CHAPTER 5 GASES AND THE KINETIC- MOLECULAR THEORY

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1 CHAPTER 5 GASES AND THE KINETIC- MOLECULAR THEORY 5.1 a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger container. b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced. 5. The particles in a gas are further apart than those are in a liquid. a) The greater empty space between gas molecules allows gases to be more compressible than liquids. b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than liquids. c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be solutions. d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density. 5. The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according to the air pressure on the reservoir. On a mountaintop, the air pressure is less, so the height of mercury it balances is shorter than at sea level. 5.4 The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury column is directly proportional to its height. 5.5 When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end manometer as the flask pressure cannot be less than the vacuum in the other arm. 5.6 The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the densities of the two liquids. hho dhg = h d h Hg HO d HO Hg = xhhg= d 1.5 g/ml 10 m 1 cm 75 mmhg HO 1.00 g/ml 1 mm 10 m = = 979 cm H O 5.7 P (mmh O) = h HO d Hg = xhhg= d 1.5 g/ml 10 m 1 cm ( 755 mmhg) HO 1.00 g/ml 1 mm 10 m = = 1.0 x 10 cm H O 5.8 a) ( atm) b) ( 99 torr) c) ( 65 kpa) 760 mmhg = 566. = 566 mmhg 1atm bar = 1.56 = 1. bar 760 torr 1atm =.607 =.60 atm kpa 5-1

2 d) ( 804 mmhg) kpa 760 mmhg = = 107 kpa 5.9 a) ( 74.8 cmhg) b) ( 7.0 atm) c) ( 8.50 atm) d) ( kpa) 10 m 1 mm 1 atm = = atm 1cm 10 m 760mmHg kpa = =.74 x 10 kpa 1atm bar = = 8.61 bar 1atm 760 torr = = 6.80 torr kpa 5.10 Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. (.5 cm 10 m 1 mm 1 torr ) 1cm 10 =.5 torr m 1mmHg 78.5 torr -.5 torr = torr 1atm P(atm) = ( torr) = = atm 760 torr 5.11 Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. ( 1.0 cm 10 m 1 mm 1 torr ) 1cm 10 = 1.0 torr m 1mmHg 765. torr torr = 75. torr kpa P(kPa) = ( 75. torr) = = 100. kpa 760 torr 5.1 The difference in the height of the Hg is directly related to pressure in atmospheres. 1mmHg 1atm P(atm) = ( 0.74 mhg) = = atm 10 mhg 760 mmhg 5.1 P(Pa) = 5.56 cm 10 mhg 1 mmhg x 10 Pa 1 cmhg 760 mmhg 10 m = = 4.75 x 10 Pa 5.14 a) P(atm) = (.75 x 10 mmhg) b) P(atm) = ( 91 psi) c) P(atm) = ( 9.15 x 10 6 Pa) 1atm = = 0.6 atm 760 mmhg 1atm = = 6. atm 14.7 psi 1atm 5 = 90.0 = 90. atm x 10 Pa 5-

3 d) P(atm) = (.44 x 10 torr) 4 1atm 760 torr =.105 =.1 atm 5.15 a) 1 atm = x 10 5 N/m The force on 1 m of ocean is x 10 5 N. F = m x g x 10 5 N = mg 5 kg m x 10 = (mass) (9.81 m/s ) s mass = x 10 4 = 1.0 x 10 4 kg 4 kg 10 g 10 m b) x 10 = x 10 m 1kg 1cm g/cm (unrounded) g 1mL 1cm Height = x 10 = = 45.7 cm Os cm.6 g 1 ml 5.16 The statement is incomplete with respect to temperature and mass of sample. At constant temperature and moles of gas, the volume of gas is inversely proportional to the pressure variable fixed a) Volume; Temperature Pressure; Moles b) Moles; Volume Temperature; Pressure c) Pressure; Temperature Volume; Moles 5.18 At constant temperature and volume, the pressure of the gas is directly proportional to the number of moles of the gas. Verify this by examining the ideal gas equation. At constant T and V, the ideal gas equation becomes P = n(rt/v) or P = n x constant a) n fixed b) P halved c) P fixed d) T doubled 5.0 a) As the pressure on a gas increases, the molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to one third of the original volume at constant temperature (Boyle s Law). b) As the temperature of a gas increases, the gas molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of.5 (at constant pressure) then the volume will increase by a factor of.5 (Charles s Law). c) As the number of molecules of gas increase, the force they exert on the container increases. This results in an increase in the volume of the container. Adding two moles of gas to one mole increases the number of moles by a factor of three, thus the volume increases by a factor of three (Avogadro s Law). 5.1 V = V 1 x (P 1 /P ) x (T /T 1 ) a) If P is 1/4 of P 1, then the volume would be increased by a factor of 4. b) V = V 1 x (101 kpa/0 kpa) x (155 K/10 = 1/4, so the volume would be decreased by a factor of 4. c) V = V 1 x (0 kpa/101 kpa) x (05 K/05 =, so the volume would be increased by a factor of. 5. a) The temperature is decreased by a factor of, so the volume is decreased by a factor of (Charles s Law). b) The temperature increases by a factor of [( ) / (50 + 7)] = 1.56, so the volume is increased by a factor of 1.56 (Charles s Law). c) The pressure is increased by a factor of 4, so the volume decreases by a factor of 4 (Boyle s Law). 5-

4 5. V = V 1 x (P 1 /P ) x (T /T 1 ) x (n /n 1 ) a) Since n = 0.5 n 1, the volume would be decreased by a factor of. b) This corresponds to both P and T remaining constant, so the volume remains constant. c) Since P = 0.5 P 1 and T = 0.5 T 1, these two effects offset one another and the volume remains constant. 5.4 Plan: The temperature must be lowered to reduce the volume of a gas. Charles s Law states that at constant pressure and with a fixed amount of gas, the volume of a gas is directly proportional to the absolute temperature of the gas. Solution: (V 1 / T 1 ) = (V / T ) at constant n and P T = T 1 (V / V 1 ).50 L T = (( ) - 7 = = -4 C 5.10 L 5.5 V = V 1 (T / T 1 ) = ( 9 L) ( ) ( + ) 7 K K = = 56 L 5.6 Plan: To find V, summarize and convert gas variables and apply the combined gas law equation. (P 1 V 1 /T 1 = P V T ) Solution: 7 K 15. kpa V = V 1 (T / T 1 ) (P 1 / P ) = ( 5.5 L) = 5.47 = 5. L 98 K kpa 5.7 V = V 1 (T / T 1 ) (P 1 / P ) = (.65 L) 7 14 K 745 torr 98 K 67 torr = = 6.44 L 5.8 Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using the ideal gas equation, n = PV/RT. The gas constant, R = L atm/mol K, gives pressure in atmospheres and temperature in Kelvin. The given pressure in torr must be converted to atmospheres and the temperature converted to Kelvin. Solution: ( 8 torr)( 5.0 L) 1atm n = PV / RT = = = mol chlorine 760 torr (( 7 + 7) 5.9 PV = nrt ( + ) 1.47 x 10 mol K 1 ml 760 torr P = nrt / V = 75.0 ml 10 L 1 atm = = 66 torr 5-4

