Chem. Hwk 10 solutions

Chem. Hwk 10 solutions - Pham Quoc Homework 10 Due Nov 7...

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Pham, Quoc – Homework 10 – Due: Nov 7 2005, midnight – Inst: McCord 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Work problems in the book also. Exam 3 is Wednesday, 11/9, go to the same rooms as last time A-K in HMA L-Z in WEL 1.308 DO bring a calculator this 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 61.5 g toluene and 66.0 g benzene. Correct answer: 0 . 441 . Explanation: m toluene = 61.5 g m benzene = 66.0 g n toulene = (61 . 5 g toluene) 1 mol 92 . 14 g · = 0 . 667 mol n benzene = (66 . 0 g benzene) 1 mol 78 . 11 g · = 0 . 845 mol The total number of moles of all species present is 0 . 667 mol + 0 . 845 mol = 1 . 51 mol The mole fraction of toluene is then X toluene = n toluene n total = 0 . 667 mol 1 . 51 mol = 0 . 441 002 (part 1 of 1) 10 points We mix 51 grams of oxygen gas with 185 grams of argon gas in a volume of 630 mL at 190 C.
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