) 60. Molar masses: He, g; Ar, g He 4.5 g He =.037 mol He g He 56. g Ar Ar g Ar =.407 mol Ar For He, = For Ar, = (.037 mol)( L atm mol K )(98 K) = = 5.07 atm (5.00 L) (.407 mol)( L atm mol K )(303 K) = = 3.50 atm (0.00 L) The helium is at a higher pressure than the argon. 6. =.0 atm =? atm = 4.3 ml = 5. ml T = 5 C = 98 K T = 50 C = 33 K 77

12 T = T (33 K)(.0 atm)(4.3 ml) = (98 K)(5. ml) =.75 atm 6. molar mass Ar = g; 9 C = 30 K; 4 C = 35 K Ar.9 g Ar = mol Ar g Ar = = ( mol)( L atm mol K )(30 K) = = 0.33 atm (.4 L) ( mol)( L atm mol K )(35 K) = = atm (.4 L) 63. =.05 atm = atm = 459 ml =? ml T = 7 C = 300. K T = 5 C = 88 K T = T (88 K)(.05 atm)(459 ml) = = 464 ml (300 K)(0.997 atm) 64. Molar mass of H O = 8.0 g;.0 ml = L; 5 C = 498 K H O 0.50 g H O 8.0 g H O = mol H O = ( mol)( L atm mol K )(498 K) = = 83 atm =.8 0 atm (0.000 L) 65. In deriving the ideal gas law, we assume that the molecules of gas occupy no volume, and that the molecules do not interact with each other. Under these conditions, there is no difference between gas molecules of different substances (other than their masses) as far as the bulk behavior of the gas is concerned. Each gas behaves independently of other gases present, and the overall properties of the sample are determined by the overall quantity of gas present. total = n where n is the number of individual gases present in the mixture. 66. As a gas is bubbled through water, the bubbles of gas become saturated with water vapor, thus forming a gaseous mixture. The total pressure in a sample of gas that has been collected by bubbling through water is made up of two components: the pressure of the gas of interest and the pressure of water vapor. The partial pressure of the gas of interest is then the total pressure of the sample minus the vapor pressure of water. 67. molar masses: He, g; Ne, 0.8 g; 5 C = 98 K.4 g He.79 g Ne He g He Ne 0.8 g Ne = 0.60 mol He = 0.38 mol Ne 78

13 helium = neon = nheliumrt nneo (0.60 mol)( L atm mol K )(98 K) = = 4. atm (.04 L) (0.38 mol)( L atm mol K )(98 K) = = 3.5 atm (.04 L) total = 4. atm atm = 7.5 atm 68. molar masses: Ne, 0.8 g; Ar, g; 7 C = 300 K Ne.8 g Ne = mol Ne 0.8 g Ne.49 g Ar Ar g Ar = mol Ar neon = nneo ( mol)( L atm mol K )(300 K) = = 0.58 atm (9.87 L) argon = n RT = = atm (9.87 L) argon ( mol)( L atm mol K )(300 K) total = 0.58 atm atm = 0.34 atm g O =.64 O ; 65. g CO =.479 mol CO ; total moles = 3.0 mol.64 O oxygen = 9. atm 3.0 mol total = 4.84 atm O.479 mol CO carbon dioxide = 9. atm 3.0 mol total = 4.37 atm CO Once the partial pressure of O had been calculated, we also could have calculated the partial pressure of CO as the difference between the total pressure (9. atm) and the partial pressure of O (4.84 atm) mm Hg =.7 atm; 6 C = 99 K; molar masses: Ne, 0.8 g; Ar, g (.7 atm)(3.00 L) n= = = mol RT ( L atm mol K )(99 K) The number of moles of an ideal gas required to fill a given-sized container to a particular pressure at a particular temperature does not depend on the specific identity of the gas. So mol of Ne gas or mol of Ar gas would give the same pressure in the same flask at the same temperature. 0.8 g Ne mass Ne = mol Ne = 3.00 g Ne Ne mass Ar = mol Ar g Ar Ar = 5.94 g Ar 7. oxygen = total water vapor = = 745 torr 79