5 5.0 Plan: Solve the ideal gas equation for moles and convert to mass using the molar mass of ClF. Solution: PV = (m / M)RT g ( 699 mmhg)( 07 ml) 9.45 mol 1atm 10 L m = PV M / RT = 760 mmhg 1 ml (( 7 45) + = = g ClF 5.1 PV = (m/ M)RT ( 75.0 g) (( 7 115) + P = mrt/ M V = = = 18 atm N O g 44.0 (.1 L) mol 5. PV = nrt The total pressure of the gas is ( )psi. (( ) psi)( 1.5 L) 1atm n = PV / RT = = = 0.4 mol SO 14.7 psi (( 7 + ) 5. Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the number of moles of gas calculated at sea level will be the same as the number of moles of gas at the higher altitude. Using the ideal gas equation, (n x R) is a constant equal to (PV/T). Given the sea-level conditions of volume, pressure and temperature, and the temperature and pressure at the higher altitude for the gas in the balloon, we can set up an equation to solve for the volume at the higher altitude. Comparing the calculated volume to the given maximum volume of 85 L will tell us if the balloon has reached its maximum volume at this altitude. Solution: (P 1 V 1 / T 1 ) = (P V / T ) ( 755 torr )( 55.0 L )(( 7 5 ) K ) 1atm V = (P 1 V 1 T / T 1 P ) = = = 7.5 x 10 ( L ( 7 + ) ( atm) 760 torr The calculated volume of the gas at the higher altitude is less than the maximum volume of the balloon. No, the balloon will not reach its maximum volume. Check: Should we expect that the volume of the gas in the balloon should increase? At the higher altitude, the pressure decreases, this increases the volume of the gas. At the higher altitude, the temperature decreases, this decreases the volume of the gas. Which of these will dominate? The pressure decreases by a factor of 0.99/0.066 = 15. If we label the initial volume V 1, then the resulting volume is 15 V 1. The temperature decreases by a factor of 96/68 = 1.1, so the resulting volume is V 1 /1.1 or 0.91 V 1. The increase in volume due to the change in pressure is greater than the decrease in volume due to change in temperature, so the volume of gas at the higher altitude should be greater than the volume at sea level. 5.4 Air is mostly N (8.0 g/mol), O (.00 g/mol), and argon (9.95 g/mol). These heavy gases dominate the density of dry air. Moist air contains H O (18.0 g/mol). The relatively light water molecules lower the density of the moist air. 5.5 The molar mass of H is less than the average molar mass of air (mostly N, O, and Ar), so air is denser. To collect a beaker of H (g), invert the beaker so that the air will be replaced by the lighter H. The molar mass of CO is greater than the average molar mass of air, so CO (g) is more dense. Collect the CO holding the beaker upright, so the lighter air will be displaced out the top of the beaker. 5.6 Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present. 5.7 P A = X A P T The partial pressure of a gas (P A ) in a mixture is directly proportional to its mole fraction (X A ). 5-5

6 5.8 Plan: Using the ideal gas equation and the molar mass of xenon, 11. g/mol, we can find the density of xenon gas at STP. Standard temperature is 0 C and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: ( 11. g/mol)( 1 atm) d = M P / RT = = = 5.86 g/l ( d = M P / RT = ( g/ mol)( 1.5 atm) L atm (( + ) ) K = = 6.4 g/l 5.40 Plan: Apply the ideal gas equation to determine the number of moles. Convert moles to mass and divide by the volume to obtain density in g/l. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: ( 1 atm)( L) n = PV / RT = = x 10 = 1.78 x 10 mol AsH ( 7 ( x 10 mol ) ( g/ mol) d = mass / volume = = =.48 g/l L 5.41 d = M P / RT M = drt / P = Therefore, the gas is Ne..71 g/ L K ( + ) (.00 atm) = = 0. g/mol 5.4 Plan: Rearrange the formula PV = (m / M)RT to solve for molar mass: M = mrt / PV. Solution: ( 06 ng) (( 7 45) K + ) 10 9 g 1 L 760 torr M = mrt / PV = µ 6 ( 88 torr)( 0.06 µ L) 1 ng 10 L 1 atm = = 51.1 g/mol 5.4 M = mrt / PV = 0.10 g ( 7 K ) 1mL 760mmHg ( 747 mmhg)( 6.8 ml) 10 L 1 atm ( + ) = = 9.8 g/mol The molar masses are N = 8 g/mol, Ne = 0 g/mol, and Ar = 40 g/mol Therefore, the gas is Ar. 5-6

7 5.44 Plan: Use the ideal gas equation to determine the number of moles of Ar and O. The gases are combined (n TOT = n Ar + n O ) into a 400 ml flask (V) at 7 C (T). Determine the total pressure from n TOT, V, and T. Solution: n = PV / RT ( 1.0 atm)( L) Moles Ar = PV / RT = = mol Ar (unrounded) (( 7 + 7) L atm (( + ) ) 501 torr 0.00 L 1atm Moles O = PV / RT = = mol O (unrounded) 760 torr K (( ) mol) (( 7 7) K + ) 1mL P mixture = nrt / V = 400 ml 10 L = 1.70 = 1. atm 5.45 Total moles = n T = PV/RT = ( + ) 66 mmhg 55 ml 1atm 10 L 760 mmhg 1 ml ( 7 5 = mol (unrounded) Moles Ne = n Ne = (0.146 g Ne) (1 mol Ne / 0.18 g Ne) = mol Ne (unrounded) Moles Ar = n T - n Ne = ( ) mol = = mol Ar 5.46 d = M P / RT At 17 C At 60.0 C d = M P / RT = d = M P / RT = L atm (( + ) ) 8.8 g/ mol 744 torr 1atm = = 1.18 g/l 760 torr K L atm (( + ) ) 8.8 g/ mol 744 torr 1atm = = 1.0 g/l 760 torr K 5.47 PV = nrt n / V = P / RT = (( ) ) 650. torr 1atm 760 torr K = = mol/l 5.48 Plan: The problem gives the mass, volume, temperature and pressure of a gas, so we can solve for molar mass using M = mrt/pv. The problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms. We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound. 5-7

