Final Key - CePrNdPmSmEuGdTbDyHoErTmYbLu...

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Unformatted text preview: CePrNdPmSmEuGdTbDyHoErTmYbLu ThPaUNpPuAmCmBkCfEsFmMdNoLr HHe LiBeBCNOFNe NaMgAlSiPSClAr KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe CsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRn FrRaAcRfDbSgBhHsMt 12 345678910 1112131415161718 192021222324252627282930313233343536 373839404142434445464748495051525354 555657727374757677787980818283848586 878889104105106107108109 5859606162636465666768697071 90919293949596979899100101102103 1.00794.0026 6.9419.012210.81112.01114.006715.999418.998420.1797 22.989824.305026.981528.085530.973832.06635.452739.948 39.098340.07844.955947.8850.941551.996154.938055.84758.933258.6963.54665.3969.72372.6174.921678.9679.90483.80 85.467887.6288.905991.22492.906495.94(98)101.07102.9055106.42107.8682112.411114.82118.710121.75127.60126.9045131.39 132.9054137.327138.9055178.49180.9479183.85186.207190.2192.22195.08196.9665200.59204.3833207.2208.9804(209)(210)(222) (223)(226)(227)(261)(262)(263)(262)(265)(266) 140.115140.9076144.24(145)150.36151.965157.25158.9253162.50164.9303167.26168.9342173.04174.967 232.0381231.0359238.0289(237)(244)(243)(247)(247)(251)(252)(257)(258)(259)(260) 1A8A 2A3A4A5A6A7A 3B4B5B6B7B8B1B2B 118 21314151617 3456789101112 Periodic Table of the Elements Dsouza, Miranda – Final 1 – Due: May 12 2006, 5:00 pm – Inst: David Laude 2 This print-out should have 60 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) 5 points The following data were collected for the re- action 2A + B → C at 25.0 ◦ C. Initial Initial Initial rate Trial [A 2 ] [B] Δ[C]/Δt mol/M mol/M M/min 1 . 1 . 1 4 . × 10- 4 2 . 3 . 3 1 . 2 × 10- 3 3 . 1 . 3 4 . × 10- 4 4 . 2 . 4 8 . × 10- 4 What is the rate-law expression for this reaction? 1. 4 . × 10- 4 /min [B] 2. 3 . × 10- 4 /min [A] 3. 4 . × 10- 3 /min [B] 4. 4 . × 10- 3 /min [A] correct Explanation: Rate a Rate b = k [A] x a [B] y a [A] x a [B] y b Compare trials 1 and 3: Rate 3 Rate 1 = µ [A] 3 [A] 1 ¶ x µ [B] 3 [B] 1 ¶ y 4 . × 10- 4 4 . × 10- 4 = µ . 1 . 1 ¶ x µ . 3 . 1 ¶ y 1 = 3 y y = 0 Compare trials 4 and 1: Rate 4 Rate 1 = µ . 2 . 1 ¶ x 2 = 2 x x = 1 Using trial 1, Rate = k [A][B] = k [A] 4 . × 10- 4 M / min = k (0 . 1 M) k = 0 . 4 × 10- 3 / min . Thus the rate-law expression is Rate = 4 . × 10- 3 / min [A] 002 (part 1 of 1) 5 points For a solution labeled “0.10 M H 2 SO 3 (aq),” p K a1 = 1 . 81 and p K a2 = 6 . 91, which of the following is TRUE? 1. The pH is 0.70. 2. The pH is about 1.5. correct 3. The pH is 1.0. 4. The pH is about 4.4. 5. [H + ] = 0.2 M Explanation: 003 (part 1 of 1) 5 points Which of the following aqueous solutions gives a pH greater than 7?...
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This note was uploaded on 05/02/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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Final Key - CePrNdPmSmEuGdTbDyHoErTmYbLu...

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