14 7. molar masses: He, g; Ar, g; 73 C = 546 K He.5 g He = mol He g Ne.9 g Ar Ar g Ar = mol Ar helium = nheliumrt (0.873 mol)( L atm mol K )(546 K) = =.45 atm (5.5 L) argon = n RT = = 0.66 atm (5.5 L) argon ( mol)( L atm mol K )(546 K) total = 0.58 atm atm = 3.07 atm 73. oxygen = total water vapor = (755 3) mm Hg = 73 mm Hg = atm T = 4 C + 73 = 97 K; = 500. ml = L (0.963 atm)(0.500 L) n= = = =.98 0 mol O RT ( L atm mol K )(97 K) atm = mm Hg; molar mass of Zn = g hydrogen = mm Hg 3 mm Hg = 75.3 mm Hg = atm = 40 ml = 0.40 L; T = 30 C + 73 = 303 K n hydrogen (0.990 atm)(0.40 L) = = = mol hydrogen RT ( L atm mol K )(303 K) mol H Zn H = mol of Zn must have reacted mol Zn g Zn Zn = 0.65 g Zn must have reacted 75. A law is a statement that precisely expresses generally observed behavior. A theory consists of a set of assumptions/hypotheses that is put forth to explain the observed behavior of matter. Theories attempt to explain natural laws. 76. A theory is successful if it explains known experimental observations. Theories that have been successful in the past may not be successful in the future (for example, as technology evolves, more sophisticated experiments may be possible in the future). 77. assume that the volume of the molecules themselves in a gas sample is negligible compared to the bulk volume of the gas sample: this helps us to explain why gases are so compressible. 78. pressure 79. kinetic energy 80

15 80. no 8. The temperature of a gas reflects, on average, how rapidly the molecules in the gas are moving. At high temperatures, the particles are moving very fast and collide with the walls of the container frequently, whereas at low temperatures, the molecules are moving more slowly and collide with the walls of the container infrequently. The Kelvin temperature is directly proportional to the average kinetic energy of the particles in a gas. 8. If the temperature of a sample of gas is increased, the average kinetic energy of the particles of gas increases. This means that the speeds of the particles increase. If the particles have a higher speed, they will hit the walls of the container more frequently and with greater force, thereby increasing the pressure. 83. The molar volume of a gas is the volume occupied by one mole of the gas under a particular set of temperature and pressure conditions (usually ST: 0 C, atm). When measured under the same conditions, all ideal gases have the same molar volume (.4 L at ST). 84. Standard Temperature and ressure, ST = 0 C, atm pressure. These conditions were chosen because they are easy to attain and reproduce experimentally. The barometric pressure within a laboratory is likely to be near atm most days, and 0 C can be attained with a simple ice bath. 85. molar masses: CaO, g; CO ; 44.0 g CaO.5 g CaO = 0.09 mol CaO g CaO From the balanced chemical equation, 0.09 mol CaO would absorb 0.09 mol CO 44.0 g CO 0.09 mol CO CO = 0.98 g CO Since one mole of an idea gas occupies.4 L at ST, 0.09 mol of CO would occupy 0.09 mol CO.4 L = L at ST 86. Molar mass of C =.0 g; 5 C = 98 K.5 g C = 0.04 C.0 g Since the balanced chemical equation shows a : stoichiometric relationship between C and O, then 0.04 of O will be needed = (0.04)( L atm mol K )(98 K) = =.50 L O (.0 atm) 87. C 8 H 8 (l) + 5O (g) 6CO (g) + 8H 0(l) molar mass C 8 H 8 = 4. g C8H8 0.0 g C 8 H 8 4. g C H = mol C 8H