8 Solution: ( 0.48 g) (( 7 101) K + ) 760 torr M = mrt/pv = = g/mol (unrounded) ( 767 torr)( 0.04 L) 1 atm The carbon accounts for [5 (1 g/mol)] = 60 g/mol, thus, the hydrogen must make up the difference (7-60) = 1 g/mol. A value of 1 g/mol corresponds to 1 H atoms. (Since fractional atoms are not possible, rounding is acceptable.) Therefore, the molecular formula is C 5 H PV = nrt n = PV / RT Molecules of air = n x Avogadro s number = (PV / RT) x Avogadro s number ( 1.00 atm)( 1.00 L) 6.0 x 10 molecules Molecules of air = mol (( 7 5) + = x 10 molecules (unrounded) Molecules N = ( x 10 molecules) (78.08%/100%) = x 10 = 1.9 x 10 molecules N Molecules O = ( x 10 molecules) (0.94%/100%) = x 10 1 = 5.15 x 10 1 molecules O Molecules CO = ( x 10 molecules) (0.05%/100%) = x = 1 x molecules O Molecules Ar = ( x 10 molecules) (0.9%/100%) = x 10 0 =. x 10 0 molecules O 5.50 a) PV = nrt L atm ( + ) 850. torr 1 L 1atm n = PV / RT = = = 0.90 mol gas 760 torr ( 7 45 b) The information given in ppm is a way of expressing the proportion, or fraction, of SO present in the mixture. Since n is directly proportional to V, the volume fraction can be used in place of the mole fraction used in equation 5.1. There are 7.95 x 10 parts SO in a million parts of mixture, so volume fraction = (7.95 x 10 / 1 x 10 6 ) = 7.95 x 10. Therefore, P SO = volume fraction x P TOT = (7.95 x 10 ) (850. torr) = = 6.76 torr Plan: We can find the moles of oxygen from the standard molar volume of gases and use the stoichiometric ratio from the balanced equation to determine the moles of phosphorus that will react with the oxygen. Solution: P 4 (s) + 5 O (g) P 4 O 10 (s) 1 molo 1 mol P g P4 Mass P 4 = ( 5.5 LO ) = = 9. g P 4.4 L O 5 mol O 1 mol P4 5.5 KClO (s) KCl(s) + O (g) Moles O = n = PV / RT Moles of O x mole ratio x molar mass KClO = g KClO ( 75 torr)( 68 ml ) 1atm 10 L mol KClO 1.55 g KClO Mass KClO = 760 torr 1 ml mol O 1 mol KClO (( 7 18) + = = 1.57 g KClO 5-8

9 5.5 Plan: To find the mass of PH, write the balanced equation, convert mass of P 4 to moles, solve for moles of H using the standard molar volume (or use ideal gas equation), and proceed with the limiting reactant problem. Solution: P 4 (s) + 6 H (g) 4 PH (g) Moles hydrogen = (8.0 L) (1 mol /.4 L) = mol H (unrounded) Moles phosphorus = (7.5 g P 4 ) (1 mol P 4 / 1.88 g P 4 ) = mol P 4 (unrounded) Dividing each of the moles by the coefficients shows that P 4 is the limiting reactant. 1molP4 4 mol PH.99 g PH Mass PH = ( 7.5 g P4 ) = = 41. g PH 1.88 g P4 1 mol P4 1 mol PH NH (g) + 5 O (g) 4 NO(g) + 6 H O(l) The moles are directly proportional to the volumes of the gases at the same temperature and pressure. Thus, the limiting reactant may be found by comparing the volumes of the gases. The volumes are divided by the coefficient in the balanced chemical equation, and the lower amount is limiting. (5.6 L NH / 4) = 8.90 L NH (40.5 L O / 5) = 8.10 L O O is the limiting reactant. 1molO 4 mol NO 0.01 g NO Mass NO = ( 40.5 LO ) = = 4.4 g NO.4 L O 5 mol O 1 mol NO 5.55 Plan: First, write the balanced equation. The moles of hydrogen produced can be calculated from the ideal gas equation and then the stoichiometric ratio from the balanced equation is used to determine the moles of aluminum that reacted. The problem specifies hydrogen gas collected over water, so the partial pressure of water must first be subtracted. Table 5. reports pressure at 6 C (5. torr) and 8 C (8. torr), so take the average of the two values. Solution: Al(s) + 6 HCl(aq) AlCl (aq) + H (g) Hydrogen pressure = total pressure - pressure from water = (751 mmhg) - [( ) torr / ] = 74.5 torr (unrounded) Moles of hydrogen = n = PV / RT Mass of aluminum = mol of hydrogen x mole ratio x molar mass of Al = PV/RT x mole ratio x molar mass of Al ( 74.5 torr)( 5.8 ml ) 1 atm 10 L mol Al 6.98 g Al Mass of Al = 760 torr 1 ml mol H 1 mol Al (( 7 7) + = = g Al 5.56 First, write the balanced equation. The problem specifies hydrogen gas collected over water, so the partial pressure of water must first be subtracted. Table 5. reports pressure at 18 C (15.5 torr). Li(s) + H O(l) LiOH(aq) + H (g) Pressure H = total pressure - vapor pressure of H O = 75 mmhg torr (1mmHg/1torr) Moles H = g Li x 1/molar mass Li x mole ratio Volume H = V = nrt / P = (g Li x 1/molar mass Li x mole ratio)rt / P 1molLi 1molH 0.84 g Li (( 7 18) g Li mol Li mmhg Volume H = ( ) mmhg 1 atm = = 1.5 L H 5-9

10 5.57 Plan: To find ml of SO, write the balanced equation, convert the given mass of P 4 S to moles, use the molar ratio to find moles of SO, and use the ideal gas equation to find volume. Solution: P 4 S (s) + 8 O (g) P 4 O 10 (s) + SO (g) Moles SO = mass P 4 S x 1/molar mass P 4 S x mole ratio Volume SO = nrt / P = (mass P 4 S x 1/molar mass P 4 S x mole ratio)rt / P 1molPS 4 molso ( g P4S ) (( 7 ) 0.09 g P4S 1 mol P4S torr 1 ml V = 75 torr 1 atm 10 L = 86.9 = 86 ml 5.58 CCl 4 (g) + HF(g) CF Cl (g) + HCl(g) Moles Freon = n = PV / RT Mass CCl 4 = mol Freon x mole ratio x molar mass of CCl 4 = (PV / RT) x mole ratio x molar mass of CCl 4 ( 1.0 atm)( 16.0 dm ) 1L 1 mol CCl g CCl4 = L atm 1dm 1 mol CFCl 1 mol CCl (( 7 7) + = = 1.0 x 10 g CCl Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the number of moles of silicon tetrafluoride gas formed. Then, using the ideal gas equation with the moles of gas, the temperature and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Solution: XeF 6 (s) + SiO (s) XeOF 4 (l) + SiF 4 (g) Mole SiF 4 = n = mass XeF 6 x 1/molar mass XeF 6 x mole ratio Pressure SiF 4 = P = nrt / V = (mass XeF 6 x 1/molar mass XeF 6 x mole ratio)rt / V 1molXeF6 1molSiF4 (.00 g XeF6 ) (( 7 5) 45. g XeF6 mol XeF + 6 P = 1.00 L = = atm SiF PbS(s) + O (g) PbO(g) + SO (g) 10 g 1 mol PbS Moles PbS = (.75 kg PbS) = mol PbS (unrounded) 1 kg 9. g PbS Moles O = n = PV / RT = (.0atm)( 8L) L atm (( + ) ) = mol O (unrounded) K Dividing each mole by the coefficient in the equation [( / ) and (11.66 / )] shows that O has the lower value. Thus, O is limiting. Volume SO = (11.66 mol O ) ( mol SO / mol O ) (.4 L SO / 1 mol SO ) = 168. = 1.7 x 10 L SO 5-10

