Spring 2009 - Nitsche - Final (solution)

Spring 2009 - Nitsche - Final (solution) - Chemistry 1A,...

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Page 1 of 17 Chemistry 1A, Spring 2009 Final Exam May 16, 2009 (180 min, closed book) Name:___________________ SID:_____________________ TA Name:________________ x There are 40 Multiple choice questions worth 6 points each. x There are 3, multi-part short answer questions. x For the multiple choice section, fill in the Scantron form AND circle your answer on the exam. x Put your written answers in the boxes provided. Full credit cannot be gained for answers outside the boxes provided. x The lecture, homework, chemquizzes, discussion or experiment that each question is based upon is listed after the question e.g. [L3, HW 1.13, CQ 7.3] Question Points Score Multiple Choice Section 240 Question 41 Buffer 18 Question 42 Thermo. 22 Question 43 Electrochem. 20 Total 300
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Page 2 of 17 Thermodynamics: ' G q = ' H q - T ' S q ' H q = 6 ' H q f (products) - 6 ' H q f (reactants) ' S q = 6 S q (products) - 6 S q (reactants) ' G q = 6 ' G q f (products) - 6 ' G q 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 q + RTln Q * *± ² 57OQ³D´µ D DFWLYLW\ Ȗ3¶3± RU Ȗ>$@¶>$@± ' G q = - RTln K ' G q = - nF ' ȯ º ' ȯ = ' ȯ º - (RT/nF) lnQ R S T R H K q ' ± q ' ² 1 ln ¨7 LN b,f m Ȇ L057 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 ² ± Decay Kinetics: [A] t = [A] 0 e -kt or N t = N 0 e - O t (in book k = O ) ln( N t /N 0 )=- O t t 1/2 = ln2/k or t 1/2 = ln2/ O Quantum: E = h Q OQ = c O deBroglie = h / p = h / mv E kin (e-) = h Q - ) = h Q - h Q 0 f ² R n Z E n 2 2 ¨[ ¨S a K p = mv Particle in a box (1-D Quantum): E n = h 2 n 2 /8mL 2 ; n = 1, 2, 3. .. Vibrational: E v = (v + ½ ) hA/2 ʌµ $ ³N¶P´ ½ Rotational: E n Q³Q ² ·´ K%µ % K¶¸ʌ 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 1 eV = 1.60218 × 10 -9 J 1Ci = 3.7 × 10 10 disintegrations/sec R f = 3.289 × 10 15 Hz or 2.179 × 10 -18 J k = 1.38066 × 10 -23 J K -1 h = 6.62608 × 10 -34 J s m e = 9.101939 × 10 -31 kg c = 2.99792 × 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 × 10 -2 L atm K -1 mol -1 1 nm = 10 -9 m 1 kJ = 1000 J 1 atm = 760 mm Hg = 760 torr § · EDU 1 L atm § ·¹¹ -
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Page 3 of 17 M ULTIPLE C HOICE 1) After the reaction of 2 moles of H 2 and 2 moles of O 2 to form H 2 A) O, which species has the greater number of moles? [CQ 2.1] H B) 2 O C) 2 H 2 D) O they all have an equal number of moles after the reaction 2) Radon has seventeen isotopes ranging from 210 Rn to 226 A) Rn. What do all of these isotopes have in common? [HW B11] the same number of electrons B) the same number of protons C) the same number of neutrons D) a and b E) b and c 3) In the photoelectric effect, increasing the intensity of the monochromatic light will ____. [L4] A) ensure that electrons will be emitted from all metals. B)
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Spring 2009 - Nitsche - Final (solution) - Chemistry 1A,...

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