16 5 mol O mol C 8 H 8 mol C H =.095 mol O 8 8 At ST, one mole of an ideal gas occupies.4 L of volume..095 mol O.4 L = 4.5 L O at ST mol 88. Molar mass of Mg = 4.3 g; ST:.00 atm, 73 K.0 g Mg = mol Mg 4.3 g As the coefficients for Mg and Cl in the balanced equation are the same, for mol of Mg reacting we will need mol of Cl. = mol Cl.4 L = 0.94 L Cl at ST C = 300 K; 6 C = 99 K; molar mass NH 4 Cl = g (.0 atm)(4. L) mol NH 3 present = n= = = = 0.74 mol NH RT 3 ( L atm mol K )(300 K) mol HCl present = (0.998 atm)(5.35 L) n= = = = 0.8 mol HCl RT ( L atm mol K )(99 K) NH 3 and HCl react on a : basis: NH 3 is the limiting reactant. NH4Cl g NH4Cl 0.74 mol NH 3 = 9.3 g NH 4 Cl produced NH NH Cl molar mass CaC = 64.0 g; 5 C = 98 K.49 g CaC 64.0 g = mol CaC From the balanced chemical equation for the reaction, mol of CaC reacting completely would generate mol of acetylene, C H = = ( mol)( L atm mol K )(98 K) = = 0.94 L (.0 atm) ( mol)( L atm mol K )(73 K) = = L at ST (.00 atm) 9. CuSO 4 5H O(s) CuSO 4 (s) + 5H O(g) 350 C = 63 K; molar mass CuSO 4 5H O = 49.7 g. CuSO45HO 5.00 g CuSO 4 =5H O g CuSO 5H O = mol CuSO 4 5H O 4 8

17 5 mol HO mol CuSO 4 5H O. CuSO 5H O = 0.00 H O 4 = (0.00)( L atm mol K )(63 K) = = 4.9 L H O (.04 atm) 9. Molar mass of Mg 3 N = g; T = 4 C = 97 K; = 75 mm Hg = atm 0.3 g Mg 3 N g = 0.0 mol Mg 3N From the balanced chemical equation, the amount of NH 3 produced will be mol NH3 0.0 mol Mg 3 N Mg N = 0.04 mol NH 3 3 = (0.04 mol)( L atm mol K )(97 K) = = 5.03 L (0.989 atm) This assumes that the ammonia was collected dry. 93. Molar masses: He, g; H,.06 g; 8 C = 30 K He 4. g He = 3.55 mol He g He H.6 g H.06 g H = 0.7 mol H total moles = 3.55 mol mol = 4.5 mol (4.5 mol)( L atm mol K )(30 K) = = = 357 L (0.985 atm) 94. Molar masses: O, 3.00 g; N, 8.0 g; T = 35 C = 308 K; = 755 mm Hg = atm O 6. g O 3.00 g O = 0.89 mol O N 35. g N 8.0 g N =.5 mol N total moles = 0.89 mol +.5 mol =.07 mol (.07 mol)( L atm mol K )(308 K) = = = 5.7 L (0.993 atm) 95. = 89 mm Hg =.00 atm = 760 mm Hg = 5. ml =? T = 95 C + 73 = 368 K T = 73 K 83

18 T = T (73 K)(89 mm Hg)(5. ml) = =.9 ml (368 K)(760 mm Hg) 96. molar masses: He, g; Ar, g He 5.0 g He =.54 mol He g He.54 mol He.4 L 4. g Ar Ar g Ar.054 mol Ar.4 L = 8. L =.054 mol Ar = 3.6 L 97. molar masses: O, 3.00 g; N, 8.0 g; CO, 44.0 g; Ne, 0.8 g O 5.00 g O 3.00 g O = mol O N 5.00 g N 8.0 g N = mol N CO 5.00 g CO 44.0 g CO = 0.36 mol CO 5.00 g Ne Ne 0.8 g Ne = mol Ne Total moles of gas = = L is the volume occupied by one mole of any ideal gas at ST. This would apply even if the gas sample is a mixture of individual gases L = 5.59 L = 5.6 L The partial pressure of each individual gas in the mixture will be related to what fraction on a mole basis each gas represents in the mixture mol O oxygen =.00 atm total = 0.5 atm O mol N nitrogen =.00 atm total = 0.56 atm N 0.36 mol CO carbon dioxide =.00 atm total = 0.63 atm CO neon =.00 atm mol Ne total = atm Ne 84