11 5.61 HgO(s) Hg(l) + O (g) Moles O = n = amount of HgO x 1/molar mass HgO x mole ratio Pressure O = P = nrt / V = (amount of HgO x 1/molar mass HgO x mole ratio)rt / V 0.0% 1 mol HgO 1molO L atm ( 40.0 g HgO) (( 7 5.0) 100% 16.6 g HgO mol HgO + 1mL P = 50 ml 10 L = = atm O 5.6 As the temperature of the gas sample increases, the most probable speed increases. This will increase both the number of collisions per unit time and the force of each collision with the sample walls. Thus, the gas pressure increases. 5.6 At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical. This is because at the same temperature, all gases have the same average kinetic energy, resulting in the same pressure The rate of effusion is much higher for a gas than its rate of diffusion. Effusion occurs into an evacuated space, whereas diffusion occurs into another gas. It is reasonable to expect that a gas will escape faster into a vacuum than it will into a space already occupied by another gas. The ratio of the rates of effusion and diffusion for two gases will be the same since both are inversely proportional to the square root of their molar masses a) Mass O > mass H b) d > d O H c) Identical d) Identical e) Average speed H > average speed O f) Effusion time H < effusion time O 5.66 The molar masses of the three gases are.016 for H (Flask A), 4.00 for He (Flask B), and for CH 4 (Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H will contain more gas molecules than 4 g of He or 4 g of CH 4. Since helium has a smaller molar mass than methane, 4 g of He will contain more gas molecules than 4 g of CH 4. a) P A > P B > P C The pressure of a gas is proportional to the number of gas molecules. So, the gas sample with more gas molecules will have a greater pressure. b) E A = E B = E C Average kinetic energy depends only on temperature. The temperature of each gas sample is 7 K, so they all have the same average kinetic energy. c) rate A > rate B > rate C When comparing the speed of two gas molecules, the one with the lower mass travels faster. d) total E A > total E B > total E C Since the average kinetic energy for each gas is the same (part b of this problem) then the total kinetic energy would equal the average times the number of molecules. Since the hydrogen flask contains the most molecules, its total kinetic energy will be the greatest. e) d A = d B = d C Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same mass, 4 g. Thus, the density of each is 4 g/5 L = 0.8 g/l. f) Collision frequency (A) > collision frequency (B) > collision frequency (C) The number of collisions depends on both the speed and the distance between gas molecules. Since hydrogen is the lightest molecule it has the greatest speed and the 5 L flask of hydrogen also contains the most molecules, so collisions will occur more frequently between hydrogen molecules than between helium molecules. By the same reasoning, collisions will occur more frequently between helium molecules than between methane molecules. 5-11

12 5.67 To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (equation 5.14). Rate H Rate UF = Molar Mass UF6 Molar Mass H = 5.0 g/mol = 1.17 = g/mol To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (equation 5.14). Rate O Molar Mass Kr = Rate Kr Molar Mass O = 8.80 g/mol = = g/mol 5.69 a) The gases have the same average kinetic energy because they are at the same temperature. The heavier Ar atoms are moving slower than the lighter He atoms to maintain the same average kinetic energy. Therefore, Curve 1 better represents the behavior of Ar. b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice. c) Fluorine gas exists as a diatomic molecule, F, with M = 8.00 g/mol. Therefore, F is much closer in size to Ar than He, so Curve 1 more closely represents F s behavior a) Curve 1 b) Curve c) Curve 5.71 Rate He Rate F = Time F Time He = Molar Mass F Molar Mass He = 8.00 g/mol = (unrounded) 4.00 g/mol Time F = (time He) = (4.55 min) = = 14.0 min Rate H 5.7 Rate Unknown = Time Unk Time H = Molar Mass Unk Molar Mass H 11.1 min.4 min = Molar Mass Unk.016 g/mol Molar Mass Unk =.016 g/mol Molar Mass unknown = = 4.4 g / mol 5.7 White phosphorus is a molecular form of the element phosphorus consisting of some number, x, of phosphorus atoms. Determine the number of phosphorus atoms, x, in one molecule of white phosphorus from the rate of effusion of the gaseous phosphorus molecules. Rate P x Rate Ne = Molar Mass Ne = Molar Mass P x Molar Mass Ne Molar Mass P = (0.404) = (unrounded) x Molar Mass P x = Molar Mass Ne / = 0.18 / = g/mol Molar Mass P x = x (Molar Mass P) = x (0.97 g/mol) x = (1.698 g/mol) / (0.97 g/mol) =.99 = 4 Thus, 4 atoms per molecule, so P x = P

13 5.74 a) 0 C = 7 K 0 C = 0 K (4.00 g He/mol) (1 kg / 10 g) = kg/mol J 8.14 ( 7 kg m /s u rms He (at 0 C) = = 1.04 x kg/ mol J = 1.0 x 10 m /s J 8.14 ( 0 kg m /s u rms He (at 0 C) = = x kg/ mol J = 1.7 x 10 m /s b) (11. g Xe/mol) (1 kg / 10 g) = 0.11 kg/mol J 8.14 ( 0 kg m /s u rms He (at 0 C) = = 9.91 m/s (unrounded) 0.11 kg/ mol J Rate He / Rate Xe = (1.740 x 10 m/s) / (9.91 m/s) = = 5.7 He molecules travel at almost 6x the speed of Xe molecules. c) E He = 1/ mv = (1/) ( kg/mol) (1.740 x 10 m/s) (1J/(kg m /s ) = =.78 x 10 J/mol E Xe = 1/ mv = (1/) (0.11 kg/mol) (9.91 m/s) (1J/(kg m /s ) = =.78 x 10 J/mol d) ( J/mol) (1 mol He / 6.0 x 10 atoms He) = x 10 1 = 6.7 x 10 1 J/He atom 5.75 Intermolecular attractions cause the real pressure to be less than ideal pressure, so it causes a negative deviation. The size of the intermolecular attraction is related to the constant a. According to Table 5.5, a = 1.9, a N Kr =. and a =.59. Therefore, CO CO experiences a greater negative deviation in pressure than the other two gases: N < Kr < CO Molecular size causes a positive deviation from ideal behavior. Thus, V Real Gases > V Ideal Gases. Therefore, the order is H < O < Cl Nitrogen gas behaves more ideally at 1 atm than at 500 atm because at lower pressures the gas molecules are farther apart. An ideal gas is defined as consisting of gas molecules that act independently of the other gas molecules. When gas molecules are far apart, they act ideally, because intermolecular attractions are less important and the volume of the molecules is a smaller fraction of the container volume At 150 C. At higher temperatures, intermolecular attractions become less important and the volume occupied by the molecules becomes less important Do not forget to multiple the area of each side by two. Surface area of can = (40.0 cm) (15.0 cm) + (40.0 cm) (1.5 cm) + (15.0 cm) (1.5 cm) =.575 x 10 cm (unrounded) Total force = (.575 x 10 cm ) (1 in/.54 cm) (14.7 lb/in ) = x 10 = 5.87 x 10 lbs 5.80 Molar mass has units of g/mol, so solve for the number of hemoglobin (Hb) moles combined with O. Moles of oxygen (from ideal gas equation) combine with Hb in a 4:1 ratio. PV = nrt ( 74 torr)( 1.5 ml ) 1atm 10 L Moles O = PV / RT = 760 torr 1 ml (( 7 7) + = x 10 5 mol O (unrounded) Moles Hb = ( x 10 5 mol O ) (1 mol Hb / 4 mol O ) = x 10 5 mol Hb (unrounded) Molar mass hemoglobin = (1.00 g Hb) / ( x 10 5 mol Hb) = x 10 4 = 6.81 x 10 4 g/mol 5-1

