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

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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 – find 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 2 Al + 6 HCl 2 AlCl 3 + 3 H 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 signficant figures. 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 × 3 mol H 2 2 mol 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 - 3 is N O O O - or N O O O - The three areas of high electron density make the molecular geometry trigonal-planar. N O O O 006 (part 1 of 1) 10 points A 6.35 L sample of carbon monoxide is col- lected at 55.0 C and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0 C?
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