19 98. Molar masses: He, g; Ne, 0.8 g He 6.5 g He =.56 He g He 4.97 g Ne Ne 0.8 g Ne = mol Ne n total = mol =.807 mol As of an ideal gas occupies.4 L at ST, the volume is given by.807 mol.4 L = L = 40.5 L. The partial pressure of a given gas in a mixture will be proportional to what fraction of the total number of moles of gas the given gas represents He = Ne =.56 He.807 mol total mol Ne.807 mol total 99. Na(s) + Cl (g) NaCl(s) molar mass Na =.99 g Na 4.8 = 0.09 mol Na.99 g Na.00 atm = atm = atm.00 atm = atm = 0.36 atm Cl 0.09 mol Na mol Na = mol Cl mol Cl.4 L =.34 L Cl at ST 00. C H (g) + 5O (g) H O(g) + 4CO (g) molar mass C H = 6.04 g.00 g C H 6.04 g = mol C H From the balanced chemical equation, = mol of CO will be produced mol CO.4 L =.7 L at ST 0. FeO(s) + CO(g) Fe(s) + CO (g) molar mass FeO = 7.85 g;.45 kg = g FeO g FeO = 0.8 mol FeO 7.85 g FeO 85

20 Since the coefficients of the balanced equation are all one, if 0.8 mol FeO reacts, then 0.8 mol CO(g) is required and 0.8 mol of CO (g) is produced. 0.8 mol.4 L = 45 L L CO(g) is required for reaction and L CO (g) are produced by the reaction ml = 0.5 L 0.5 L.4 L = mol H From the balanced chemical equation, one mole of zinc is required for each mole of hydrogen produced. Therefore, mol of Zn will be required g Zn mol Zn = g Zn 03. kelvin (absolute) 04. twice 05. consist of tiny particles, which are so small that the fraction of the bulk volume of the gas occupied by the particles is negligible. The particles of a gas are in constant random motion and collide with the walls of the container (giving rise to the pressure of the gas). The particles of a gas do not attract or repel each other. The average kinetic energy of the particles of a gas is reflected in the temperature of the gas sample. 06. a. = k; = b. = kt; /T = /T c. = an; /n = /n d. = e. /T = /T 07. sum 08. First determine what volume the helium in the tank would have if it were at a pressure of 755 mm Hg (corresponding to the pressure the gas will have in the balloons) atm = 6384 mm Hg = (5. L) 6384 mm Hg 755 mm Hg = 3 L Allowing for the fact that 5. L of He will have to remain in the tank, this leaves 3 5. = 87.8 L of He for filling the balloons L He balloon = 5 balloons.50 L He 86

21 09. A decrease in temperature would tend to make the volume of the weather balloon decrease. As the overall volume of a weather balloon increases when it rises to higher altitudes, the contribution to the new volume of the gas from the decrease in pressure must be more important than the decrease in temperature (the temperature change in kelvins is not as dramatic as it seems in degrees Celsius). 0. According to the balanced chemical equation, when of (NH 4 ) CO 3 reacts, a total of 4 moles of gaseous substances is produced. molar mass (NH 4 ) CO 3 = g; 453 C = 76 K 5.0 g = g As 0.54 of (NH 4 ) CO 3 reacts, 4(0.54) =.6 mol of gaseous products result. = (.6 mol)( L atm mol K )(76 K) = = 4 L (.04 atm). CaCO 3 (s) CaO(s) + CO (g) 774 torr =.08 atm; 55 C + 73 = 38 K; molar mass CaCO 3 = 00. g 0.0 g CaCO g = mol CaCO 3 From the balanced equation, mol CO will be produced. = ( mol)( L atm mol K )(38 K) = =.64 L CO (.08 atm). CaCO 3 (s) + H + (aq) Ca + (aq) + H O(l) + CO (g) molar mass CaCO 3 = 00. g; 60 C + 73 = 333 K 0.0 g CaCO g = mol CaCO 3 = mol CO also carbon dioxide = total water vapor carbon dioxide = 774 mm Hg 49.4 mm Hg = 64.6 mm Hg = 0.8 atm wet = ( mol)( L atm mol K )(333 K) = = 3.3 L wet CO (0.8 atm) dry = 3.3 L 64.6 mm Hg 774 mm Hg =.68 L 3. S(s) + 3O (g) SO 3 (g) 350. C + 73 = 63 K; molar mass S = 3.07 g S 5.00 g = mol S 3.07 g S 87