14 5.81 Reaction 1: NaHCO (s) Na CO (s) + H O(l) + CO (g) Moles CO = (1.00 g NaHCO ) (1 mol NaHCO / g NaHCO ) (1 mol CO / mol NaHCO ) = x 10 mol CO (unrounded) ( x10 mol) (( ) K + ) 1mL V = nrt / P = atm 10 L = = 7 ml CO Reaction 1 Reaction : NaHCO (s) + H + (aq) H O(l) + CO (g) + Na + (aq) Moles CO = (1.00 g NaHCO ) (1 mol NaHCO / g NaHCO ) (1 mol CO /1 mol NaHCO ) = x 10 mol CO (unrounded) ( x 10 mol) (( ) K + ) 1mL V = nrt / P = atm 10 L = = 474 ml CO Reaction 5.8 Rearrange PV = nrt to R = PV / nt This gives: P i V i / n i T i = P f V f / n f T f P i = 1.01 atm P f = 1.01 atm (Thus, P has no effect, and does not need to be included.) T i = 05 K T f = 50 K n i = n i n f = 0.75 n i V i = 600. L V f =? ( 0.75ni ( 50 ( 600. L) ) V f = (n f T f V i ) / (n i T i ) = n 05 K = = 69 L i 5.8 a) Ideal gas equation: PV = nrt 10 g 1molCl L atm ( kg Cl ) (( 7 5) 1 kg g Cl + P IGL = nrt/v = L =.4791 =.5 atm na b) P + ( V nb) = nrt V nrt n a P VDW = V nb V n = ( kg) (10 g/1 kg) (1 mol Cl / g Cl ) = mol Cl (unrounded) atm L ( mol Cl ) ( ) ( ) K mol Cl mol P VDW = L L ( mol Cl ) ( L) mol = 1.5 = 1. atm 5-14

15 ( g) (( ) K + ) 1mL 5.84 a) Molar Mass I = mrt / PV = ( atm)( ml) 10 L = = 6.09 g I / mol ( g) (( ) K + ) 1mL Molar Mass II = mrt / PV = ( atm)( ml) 10 L = 5.9 = 5.9 g II / mol ( g) (( ) K + ) 1mL Molar Mass III = mrt / PV = ( atm)( ml) 10 L = = g III / mol b)% H in I = 100% % = 14.7% H % H in II = 100% % = 18.90% H % H in III = 100% % = 17.0% H Assume 100 grams of each so the mass percentages are also the grams of the element. I (85.6 g B) (1 mol B / g B) = mol B (unrounded) (14.7 g H) (1 mol H / g H) = mol H (unrounded) Dividing by the smaller value (7.9169) gives B = 1 and H = The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying by 5 gives B = 5 and H = 9. The empirical formula is B 5 H 9, which has a formula mass of 6.1 g/mol. The empirical formula mass is near the molecular mass from part a. Therefore, the empirical and molecular formulas are both B 5 H 9. II (81.10 g B) (1 mol B / g B) = mol B (unrounded) (18.90 g H) (1 mol H / g H) = mol H (unrounded) Dividing by the smaller value (7.501) gives B = 1 and H =.499 The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is. Multiplying by gives B = and H = 5. The empirical formula is B H 5, which has a formula mass of 6.66 g/mol. Dividing the molecular formula mass by the empirical formula mass gives the relationship between the formulas: (5.9) / (6.66) =. The molecular formula is two times the empirical formula, or B 4 H 10. III (8.98 g B) (1 mol B / g B) = mol B (unrounded) (17.0 g H) (1 mol H / g H) = mol H (unrounded) Dividing by the smaller value (7.676) gives B = 1 and H =. The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying by 5 gives B = 5 and H = 11. The empirical formula is B 5 H 11, which has a formula mass of g/mol. The empirical formula mass is near the molecular mass from part a. Therefore, the empirical and molecular formulas are both B 5 H

16 Rate SO c) Rate IV = Molar Mass IV Molar Mass SO 50.0 ml 1.04 min Molar Mass Unk = = (unrounded) 50.0 ml g/mol 1.00 min Molar Mass IV = = 7.68 g / mol IV % H in IV = 100% % = 1.86% H Assume 100 grams of each so the mass percentages are also the grams of the element. (78.14 g B) (1 mol B / g B) = mol B (unrounded) (1.86 g H) (1 mol H / g H) = mol H (unrounded) Dividing by the smaller value (7.849) gives B = 1 and H =.000 The empirical formula is BH, which has a formula mass of 1.8 g/mol. Dividing the molecular formula mass by the empirical formula mass gives the relationship between the formulas: (7.68) / (1.8) =. The molecular formula is two times the empirical formula, or B H a) V f = V i (T f / T i ) (P i / P f ) V f = V i ((7+100 / (7+00 ( atm / 1 atm) V f = V i ((7 / (47 ( atm / 1 atm) = V i (unrounded) b) V f = V i (T f / T i ) (P i / P f ) V f = V i ((57 / (7 (1 atm / atm) = V i (unrounded) c) V f = V i (T f / T i ) (P i / P f ) V f = V i ((400 / (00 ( atm / 6 atm) = V i d) V f = V i (T f / T i ) (P i / P f ) V f = V i ((4 / (57 (0. atm / 0.4 atm) = V i (unrounded) Increase Decrease Unchanged Decrease 5.86 a) Partial pressures are calculated from Dalton s Law of Partial Pressures: P A = X A (P total ) Convert each mole percent to a mole fraction by dividing by 100%. P Nitrogen = X Nitrogen P Total = (0.786) (1.00 atm) (760 torr / 1 atm) = = 597 torr N P Oxygen = X Oxygen P Total = (0.09) (1.00 atm) (760 torr / 1 atm) = = 159 torr O P Carbon Dioxide = X Carbon Dioxide P Total = (0.0004) (1.00 atm) (760 torr / 1 atm) = 0.04 = 0. torr CO P Water = X Water P Total = (0.0046) (1.00 atm) (760 torr / 1 atm) =.496 =.5 torr O b) Mole fractions can be calculated by rearranging Dalton s Law of Partial Pressures: X A = P A /P total and multiply by 100 to express mole fraction as percent. P Total = ( ) torr = 760 torr N : [(569 torr) / (760 torr)] x 100% = = 74.9 mol% N O : [(104 torr) / (760 torr)] x 100% = = 1.7 mol% O CO : [(40 torr) / (760 torr)] x 100% = 5.6 = 5. mol% CO H O: [(47 torr) / (760 torr)] x 100% = = 6. mol% CO c) Number of molecules of O can be calculated using the Ideal Gas Equation and Avogadro s number. ( 104 torr)( 0.50 L ) 1atm 6.0 x 10 O molecules Molecules O = 760 torr 1 mol O (( 7 7) + = x 10 1 = 1.6 x 10 1 molecules O 5-16

