Chem301 - Exam 3 - Version 218 Exam 3 mccord (50970) 1 This...

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Unformatted text preview: Version 218 Exam 3 mccord (50970) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. McCord CH301 This exam is only for Dr. McCords CH301 classes. 001 10.0 points Which of the following can form intermolecu- lar hydrogen bonds? 1. PH 3 2. (CH 3 ) 2 NH correct 3. CH 3 COCH 3 4. H 2 CO Explanation: Only molecules with H attached to the electronegative atoms N, O, and F can hy- drogen bond. Of the molecules given only (CH 3 ) 2 NH 2 has the H directly bonded to N, so it can undergo hydrogen bonding. 002 10.0 points A liquid in a test tube has a curved surface such that the edges touching the glass are higher than the surface at the center. This must mean that the cohesive forces are less than the adhesive forces. 1. True correct 2. False Explanation: This is the typical meniscus shape of wa- ter in glass - a concave shape. The liquid tries to maximize its contact with the glass because the adhesive forces are stronger than the cohesive forces. 003 10.0 points For a substance that remains a gas under the conditions listed, deviation from the Ideal Gas Law would be most pronounced at 1.- 100 C and 2.0 atm. 2. 100 C and 2.0 atm. 3. 100 C and 4.0 atm. 4.- 100 C and 4.0 atm. correct 5. C and 2.0 atm. Explanation: Deviations from the Ideal Gas Law are most pronounced at the boundaries of gas forma- tion: low temperature and high pressure. 004 10.0 points The density of the vapor of allicin, a compo- nent of garlic, is 1 . 14 g L 1 at 125 C and 175 Torr. What is the molar mass of allicin? 1. 273 g mol 1 2. 162 g mol 1 correct 3. 50 . 8 g mol 1 4. 869 g mol 1 5. 21 . 6 g mol 1 Explanation: T = 125 C + 273.15 K = 398 . 15 K P = (175 Torr) 1 atm 760 Torr = 0 . 230263 atm = 1 . 14 g / L The ideal gas law is P V = n R T n V = P R T with unit of measure mol/L on each side. Multiplying each by molar mass (MM) gives n V MM = P R T MM = , with units of g/L. Version 218 Exam 3 mccord (50970) 2 MM = RT P = (1 . 14 g / L) ( . 08206 L atm mol K ) . 230263 atm (398 . 15 K) = 161 . 755 g / mol 005 10.0 points A 750.0 mL metal bulb is filled with 0.421 g of CH 4 and an unknown mass of NH 3 . If the total pressure in the bulb is 3.77 atm at 315 C, then how much NH 3 is also in the bulb? 1. 0.0323 grams 2. 2.39 grams 3. 0.549 grams correct 4. 0.00713 grams 5. 14.5 grams 6. 1.17 grams 7. 0.996 grams Explanation: n CH 4 = (0 . 421 g) mol 16 g = 0 . 0263 mol V = 750 mL = 0 . 75 L T = 315 C + 273 = 588 K Applying the ideal gas law equation, P V = n R T n tot = P V R T = (3 . 77 atm) (0 . 75 L) (0 . 08206 L atm mol K ) (588 K) = 0 . 0585995 mol n tot = n NH 3 + n CH 4 n NH 3 = n tot- n CH 4 = 0 . 0585995 mol- . 0263 mol = 0 . 0322995 mol m NH 3 = n NH e MM NH 3 = (0 . 0322995 mol) 17 g NH 3 1 mol NH 3 = 0 . 549091 g NH 3 006 10.0 points The DOMINANT intermolecular force that...
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This note was uploaded on 12/05/2011 for the course CHEM 301 taught by Professor Wandelt during the Fall '08 term at University of Texas at Austin.

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Chem301 - Exam 3 - Version 218 Exam 3 mccord (50970) 1 This...

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