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10 Gases Solutions to Exercises (c) Plan/Solve. qt air L air L O₂ available. mol O₂ available = PV/RT. mol O₂/hr (from part (a)) total mol O₂ consumed. Compare O₂ available and O₂ consumed. 1 qtair X 0.946 L 1qt X 0.21 in air = 0.199 L O₂ available n = 1atm X 0.08206 mol- K atm X 297 K = 8.17 10⁻³ = 8 X 10⁻³ mol O₂ available roach uses 1.71 X 10⁻⁴ mol X 48 hr = 8.21 X 10⁻³ = 8 10⁻³ mol O₂ consumed 1hr Not only does the roach use 20% of the available O₂, it needs all the O₂ in the jar. 10.46 Change mass to kg; 1 hr = 60 min; pay attention to units. (a) 185 lb X 2.2046 1kg lb 47.5 kg mL min O₂ X 60 min = 2.39 10⁵ mL (b) 165 lb 2.2046 1kg 65.0 kg mL min O₂ X 60 min = 2.92 X 10⁵ mL 10.47 (a) Analyze. Given: 119 tons Hg, 1 atm, 298 K. Find: vol Hg(g). Plan. Change tons to grams; use V = gRT/MMxP to calculate volume. Solve. 119 tons X 2000 1 lb X 453.59 = 1.07954 10⁸ = 1.08 g Hg ton V = 1.07954 X 10⁸ g Hg 1 mol Hg X 0.08206 atm 200.59 Hg mol K 1 atm = 1.3161 10⁷ = 1.32 10⁷ L (b) Analyze. Given: total vol atmospheric gases (air) = 51 X 10¹² m³; 245 ppb Hg(g) by volume. Find: mol Hg(g) in atmospheres. Plan. Change m³ to L; use definition of ppb to get L Hg(g); use n = PV/RT to calculate mol Hg. Assume STP. Solve. 51 X 10¹² m³ X 10³ 1m³ dm³ X 1dm³ 1L = 5.1 10¹⁶ L air 5.1 10¹⁶ L air X 245 Lair = 1.2495 10¹⁰ = 1.2 10¹⁰ 1 n = 1 atm X 1.2495 X 10¹⁰ LHg(g) mol K = 5.5775 10⁸ = 5.6 10⁸ mol Hg(g) K 0.08206 atm Check. Note that calculating amount of Hg from volume Hg(g) depends on the the assumed temperature of the atmosphere. 10.48 mass = 1800 g = 1.8 X 10⁻⁶ g; V = 1 m³ = 1 10³ L; T = 273 + 10° = 283 K (a) P = g RT P = 1.8 X 10⁻⁶ g Hg X 1 mol Hg 0.08206 atm 1 x 10³ L 283K MMxV K = 2.1 X 10⁻¹⁰ atm 286