17 4 1.7 x 10 Rn atoms 15 1 mol Ra 6.0 x 10 Ra atoms 1.0 x 10 Ra atoms 600 s 4 h 6 g Ra 1 mol Ra s h day = x Rn atoms / day (unrounded) 15 1molRn x 10 Rn atoms ( x 10 Rn atoms V = nrt / P = ( 1.00 atm) = x 10 7 = 1. x 10 7 L Rn 5.87 Atoms Rn = ( 1.0 g Ra) mmhg 08 ml 98 K 1atm 5.88 a) V f = P i V i T f / T i P f = = 99. = 4 x 10 ml 86 K 1 atm 760 mmhg mmhg 08 ml 77% N 1atm 10 L b) n = PV / RT = L atm 100% 760 mmhg 1 ml ( 86 K ) = = 0.01 mol N 5.89 a) Moles = (5.14 x t) (1000 kg / t) (10 g / kg) (1 mol / 8.8 g) = x 10 0 = 1.78 x 10 0 mol gas 0 ( x 10 mol) (( 7 5) + b) V = nrt / P = = x 10 1 = 4 x 10 1 L 1atm ( 1 ) c) Height = 8 ( x 10 km ) x 10 L 10 m 1 km = = 9 km 1L 10 m The pressure exerted by the atmosphere is 1 atm only at the surface of the earth. Above the earth s surface, the pressure is lower than this value, so the air expands and extends much further up than 9 km The balanced equation and reactant amounts are given, so the first step is to identify the limiting reactant g Cu 1 mol Cu mol NO Moles NO from Cu = ( 4.95 cm ) = mol NO (unrounded) cm 6.55 g Cu 1 mol Cu 68.0% HNO 1cm 1.4g1 mol HNO molno Moles NO from HNO = ( 0.0 ml) 100% 1 ml ml 6.0 g 4 mol HNO = mol NO (unrounded) Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after the reaction goes to completion. Use the calculated number of moles of NO and the given temperature and pressure in the ideal gas equation to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the only gas involved in the reaction. ( mol NO ) (( 7. 8.) K + ) 760 torr V = nrt / P = ( 75 torr) 1 atm = = 5.7 L NO 5-17

18 1.0 atm 100 ml 10 L 5.91 a) n = PV / RT = = = mol air 1mL (( 7 + 7) b) Molecules = ( mol air) ( 6.0 x 10 molecules air / mol air) =.891 x 10 =.8 x 10 molecules air NaBr(aq) + NaBrO (aq) + H SO 4 (aq) Br (g) + Na SO 4 (aq) + H O(g) 1molNaBr molbr Moles Br from NaBr = ( 75 g NaBr) = mol Br (unrounded) g NaBr 5 mol NaBr 1molNaBrO molbr Moles Br from NaBrO = ( g NaBrO ) g NaBrO 1 mol NaBrO =.9167 mol Br (unrounded) The NaBr is limiting. ( mol) (( 7 00) + Volume = nrt / P = = 88.5 = 88. L Br atm 5.9 The reaction is: NaN (s) Na(s) + N (g) The vapor pressure of water (Table 5.) at 6 C is 5. torr (= 5. mmhg). Volume = nrt/ P = 1molNaN moln 50.0 g NaN (( 7 6) 65.0 g NaN mol NaN mmhg (( ) mmhg ) 1atm = = 9.9 L N 5.94 It is necessary to determine both the empirical formula and the molar mass of the gas. Empirical formula: Assume grams of sample to make the percentages of each element equal to the grams of that element. Thus, 64.81% C = g C, 1.60% H = 1.60 g H, and 1.59% O = 1.59 g O. Moles C = (64.81 g C) (1 mol C / 1.01 g C) = mol C (unrounded) Moles H = (1.60 g H) (1 mol H / g H) = mol H (unrounded) Moles O = (1.59 g O) (1 mol O / g O) = mol O (unrounded) Divide by the smallest number of moles (O): C = (5.964 mol) / ( mol) = 4 H = ( mol) / ( mol) = 10 O = ( mol) / ( mol) = 1 Empirical formula = C 4 H 10 O (empirical formula mass = 74.1 g/mol) Molar Mass: (.57 g) (( 7 5) + M = mrt / PV = = g/mol (unrounded) ( 0.40 atm)(.00 L) Molecular formula: Since the molar mass and the empirical formula mass are similar, the empirical and molecular formulas must both be: C 4 H 10 O 5-18

19 5.95 The empirical formula for aluminum chloride is AlCl (M = 1. g/mol). Calculate the molar mass of the gaseous species from the ratio of effusion rates. This molar mass, divided by the empirical weight, should give a whole number multiple that will yield the molecular formula. Rate Unk Rate He = 0.1 = Molar Mass He Molar Mass Unk 4.00 g/mol 0.1 = Molar Mass Unk Molar Mass Unknown = g/mol (unrounded) The whole number multiple is /1., which is about. Therefore, the molecular formula of the gaseous species is x (AlCl ) = Al Cl a) C 8 H 18 (l) + 5 O (g) 16 CO (g) + 18 H O (g) 1molC8H18 4 mol gas Moles products = ( 100. g C8H18) = mol gas (unrounded) 114. g C8H18 mol C8H18 ( mol gas) (( 7 50) K + ) 760 torr V = nrt / P = ( 75 torr) 1 atm = = 787 L gas 1molC8H18 5 molo b) Moles O = ( 100. g C8H18) = mol O (unrounded) 114. g C8H18 mol C8H18 Moles other gases = ( mol O ) [( )%] / (1%) = mol gas (unrounded) ( mol gas) (( 7 50) K + ) 760 torr V = nrt / P = ( 75 torr) 1 atm = L residual air (unrounded) Total volume of gaseous exhaust = L L = =.96 x 10 L 5.97 Plan: First, write the balanced equation for the reaction: SO + O SO. The total number of moles of gas will change as the reaction occurs since moles of reactant gas forms moles of product gas. From the volume, temperature and pressures given, we can calculate the number of moles of gas before and after the reaction. For each mole of SO formed, the total number of moles of gas decreases by 1/ mole. Thus, twice the decrease in moles of gas equals the moles of SO formed. Solution: Moles of gas before and after reaction ( 1.95 atm)(.00 L) Initial moles = PV / RT = = mol (unrounded) ( 900. ( 1.65 atm)(.00 L) Final moles = PV / RT = = mol (unrounded) ( 900. Moles of SO produced = x decrease in the total number of moles = x ( mol mol) = = 1.6 x 10 mol Check: If the starting amount is total moles of SO and O, then x + y = mol, where x = mol of SO and y = mol of O. After the reaction: (x - z) + (y - 0.5z) + z = mol Where z = mol of SO formed = mol of SO reacted = (mol of O reacted). 5-19

