Finalvr88 - 1 atm = 1.01325 10 5 Pa 1 bar = 10 5 Pa 1 atm =...

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Unformatted text preview: 1 atm = 1.01325 10 5 Pa 1 bar = 10 5 Pa 1 atm = 14.7 psi sea level = 1 atm 1 A= 10- 10 m 1 L atm = 101.325 J 1 L bar = 100 J R = 0 . 08206 L atm/mol K R = 8 . 314 J/mol K R = 62 . 36 L torr/mol K N A = 6 . 022 10 23 mol- 1 k = 1 . 381 10- 23 J/K E = h c = = h p = h mv 1 2 mv 2 = h- = R 1 n 2 1- 1 n 2 2 R = 3 . 29 10 15 Hz E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , E n =- h R n 2 R = m e e 4 8 h 3 2 P = dhg PV = nRT P total = P A + P B + P C + x A = P A /P total P + a n 2 V 2 ( V- nb ) = nRT rate of effusion (or speed) r T M U = q + w H = U + PV U = q V = C V T = n C V, m T H = q P = C P T = n C P, m T w =- P V w =- nRT U = H- P V U = H- nRT H rxn = H 1 + H 2 + H 3 + ... H rxn = X nB.E. (react)- X nB.E. (prod) H rxn = X n H f (prod)- X n H f (react) S rxn = X nS (prod)- X nS (react) G rxn = X n G f (prod)- X n G f (react) H r ( T 2 ) = H r ( T 1 ) + C P T C P = X nC P, m (prod)- X nC P, m (react) w =- nRT ln V 2 V 1 S = nR ln V 2 V 1 S = C V ln T 2 T 1 S = C P ln T 2 T 1 H vaporization 85 J/mol K d S = d q rev /T S = k ln W G = H- TS G = H- T S Cheung, Anthony Final 1 Due: Dec 18 2006, 6:00 pm Inst: McCord 2 This print-out should have 55 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points How many non-bonding electron pairs are around I in IF- 2 ? 1. 2 2. 4 3. 3 correct 4. 5. 1 Explanation: Iodine has seven valence electrons. Flourine has seven valence electrons. Each flourine atom will form a covalent bond by sharing one of iodines electrons. This leaves iodine with 5 electrons plus the additional electron that gives the molecule a net negative charge. That makes six non-bonding electrons around iodine, or three pairs. I F F 002 (part 1 of 1) 10 points What is the shape of IF + 4 ? 1. trigonal bipyramidal 2. square planar 3. tetrahedral 4. T-shaped 5. seesaw correct Explanation: There are five regions of electron density (including one lone pair) around the central atom: I F F F F + 003 (part 1 of 1) 13 points A metal bulb is filled with 10.2 g of CHCl 3 and 1.67 g of CH 4 (two gases). What is the total pressure in the bulb if the temperature is 345 C and the volume is 50.0 mL? 1. 0.107 atm 2. 12.1 atm 3. 1450 atm 4. There is no pressure because the metal bulb exerts more force than the gas does. 5. 107 atm 6. 12,030 atm 7. 193 atm correct Explanation: n CHCl 3 = 10 . 2 g mol 119 . 5 g = 0 . 0854 mol n CH 4 = 1 . 67 g mol 16 g = 0 . 104 mol n total = 0 . 1898 mol V = 50 mL = 0 . 05 L T = 345 C + 273 . 15 = 618 . 15 K Applying the ideal gas law equation, P V = n R T P = n R T V P = (0 . 1898)(0 . 08206)(618 . 15) . 05 = 192 . 554 atm...
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This note was uploaded on 01/23/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.

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Finalvr88 - 1 atm = 1.01325 10 5 Pa 1 bar = 10 5 Pa 1 atm =...

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