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HW8answers - Cheung Anthony Homework 8 Due Nov 1 2006...

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Cheung, Anthony – Homework 8 – Due: Nov 1 2006, midnight – Inst: McCord 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Toluene (C 6 H 5 CH 3 ) is a liquid compound similar to benzene (C 6 H 6 ). Calculate the mole fraction of toluene in the solution that contains 99.3 g toluene and 130 g benzene. Correct answer: 0 . 393 . Explanation: m toluene = 99.3 g m benzene = 130 g n toulene = (99 . 3 g toluene) 1 mol 92 . 14 g · = 1 . 08 mol n benzene = (130 g benzene) 1 mol 78 . 11 g · = 1 . 66 mol The total number of moles of all species present is 1 . 08 mol + 1 . 66 mol = 2 . 74 mol The mole fraction of toluene is then X toluene = n toluene n total = 1 . 08 mol 2 . 74 mol = 0 . 393 002 (part 1 of 1) 10 points We mix 51 grams of oxygen gas with 59 grams of argon gas in a volume of 931 mL at 164 C. What will be the final pressure of the gas mixture? Correct answer: 118 . 402 atm. Explanation: n O 2 = 51 grams · mol 32 g = 1 . 59375 mol n Ar = 59 grams · mol 39 . 948 g = 1 . 47692 mol n total = 3 . 07067 mol V = 931 mL = 0 . 931 L T = 164 C = 437 K P V = n R T P = n R T V = (3 . 07067 mol) ( 0 . 08206 L · atm mol · K ) (437 K) 0 . 931 L = 118 . 402 atm 003 (part 1 of 1) 10 points What is the density of nitrogen gas at STP? 1. 4.00 g/L 2. 1.25 g/L correct 3. 0.625 g/L 4. 2.50 g/L 5. 0.500 g/L Explanation: 004 (part 1 of 3) 10 points A chemist has synthesized a greenish-yellow gaseous compound that contains only chlorine and oxygen and has a density of 7.71 g/L at 36.0 degrees Celsius and 2188.8 mm Hg. What is the molar mass of the compound? 1. 25.8 g/mol 2. 67.9 g/mol correct 3. 86.9 g/mol 4. 109 g/mol 5. 51.5 g/mol Explanation: T = 36 C + 273 = 309 K P = 2188 . 8 mm Hg atm 760 mm Hg = 2 . 88 atm P V = n R T n V = P R T = (2 . 88 atm) (309 K)(0 . 08206 L · atm molK ) = 0 . 1136 mol L MW = 7 . 71 g / L 0 . 1136 mol L = 67 . 87 g / mol 005
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