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Unformatted text preview: Version 407 – Exam 3 – McCord – (53110) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. McCord CH301mwf This exam is only for McCord’s MWF CH301 class. PLEASE carefully bubble in your UTEID and Version Number! 001 10.0 points Rank the compounds CH 4 LiF C 3 H 8 NaF in terms of increasing melting point. 1. LiF < NaF < C 3 H 8 < CH 4 2. C 3 H 8 < CH 4 < NaF < LiF 3. C 3 H 8 < CH 4 < LiF < NaF 4. CH 4 < C 3 H 8 < NaF < LiF correct 5. NaF < LiF < C 3 H 8 < CH 4 Explanation: 002 10.0 points Hydrogen bonds are really just a very strong form of dipoledipole interaction. 1. True correct 2. False Explanation: Permanent dipoledipole interactions are stronger than London forces and occur be tween polar covalent molecules due to charge separation. Hbonds are a special case of very strong dipoledipole interactions. They only occur when H is bonded to small, highly electroneg ative atoms – F, O or N only. 003 10.0 points A gas has a volume of 2.00 liters at a tempera ture of 127 ◦ C. What will be the volume of the gas if the temperature is increased to 327 ◦ C? (Assume the pressure remains constant.) 1. 3.00 liters correct 2. 2.00 liters 3. 4.00 liters 4. 6.00 liters Explanation: T 1 = 127 ◦ C + 273 = 400 K V 1 = 2 . 00 L T 2 = 327 ◦ C + 273 = 600 K Charles’ Law relates the volume and ab solute (Kelvin) temperature of a sample of gas: V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (2 . 00 L) (600 K) 400 K = 3 . 00 L 004 10.0 points The density of citronellal, a mosquito repel lant, is 1 . 45 g · L − 1 at 365 ◦ C and 50.0 kPa. What is the molar mass of citronellal? 1. 37 . 5 g · mol − 1 2. 95 . 7 g · mol − 1 3. 88 . 0 g · mol − 1 4. 154 g · mol − 1 correct 5. 73 . 2 g · mol − 1 Explanation: T = 365 ◦ C + 273.15 K = 638 . 15 K P = (50 kPa) 1 atm 101 . 325 kPa = 0 . 493462 atm ρ = 1 . 45 g / L The ideal gas law is P V = nRT n V = P RT Version 407 – Exam 3 – McCord – (53110) 2 with unit of measure mol/L on each side. Multiplying each by molar mass (MM) gives n V · MM = P RT · MM = ρ , with units of g/L. MM = ρ RT P = (1 . 45 g / L) ( . 08206 L · atm mol · K ) . 493462 atm × (638 . 15 K) = 153 . 875 g / mol 005 10.0 points If the compression factor Z is less than one for a given gas, which van der Waals coefficient is the dominant one? 1. Impossible to tell 2. b 3. a correct Explanation: Z factors of less than 1 have dominant at tractive forces. 006 10.0 points When two samples of ideal gases have the same ? , their molecules must have the same ? . 1. temperature; speed 2. pressure; average kinetic energy 3. mass; density 4. mass; average kinetic energy 5. pressure; mass 6. density; average kinetic energy 7. volume; average kinetic energy 8. density; mass 9. temperature; average kinetic energy cor rect 10. volume; mass Explanation: From kinetic molecular theory: U ∝ radicalbigg T MW For samples of an ideal gas temperature and...
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This note was uploaded on 08/30/2010 for the course CH 52385 taught by Professor Vandenbout during the Spring '09 term at University of Texas.
 Spring '09
 Vandenbout

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