exam2a - Thomas, Phillip Exam 2 Due: Nov 7 2007, 11:00 pm...

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Thomas, Phillip – Exam 2 – Due: Nov 7 2007, 11:00 pm – Inst: Brodbelt 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Brodbelt CH301 Exam 2 November 7, 2007 This exam has 31 problems. Programmable calculators and cell phones are prohibited. Bubble in your correct EID and version code! 001 (part 1 oF 1) 10 points IF a sample oF gas is heated From 100 C to 300 C at constant pressure, the volume will ? by a Factor oF ? . 1. increase; about 1.5 correct 2. increase; about three 3. decrease; about three 4. No other answer is correct, because the volume cannot change when the pressure is constant. 5. decrease; about 1.5 Explanation: T 1 = 100 C + 273 = 373 K T 2 = 300 C + 273 = 573 K The temperature increases by a Factor oF 573 373 = 1 . 5. According to Charles’ Law, the volume oF the gas would increase by the same Factor. 002 (part 1 oF 1) 10 points Give the bond angle surrounding each central atom: nitrogen, middle carbon, right carbon. H N C HO C H H H 1. 109.5,109.5,109.5 2. 120,120,120 3. 120,180,120 4. 180,120,109.5 5. 120,90,109.5 6. 120,120,109.5 correct 7. 120,109.5,109.5 8. 180,109.5,109.5 Explanation: 003 (part 1 oF 1) 10 points Consider 2Al + 6HCl 2AlCl 3 + 3H 2 , the reaction oF Al with HCl to produce hy- drogen gas. What is the pressure oF H 2 iF the hydrogen gas collected occupies 14 L at 300 K and was produced upon reaction oF 4 . 5 mol oF Al? Assume the yield oF the reaction is 65.0 percent. Assume the volume, tempera- ture, moles, and pressure are actually known to three signfcant fgures. 1. 0.233 atm 2. 7.72 atm correct 3. 5.28 atm 4. 0.0763 atm 5. 8.91 atm 6. 11.9 atm Explanation: V = 14 L T = 300 K n H 2 = 4 . 5 mol The 65% yield will be used at the END oF the problem. Now we calculate the number oF moles (100% yield) oF H 2 produced: ?mol H 2 = 4 . 5 mol Al × 3mol H 2 2mol Al = 6 . 75 mol H 2
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Thomas, Phillip – Exam 2 – Due: Nov 7 2007, 11:00 pm – Inst: Brodbelt 2 P V = n R T P H 2 = n H 2 R T V = (6 . 75 mol) ( 0 . 08206 L · atm mol · K ) (300 K) 14 L = 11 . 8694 atm(100%yield) = 11 . 87 × 65% = 7 . 72 atm 004 (part 1 of 1) 10 points What is the hybridization at the atom indi- cated by the arrow in the following molecule? H H H H C O · ·· · O · · ·· H O · · ·· C O · · · · C H H H Acetylsalicylic acid (Aspirin) 1. sp 3 d 2 2. sp 3 3. sp 2 correct 4. sp 3 d 5. sp Explanation: 005 (part 1 of 1) 10 points Use VSEPR theory to predict the molecular shape of the ion NO - 3 . 1. trigonal-bipyramidal 2. octahedral 3. trigonal-pyramidal 4. linear 5. bent or angular 6. tetrahedral 7. None of these 8. trigonal-planar correct Explanation: geometry of NO - 3 = ? The Lewis structure for NO
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This note was uploaded on 06/07/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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exam2a - Thomas, Phillip Exam 2 Due: Nov 7 2007, 11:00 pm...

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