22 3 mol O mol S mol S = mol O = (0.339 mol)( L atm mol K )(63 K) = =.8 L O (5.5 atm) 4. KClO 3 (s) KCl(s) + 3O (g) molar mass KClO 3 =.6 g; 5 C + 73 = 98 K; 630. torr = 0.89 atm KClO g KClO 3.6 g KClO = mol KClO 3 3 mol O mol KClO 3 mol KClO = 0.6 O 3 3 = (0.6)( L atm mol K )(98 K) = = 8. L O (0.89 atm) 5. molar mass He = g He 0.0 g He =.498 mol He g He.498 mol.4 L = 56.0 L He 6. a. 75 mm Hg 0,35 a 760 mm Hg = a b. 458 ka atm 0.35 ka c. 760 mm Hg.43 atm atm d. 84 torr = 84 mm Hg = 4.5 atm = mm Hg 7. a atm b a 760 mm Hg atm = 686 mm Hg 760 mm Hg 0,35 a = mm Hg c. 760 mm 445 ka 0.35 ka = mm Hg d. 34 torr = 34 mm Hg 8. a. 645 mm Hg 0,35 a 760 mm Hg = a 88

23 b. ka = 0 3 a =. 0 5 a 0,35 a c atm = a atm d. 3 torr 0,35 a 760 torr 9. a. 00 mm Hg =.38 atm = a 4.56 atm = 3 L.38 atm = 46 L b. 5. mm Hg = atm 634 ml = atm = 0.7 atm 66 ml c. 5 torr = a = 68. ka 68. ka = 443 L.05 ka = L 0. a..00 mm Hg =.00 torr.00 torr = 55 ml = 8 ml.00 torr b..0 atm = 0.35 ka.0 ka =.3 L 0.35 ka =.3 0 L c..0 mm Hg = 0.33 ka.0 ka =.3 L 0.33 ka = 9.8 L. Assume the pressure at sea level to be atm (760 mm Hg). Calculate the volume the balloon would have if it rose to the point where the pressure has dropped to 500 mm Hg. If this calculated volume is greater than the balloon s specified maximum volume (.5 L), the balloon will burst. 760 mm Hg.0 L = 3.0 L >.5 L. The balloon will burst. 500 mm Hg..5 L = ml 755 mm Hg ml 450 ml = mm Hg 3. C + 73 = 95 K; 00 C + 73 = 373 K 79 ml 373 K = 9 ml 95 K 89

24 4. a. 74 C + 73 = 347 K; 74 C + 73 = 99 K 00. ml 99 K = 57.3 ml 347 K b. 00 C + 73 = 373 K 600 ml 373 K = 448 K (75 C) 500 ml c. zero (the volume of any gas sample becomes zero at 0 K) 5. a. 0 C + 73 = 73 K 73 K 44.4 L = 54 K (68 C).4 L b. 7 C + 73 = K; 5 C + 73 = 98 K ml 98 K K = 0.30 ml c. 40 C + 73 = 33 K 33 K 000 L 3.3 L = K (6940 C) 6. C + 73 = 85 K; 9 C + 73 = 465 K 75. ml 465 K = 3 ml 85 K g O = 0.60 mol; 5.0 g O = L 0.60 mol = 30.3 L 8. For a given gas, the number of moles present in a sample is directly proportional to the mass of the sample. The problem therefore can be solved even though the gas is not identified (so that its molar mass is not known). 3. g 0.4 L 93. L =.59 g 9. a. = 4 ml = 0.4 L (. atm)(0.4 L) T = nr = (0.43 mol)( L atm mol K ) = 84.9 K b. =.3 ml = L = ( mol)( L atm mol K )(93 K) = =.5 atm (0.003 L) c. = 755 mm Hg = atm; T = 3 C + 73 = 404 K 90

