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Unformatted text preview: g = 9 . 81 m/s 2 c = 3 × 10 8 m/s d water = 1 . 00 g/mL 1 atm = 1.01325 × 10 5 Pa N A = 6 . 022 × 10 23 mol 1 k = 1 . 381 × 10 23 J/K R = 0 . 08206 L · atm/mol · K R = 8 . 314 J/mol · K R = 62 . 36 L · torr/mol · K 1 L · atm = 101.325 J E = hν c = λ/ν λ = h p = h mv 1 2 mv 2 = hν Φ Tλ max = 1 5 c 2 c 2 = 0 . 0144 K · m ν = R µ 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 , ··· P = dhg PV = nRT P total = P A + P B + P C + ··· x A = P A /P total µ P + a n 2 V 2 ¶ ( V nb ) = nRT v rms = µ 3 RT M ¶ 1 / 2 rate of effusion (or speed) ∝ r T M Δ U = q + w H = U + PV Δ U = q V = C V Δ T = nC V, m Δ T Δ H = q P = C P Δ T = nC P, m Δ T w = P Δ V w = Δ nRT Δ U = Δ H P Δ V Δ U = Δ H Δ nRT Δ H rxn = Δ H 1 + Δ H 2 + Δ H 3 + ... Δ H ◦ rxn = X nB.E. (react) X nB.E. (prod) Δ H ◦ rxn = X n Δ H ◦ f (prod) X n Δ H ◦ f (react) Δ S ◦ rxn = X nS ◦ (prod) X nS ◦ (react) Δ G ◦ rxn = X n Δ G ◦ f (prod) X n Δ G ◦ f (react) Δ H ◦ r ( T 2 ) = Δ H ◦ r ( T 1 ) + Δ C P Δ T Δ C P = X nC P, m (prod) X nC P, m (react) w = nRT ln V 2 V 1 Δ S = nR ln V 2 V 1 Δ S = C V ln T 2 T 1 Δ S = C P ln T 2 T 1 Δ H vaporization ≈ 85 J/mol K d S = d q rev /T S = k ln W G = H TS Δ G = Δ H T Δ S Kolar, Leah – Final 1 – Due: May 10 2007, 1:00 pm – Inst: McCord 2 This printout should have 56 questions. Multiplechoice 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 electrons are in the fourth princi pal shell ( n = 4) of a cadmium (Cd) atom? 1. 10 2. 48 3. 18 correct 4. 12 5. 9 6. 2 Explanation: Cd is in the 10 th column and second row of the transition metals. Therefore Cd has 10 4 d electrons, 2 4 s electrons, and 6 4 p electrons for a total of 18 n = 4 electrons. 002 (part 1 of 1) 10 points The molecular shape of SF 5 is 1. square pyramidal. correct 2. trigonal bipyramidal. 3. tetrahedral. 4. octahedral. 5. square planar. Explanation: 003 (part 1 of 1) 10 points Which of the following valenceshell config urations 4 s 4 p I) ↑↓ ↑ ↑ II) ↑↓ ↑↓ III) ↑ ↑ ↑ ↑ IV) ↑↓ ↑ ↓ is/are possible for a neutral atom for an ele ment? 1. IV only 2. None of the configurations 3. II only 4. I only correct 5. More than one of the configurations 6. III only Explanation: The atom with a 4 s 2 4 p 2 valenceshell con figuration is germanium (Ge). The ground state configuration is given by ↑↓ ↑ ↑ The other configurations represent excited states. 004 (part 1 of 1) 10 points A molecule has two lone nonbonded pairs of electrons on the central atom and four atoms bonded to the central atom. What is its molecular shape and its hybridization?...
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This test prep was uploaded on 03/19/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.
 Spring '07
 Fakhreddine/Lyon
 Chemistry

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