chem1A-sp11-mt1-Nitsche-soln

chem1A-sp11-mt1-Nitsche-soln - Chemistry 1A, Spring 2011...

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Page 1 of 16 Chemistry 1A, Spring 2011 Midterm 1 February 7, 2011 (90 min, closed book) Name:___________________ SID:_____________________ TA Name:________________ There are 40 multiple choice questions worth 3 points each. There is only one correct answer for each question unless otherwise specified in the question. Only answers on the Scantron form will be graded. The exam question sheets will NOT be collected. Use the back side and page margins as scratch paper. The periodic table can be torn off.
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Page 2 of 16 800 600 400 200 IR Red Green Blue UV Wavelength (nm) Color and Wavelength of Light
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Page 3 of 16 Thermodynamics: G ° = H ° - T S ° H ° = Σ H ° f (products) - Σ H ° f (reactants) S ° = Σ S ° (products) - Σ S ° (reactants) G ° = Σ G ° f (products) - Σ G ° f (reactants) S = k B lnW S = q rev /T E = q + w w = - P ext V for aA + bB cC + dD b a d c B A D C Q ] [ ] [ ] [ ] [ = At equilibrium, Q = K G = G ° + RTln Q G = G° + RTln(a); a = activity = γP/P° or γ[A]/[A]° G ° = - RTln K G ° = - nF Є º Є = Є º - (RT/nF) lnQ R S T R H K ° + ° = 1 ln T = ik b,f m Π = iMRT P total = P A + P B = X A P A ° + X B P B ° Acid Base: pH = - log[H 3 O + ] pX = - log X ] [ ] [ log HA A pK pH a + = Kinetics: [A] t = [A] 0 e -kt ln[A] t = ln[A] 0 – kt t 1/2 = ln2/k 1/[A] t = 1/[A] 0 + kt k = A e (-Ea/RT) ln(k 1 /k 2 ) = E a /R ( 1/T 2 – 1/T 1 ) t 1/2 = 1/[A] 0 k t 1/2 = [A] 0 /kt Quantum: E = h ν λν = c λ deBroglie = h / p = h / mv E kin (e-) = h ν - Φ = h ν - h ν 0 = R n Z E n 2 2 x ∆p ~ h p = mv E n = h 2 n 2 /8mL 2 ; n = 1, 2, 3. .. E v = (v + ½) hA/2π; A =(k/m) ½ E n = n(n + 1) hB; B = h/8π 2 I; I = 2mr 2 m = m A m B /(m A + m B ) Ideal Gas: PV = nRT RT E kin 2 3 = M 3RT v rms = Constants: N 0 = 6.02214 x 10 23 mol -1 R = 2.179874 x 10 -18 J R = 3.28984 x 10 15 Hz k = 1.38066 x 10 -23 J K -1 h = 6.62608 x 10 -34 J s m e = 9.101939 x 10 -31 kg c = 2.99792 x 10 8 m s -1 T (K) = T (C) + 273.15 F = 96,485 C / mol 1 V = 1 J / C Gas Constant: R = 8.31451 J K -1 mol -1 R = 8.20578 x 10 -2 L atm K -1 mol -1 1 nm = 10 -9 m 1 kJ = 1000 J 1 atm = 760 mm Hg = 760 torr ≈ 1 bar 1 L atm 100 J
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Page 4 of 16 S ECTION 1: A TOMS , M OLECULES AND M OLES 1. The complete combustion of liquid propyl alcohol (C 3 H 7 OH) in oxygen yields just carbon dioxide and water. When this reaction is balanced using smallest integer coefficients, the coefficient for O 2 A. is 5 B. 9 C. 10 D. 20 E. none of these ANS: B 2. The ratio of the number of bismuth atoms to the number of oxygen atoms in Bi 2 (SO 4 ) 3 A. is 1:6 B. 2:7 C. 2:3 D. 2:1 E. none of these ANS: A 3. A ternary compound contains one atom of scandium and three atoms of nitrogen for every nine atoms of oxygen. A possible formula for the compound is A. Sc(NO 3 ) B. 3 Sc 2 (N 2 O 3 ) C. 3 Sc(NO 2 ) D. 4 Sc(NO 4 ) E. 3 none of these ANS: A 4. Assume that the following reaction between phosphorus and chlorine gas goes as far as possible to give the product, gaseous phosphorus pentachloride. P 4 ( s ) + 10 Cl 2 ( g ) 4 PCl 5 ( g ) If 0.231 mol of P 4 is reacted with an excess of Cl 2 , how much PCl 5 A.
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chem1A-sp11-mt1-Nitsche-soln - Chemistry 1A, Spring 2011...

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