25 = (0.473 mol)( L atm mol K )(404 K) = = 5.8 L = ml (0.993 atm) 30. a. =. ml = 0.0 L (.034 atm)(0.0 L) T = = = 6.8 K nr ( mol)( L atm mol K ) b. =.73 ml = L = ( mol)( L atm mol K )(8 K) = = atm ( L) c. =.3 mm Hg = atm; T = 5 C + 73 = 45 K = (0.773 mol)( L atm mol K )(45 K) = = L (0.006 atm) 3. molar mass N = 8.0 g; T = 6 C + 73 = 99 K N n = 4. g N 8.0 g N = mol N = (0.507 mol)( L atm mol K )(99 K) = =.4 atm (0.0 L) 3. 7 C + 73 = 300 K The number of moles of gas it takes to fill the 00. L tanks to 0 atm at 7 C is independent of the identity of the gas. (0 atm)(00. L) n= = = = 487 mol RT ( L atm mol K )(300 K) 487 mol of any gas will fill the tanks to the required specifications. molar masses: CH 4, 6.0 g; N, 8.0 g; CO, 44.0 g for CH 4 : (487 mol)(6.0 g/mol) = 779 g = 7.79 kg CH 4 for N : (487 mol)(8.0 g/mol) = 3,636 g = 3.6 kg N for CO : (487 mol)(44.0 g/mol) =,48 g =.4 kg CO 33. molar mass He = g He n = 4.00 g He = mol He g He (.00 atm)(.4 L) T = = = 73 K = 0 C nr (0.999 mol)( L atm mol K ) 9

26 34. molar mass of O = 3.00 g; 55 mg = g O n = g 3.00 g O = mol = 00. ml = 0.00 L; T = 6 C + 73 = 99 K = (0.007 mol)( L atm mol K )(99 K) = = 0.4 atm (0.00 L) 35. =.0 atm = 0 torr = 0.89 atm =.0 L =? T = 3 C + 73 = 96 K T (4 K)(.0 atm)(.0 L) = T = (96 K)(0.89 atm) =.8 L T = 3 C = 4 K 36. =.3 atm =.89 atm = 00 ml = 0.00 L T = 300 K T =? T (300 K)(.89 atm)(0.500 L) T = = (.3 atm)(0.00 L) =.5 03 K = 500 ml = L Note that the calculation could have been carried through with the two volumes expressed in milliliters because the universal gas constant does not appear explicitly in this form of the ideal gas equation. 37. molar mass of O = 3.00 g; 5 C + 73 = 98 K O 50. g O 3.00 g O =.56 mol O total number of moles of gas =.0 mol N +.56 mol O =.56 mol (.56 mol)( L atm mol K )(98 K) = = = 3 atm (5.0 L) 38. molar masses: N, 8.0 g; He, g; ST:.00 atm, 73 K N. g N 8.0 g N = 0.43 mol N 4.05 g He He g He =.0 He Total moles of gas = 0.43 mol +.0 =.44 mol = (.44 mol)( L atm mol K )(73 K) = = 3.3 L (.00 atm) 9