20 Subtracting the two equations gives: x - (x - z) + y - (y - 0.5z) - z = z = mol SO The approach of setting up two equations and solving them gives the same result as above The reaction is: NCl (l) N (g) + Cl (g) The decomposition of all the NCl means that the final pressure must be due to the N and the Cl. The product gases are present at a 1: ratio, and the total moles are 4. a) Partial pressure N = P nitrogen = X nitrogen P total = (1 mol N / 4 mol total) (754 mmhg) = = 188 mmhg N Partial pressure Cl = P nitrogen = X nitrogen P total = ( mol Cl / 4 mol total) (754 mmhg) = = 566 mmhg Cl b) The mass of NCl may be determined several ways. Using the partial pressure of Cl gives: PV = mrt / M ( mmhg Cl )(.50 L)( 10.6 g/ mol) 1atm mol NCl m = PV M / RT = 760 mmhg mol Cl (( 7 95) + = = 4.94 g NCl 5.99 The reaction is: NH 4 NO (s) N (g) + O (g) + 4 H O(g) 10 g 1molNH4NO 7 mol Gas Moles gas = ( 15.0 kg NH4NO) 1 kg g NH4NO mol NH4NO = mol gas (unrounded) ( mol Gas) (( 7 07) + V = nrt / P = = x 10 4 =.1 x 10 4 L 1.00 atm The balanced equation is: I (aq) + S O (aq) I (aq) + S 4 O 6 (aq) 10 L mol I Initial moles of I = ( 0.00 ml) 1mL L =.04 x 10 4 mol I initial Moles I reacting with S O 10 L mol SO 1molI = ( 11.7 ml) 1mL L molso = x 10 5 mol I reacting with S O (not reacting with SO ) Moles I reacting with SO =.04 x 10 4 mol x 10 5 mol = x 10 4 mol I reacted with SO (unrounded) The balanced equation is: SO (aq) + I (aq) + H O(l) HSO 4 (aq) + I (aq) + H + (aq) Moles SO = ( x 10 4 mol I ) (1 mol SO / 1 mol I ) = x 10 4 mol SO (unrounded) ( 700. torr)( 500. ml ) 1atm 10 L Moles air = = mol air (unrounded) 760 torr 1 ml (( 7 + 8) Volume % SO = mol % SO = [( x 10 4 mol SO ) / ( mol air)] x 100% = = 0.797% 5-0

21 5.101 a) The balanced equation for the reaction is Ni(s) + 4 CO(g) Ni(CO) 4 (g). Although the problem states that Ni is impure, you can assume the impurity is unreactive and thus not include it in the reaction. The mass of Ni can be calculated from the stoichiometric relationship between moles of CO (using the ideal gas equation) and Ni. Mass Ni = mol Ni x molar mass ( kpa)(.55 m ) 1atm 1L 1molNi 58.69gNi Mass Ni = L atm kpa 10 m 4 mol CO 1mol Ni (( 7 50) + = = 1.95 x 10 g Ni b) Assume the volume is 1 m. Use the ideal gas equation to solve for moles of Ni(CO) 4, which equals the moles of Ni, and convert moles to grams using the molar mass. 1 atm m 1L 1molNi 58.69g Ni Mass Ni = L atm 10 m 1molNi(CO) 4 1molNi (( 7 155) + = x 10 4 =.5 x 10 4 g Ni The pressure limits the significant figures. c) The amount of CO needed to form Ni(CO) 4 is the same amount that is released on decomposition. The vapor pressure of water at 5 C (4. torr) must be accounted for (see Table 5.). Use the ideal gas equation to calculate the volume of CO, and compare this to the volume of Ni(CO) 4. The mass of Ni from a m (part b) can be used to calculate the amount of CO released. P CO = P total - P water = 769 torr - 4. torr = 76.8 torr (unrounded) V = nrt / P = 4 1 mol Ni 4 mol CO L atm x 10 g Ni (( 7 5) g Ni 1 mol Ni torr 10 m 76.8 torr 1 atm 1 L = = 6 m CO The answer is limited to two significant figures because the mass of Ni comes from part b Assume 100 grams of sample, thus.01% Si is equivalent to.01 grams Si, and 66.99% F is equivalent to grams of F. Moles Si = (.01 g Si) (1 mol Si / 8.09 g Si) = mol Si (unrounded) Moles F = (66.99 g F) (1 mol F / g F) = mol F (unrounded) Dividing by the smaller value ( ) gives an empirical formula of SiF (empirical formula mass = 85 g/mol). (.60 g) (( 7 7) + M = mrt / PV = = g/mol (unrounded) ( 1.50 atm)( 0.50 L) The molar mass is twice the empirical formula mass, so the molecular formula must be twice the empirical formula, or x SiF = Si F a) A preliminary equation for this reaction is 4 C x H y N z + n O 4 CO + N + 10 H O. Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To form 4 volumes of CO would require 4 volumes of O and to form 10 volumes of H O would require 5 volumes of O. Thus, 9 volumes of O was required. b) Since the volume of a gas is proportional to the number of moles of the gas we can equate volume and moles. From a volume ratio of 4 CO : N :10 H O we deduce a mole ratio of 4 C:4 N:0 H or 1 C:1 N:5 H for an empirical formula of CH 5 N. 5-1