27 39. The pressures must be expressed in the same units, either mm Hg or atm. hydrogen = total water vapor.03 atm = mm Hg hydrogen = mm Hg 4. mm Hg = mm Hg 4. mm Hg = atm hydrogen =.03 atm atm = atm 40. N (g) + 3H (g) NH 3 (g) molar mass of NH 3 = 7.03 g; C + 73 = 84 K NH g NH g NH = 0.94 mol NH 3 to be produced 3 N 0.94 mol NH 3 mol NH = 0.47 mol N required 3 mol H 0.94 mol NH 3 mol NH = 0.44 H required 3 3 nitrogen = hydrogen = (0.47 mol)( L atm mol K )(84 K) = = 3.43 L N (0.998 atm) (0.44)( L atm mol K )(84 K) = = 0.3 L H (0.998 atm) 4. C 6 H O 6 (s) + 6O (g) CO (g) + 6H O(g) molar mass of C 6 H O 6 = 80. g; 8 C + 73 = 30 K C6HO g C 6 H O g C H O = mol C 6H O mol O mol C 6 H O 6 C H O = mol O 6 6 oxygen = (0.667 mol)( L atm mol K )(30 K) = = 4. L (0.976 atm) Because the coefficients of CO (g) and H O(g) in the balanced chemical equation happen to be the same as the coefficient of O (g), the calculations for the volumes of these gases produced are identical: 4. L of each gaseous product is produced. 4. Cu S(s) + 3O (g) Cu O(s) + SO (g) molar mass Cu S = 59. g; 7.5 C + 73 = 30 K CuS 5 g Cu S 59. g Cu S = mol Cu S 93

28 3 mol O mol Cu S mol Cu S = mol O oxygen = (0.355 mol)( L atm mol K )(30 K) = = 5.8 L O (0.998 atm) mol SO mol Cu S mol Cu S = mol SO sulfur dioxide = (0.570 mol)( L atm mol K )(30 K) = = 3.9 L SO (0.998 atm) 43. NaHCO 3 (s) Na CO 3 (s) + H O(g) + CO (g) molar mass NaHCO 3 = 84.0 g; 9 C + 73 = 30 K; 769 torr =.0 atm NaHCO3.00 g NaHCO g NaHCO = mol NaHCO 3 HO mol NaHCO 3 mol NaHCO = mol H O 3 3 Because H O(g) and CO (g) have the same coefficients in the balanced chemical equation for the reaction, if mol H O is produced, then mol CO must also be produced. The total number of moles of gaseous substances produced is thus = 0.09 mol. total = (0.09 mol)( L atm mol K )(30 K) = = 0.9 L (.0 atm) 44. One mole of any ideal gas occupies.4 L at ST. 35 mol N.4 L = L 45. = atm =.00 atm = 5 L =? T = 5 C + 73 = 98 K T = 73 K T (73 K)(0.987 atm)(5 L) = T = (98 K)(.00 atm) = 3 L 46. molar masses: He, g; Ar, g; Ne, 0.8 g He 5.0 g He =.49 mol He g He.0 g Ar Ar g Ar = mol Ar 94

29 3.5 g Ne Ne 0.8 g Ne = mol Ne Total moles of gas = =.447 mol.4 L is the volume occupied by one mole of any ideal gas at ST. This would apply even if the gas sample is a mixture of individual gases..447 mol.4 L = 3 L total volume for the mixture The partial pressure of each individual gas in the mixture will be related to what fraction on a mole basis each gas represents in the mixture. He =.00 atm Ar =.00 atm Ne =.00 atm.49 mol He.447 mol total mol Ar.447 mol total = 0.86 atm = 0.07 atm mol Ne = 0. atm.447 mol total 47. CaCO 3 (s) CaO(s) + CO (g) molar mass of CaCO 3 = 00. g CaCO3 7.5 g CaCO g CaCO = 0.75 mol CaCO 3 3 From the balanced chemical equation, if 0.75 mol of CaCO 3 reacts, then 0.75 mol of CaCO 3 will be produced mol H.4 L = 6.6 L 48. The solution is only 50% H O. Therefore 5 g solution = 6.5 g H O molar mass of H O = 34.0 g; T = 7 C = 300 K; = 764 mm Hg =.0 atm 6.5 g H O 34.0 g =.84 mol H O O.84 mol H O mol H O = 0.90 mol O = (0.90 mol)( L atm mol K )(300 K) = =.4 L (.0 atm) 95

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