22 5.104 The final pressure will be the sum of the air pressure (0.980 atm) and the pressure generated by the carbon dioxide. 1molCO 10.0 g CO ( ) K g CO + Carbon dioxide pressure = nrt / V = ( L) = atm (unrounded) P total = (0.980 atm) + ( atm) = = 0. atm a) x 10 6 blue particles and x 10 6 black particles b) x 10 6 blue particles and x 10 6 black particles c) Final pressure in C = (/) (750 torr) = 500 torr d) Final pressure in B = (/) (750 torr) = 500 torr At this altitude, the atmosphere is very thin. The collision frequency is very low and thus there is very little transfer of kinetic energy. Satellites and astronauts would not become hot in the usual sense of the word To find the factor by which a diver s lungs would expand, find the factor by which P changes from 15 ft to the surface, and apply Boyle s Law. To find that factor, calculate P seawater at 15 ft by converting the given depth from ft-seawater to mmhg to atm and adding the surface pressure (1.00 atm). P (H O) = (15 ft) (1 in / ft) (.54 cm / 1 in) (10 m / 1 cm) (1 mm / 10 m) =.81 x 10 4 mmh O P (Hg) = (.81 x 10 4 mmh O) [(1.04 g H O/mL) / (1.5 g Hg /ml)] (1 atm / 760 mmhg) = atm (unrounded) P total = (1.00 atm) + ( atm) = atm (unrounded) Use Boyle s Law to find the volume change of the diver s lungs: V f = V i P i / P f = V i ( atm) / (1.00 atm) = = 4.86 V i To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and the pressure at 15 ft, P 15, to find the safest ascended pressure, P safe. P 15 / P safe = 1.5 P safe = P 15 / 1.5 = ( atm) / 1.5 =.415 atm (unrounded) Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance from the initial depth to find how far the diver could ascend. P (H O) = [( )atm] x (760 mmhg / 1 atm) x [(1.5 g Hg/mL) / (1.04 g H O/mL)] x (10 m / 1 mm) x (1.094 yd / 1 m) x ( ft / 1 yd) = ft (unrounded) Therefore, the diver can safely ascend 5.5 ft to a depth of ( ) = = 7 ft The balanced equation is: CaF (s) + H SO 4 (aq) HF(g) + CaSO 4 (s) ( 875 torr)( 8.6 L) 1atm T = PV / nr = 1molCaF 760 torr molhf L atm 15.0 g CaF g CaF 1 mol CaF = K The gas must be heated to 15 K. 5-

23 5.109 First, write the balanced equation: H O (aq) H O(l) + O (g) According to the description in the problem, a given volume of peroxide solution (0.100 L) will release a certain number of volumes of oxygen gas (0). Assume that 0 is exact L solution will produce (0 x L) =.00 L O gas You can convert volume of O (g) to moles of gas using the ideal gas equation. ( 1.00 atm)(.00 L) Moles O = PV / RT = = 8.97 x 10 mol O (unrounded) ( 7 The grams of hydrogen peroxide can now be solved from the stoichiometry using the balanced chemical equation. Mass H O = (8.97 x 10 mol O ) ( mol H O / 1 mol O ) (4.0 g H O / 1 mol H O ) = = 6.07 g H O The moles may be found using the ideal gas equation. Moles times Avogadro s number gives the molecules. 8 ( 10 mmhg)( 1 ml) 1atm 10 L 6.0 x 10 molecules Molecules = = 760 mmhg 1 ml mol ( 500 = x 10 8 = 10 8 molecules (The 10 8 mmhg limits the significant figures.) 8.14J/mol K 7K 10 g kg m /s a) u rms = = = 461 m/s.00 g/mol 1 kg J b) Collision frequency = (u rms / mean free path) = ( m/s) / (6. x 10 8 m) = 7.87 x 10 9 = 7.9 x 10 9 s Molar volume uses exactly one mole of gas along with the temperature and pressure given in the problem ( 70. V / n = RT / P = = = 0.67 L/mol 90 atm 5.11 vacuum with N (74.0 cm) (66.0 cm) The diagram above describes the two Hg height levels within the barometer. To find the mass of N, find P and V (T given) and use the ideal gas equation. The P is directly related to the change in column height of Hg. The N volume of the space occupied by the N (g) is calculated from the dimensions of the barometer. Pressure of the nitrogen = ( ) cmhg (10 m / 1 cm) (1 mm / 10 m) (1 atm / 760 mmhg) = atm (unrounded) (Notice that the subtraction limits the results to two significant figures.) Volume of the nitrogen = (1.00 x ) cm (1.0 cm ) (1 ml / 1 cm ) (10 L / 1 ml) = = L (unrounded) The moles of nitrogen may be found using the ideal gas equations, and the moles times the molar mass of nitrogen will give the mass of nitrogen. ( atm)( L) 8.0 g N Mass of nitrogen = 1mol (( 7 4) + = 4.95 x 10 = 4.94 x 10 g N 5-

24 5.114 Molar concentration is moles per liter. Use the original volume (10.0 L) to find the moles of gas, and the final volume (0.750 L) to get the molar concentration. ( 75 torr)( 10.0 L) 1atm Moles NH = PV / RT = 760 torr (( 7 1) + = mol (unrounded) Molar concentration = mol / volume, in liters, of solution. M = ( mol) / (0.750 L) = = M Determine the total moles of gas produced. The total moles times the fraction of each gas gives the moles of that gas which may be converted to metric tons atm ( 1.5 x 10 m /d) 1 L 65.5 d Moles total = n = PV / RT = L atm 10 m 1y ( 98 =.95 x 10 7 mol/year (unrounded) Mass CO = (0.4896) (.95 x 10 7 mol/year) (44.01 g CO /mol) (1 kg/10 g) (1 T/10 kg) = = 4.8 x 10 T CO /y Mass CO = (0.0146) (.95 x 10 7 mol/year) (8.01 g CO/mol) (1 kg/10 g) (1 T/10 kg) = = 9.16 x 10 T CO/y Mass H O = (0.710) (.95 x 10 7 mol/year) (18.0 g H O/mol) (1 kg/10 g) (1 T/10 kg) = = 1.50 x 10 T H O/y Mass SO = (0.1185) (.95 x 10 7 mol/year) (64.07 g SO /mol) (1 kg/10 g) (1 T/10 kg) = = 1.70 x 10 T SO /y Mass S = (0.000) (.95 x 10 7 mol/year) (64.14 g S /mol) (1 kg/10 g) (1 T/10 kg) = = 4 x 10 1 T S /y Mass H = (0.0047) (.95 x 10 7 mol/year) (.016 g H /mol) (1 kg/10 g) (1 T/10 kg) = =.1 x 10 1 T H /y Mass HCl = (0.0008) (.95 x 10 7 mol/year) (6.46 g HCl/mol) (1 kg/10 g) (1 T/10 kg) = = 6 x 10 1 T CO /y Mass H S = (0.000) (.95 x 10 7 mol/year) (4.09 g H S/mol) (1 kg/10 g) (1 T/10 kg) = = x 10 1 T CO /y The balanced chemical equation is: H (g) + O (g) H O(l) Moles H = (8.0 mol H O) ( mol H / mol H O) = 8.0 mol H Moles O = (8.0 mol H O) (1 mol O / mol H O) = 14.0 mol O a) P = nrt / V ( 8.0 mol H ) (( 7..8) + P H = = 4.17 = 4.1 atm H 0.0 L P O = 14.0 mol O K ( + ) ( 0.0 L) = = 17.1 atm O 5-4