AP Equations - 2000 3WBP The College Board Advanced...

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Unformatted text preview: 2000 3WBP The College Board Advanced Placement Examination CHEMISTRY SECTION II This green insert may be used for reference and/or scratchwork as you answer the free-response questions, but be sure to show all your work and your answers in the [Link booklet. N 0 credit will be given for work shown on this green insert. Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce 91111 the questions in this green insert, 85 33 $3 98 SE :8 5.3 SE 5.5 35 8.5 8.3“ :28 at“: me «2 z: 2: 8 Ma 8 8 3 S 8 8 3 cm SE 3a: 8%: $22 8%. OS: 8.3 2.5 SE «52 5: fie: 3.9: 2.3. aaaaaaaaamaannggg E E % 8 S 8 no we 8 S s 8 an an Es a: $~ 63 @u see Amos ace cos 8.5 8.02 3% .am.....aamn a: _: o: S. me SE 2: m2 2: 3 mm a fis 65 88 3.8“ 28 was" $8” 862 made S2 32 a9: 3.9: 3.9: a}: 3.3 2.5 8.3 maaaaamaauamaaaana on 3 a mm a a on E w» t. 3 m» E , fl Nb R on mm 35 3.02 8.5 35 En: N3: :A: 52.2 «$2 8.8. :2 as $3 3.? «a; 3.3 No.5 3.3 n-aaaaaaaaa.-aaa.. «m mm mm 3 on a. a. s. 3 Q 3 S we 3. 3. an am pm 95 8.3 82. «a: an? 9% 38 2.8 3% 8% 3.3 98% 8% 3.8 8.: 3.3 8.9. 2.3 naaaaaamaaaaanaann 3. mm 3“ mm mm 8 on ma g R on 2 S 8 «a a on a mag m3? 8% «8.8 3% 38 on: 3% Emma. E h o; m E .2 . d : $2 8.2 8.2 83. Ed :fi: mad 3% «Z Z O m on a b o w v m 2.00.— wHZH—Zm—Am— HEB “HO mama? Una—Canaan 3E“: an» 320 EXAMINATION. m E. TION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 3-5 MAY BE USEFUL IN ANSWERING .MOOM— Envy—h EU<EHD $02 On— THE QUESTIONS IN THIS SECTION OF THE INFO GO ON TO THE NEXT PAGE. STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25°C Half-reaction ng“ + 2 e- Ag" + e“ Hg2+ + 2 e- 2 Hg2+ + 2 e" Br2(l) + 2 e- 02(g) + 4 H+ + 4 e- 2 H200) 2 C1“ ._) ._) -—) —> _) —> ._) —). a —> .—) _) *—> __, __, _) _, _, ._) —> —> —> —> ——> —> _—> _) _) _.) _) _, _, _) ~—> __> _) _) _) .9 —-> GO ON TO THE NEXT PAGE. ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANT S ATOMIC STRUCTURE AB = hv I E = Av E = energy h v = frequency 1‘ = 7'; k = wavelength p = mv p = momentum ‘ -2.178 x 10'18 . v = velocity E" = n2 Joule n = principal quantum number m = mass EQUILIBRIUM . Speed of light, c = 3.0 x 108 m s'1 K = [IN [A'] , _,4 a [HA] Planck s constant, h = 6.63 x 10 I s Kb = [OH—1 [HB+] Boltzmann's constant, k = 1.38 x 10"23 J K’1 [B] . _ 23 —1 Kw = [OH-1 [Ht] = 1.0 x 10-14 @ 25,, C Avogadro s number — 6.022 X 10 molecules mol = X“ X K,, Electron charge, e = —1.602 x 10'19 coulomb _ + _ . — pH ‘ _ log [H 1’ POH " _1°g[OH V] l electron volt per atom = 96.5 kJ mol—1 14 : pH + pOH pH = pKa + log + Equilibrium Constants [HB ] ——————— pOH = pr + log Ka (weak acid) Kb (weak base) Kw (water) K P (gas pressure) [Bl pKfl = —log K“, pr = —log Kb A KP = KC(RT) ", where An = moles product gas '— moles reactant gas K (molar concentrations) 0 . S° = standard entropy THERMOCHEMISTRY o H = standard enthalpy AS° = 2 S ° Products —2 S" reactants 0° = Standard free energy AH" = 2 AH): products -2 AH; reactants E ° = standard reduction potential T = temperature AG° = 2 AG; products —2 AG; reactants n = moles AG“ = AH° - TAS" m = mass 4 = heat = —RT 1“ K = 4303 RT log K c = specific heat capacity = _n g 5° C p = molar heat capacity at constant pressure AG = AGO + In Q ___ AGO + log Q 1 faraday g; = coulombs q = chT I CF = A”. . AT GO ON TO THE NEXT PAGE. GASES, LIQUIDS, AND SOLUTIONS PV = nRT 2 e . P = pressure (P + ill—20) (V - nb) = nRT V = volume 1 A T = temperature ‘ ~ mo es PA = Pm, x X A, where X A = m n =.numl3er of moles PM, = PA + P, + PC + ' D = denSIty _ m m = mass n — 3!- v = velocity-~ K = °C + 273 PIVI = PZVZ umm = root-mean-square speed 7} T2 KE = kinetic energy D = g r = rate of effusion M = molar mass "ms _ fl = fl 1t = osmoti m M _ I c pressure KE per molecule = l m02 1 '= van thHOff fawn . . 2 K f = molal freezmg-pomt depressmn constant KE per mole = 3- RT Kb = molal boiling—point elevation constant r M Q = reaction quotient % = F: I = current (amperes) molarity, M = moles solute per liter solution q = Charge (comombs) t = time (seconds) molality = moles solute per kilogram solvent AT! .—_ in x molality E° = standard reduction potential K = equilibrium constant ATb = iKb x molality Gas constant, R = 8.31 J mol‘1 K"l = 0.0821 L atm 11101" K‘1 = 8.31 volt coulomb mol"l K OXIDATION-REDUCTION; ELECTROCHEMISTRY Boltzmann's constant, k = 1.33 x 10-23 J K“ K, for H20 = 1.86 K kg mol’l Kb for H20 = 0.512 Kkg morl -1. c d Q=w, whereaA+bB—>cC+dD a b [A] [B] STP = 0.000°c and 1.000 atm 1 = —q— , Faraday’s constant, % = 96,500 coulombs per mole - t RT 0 0592 of electrons Ecell = gen _ —1“ Q = Egell ‘ I 10% Q @ 25°C 119; n ‘ log K = "E 0.0592 GO ON TO THE NEXT PAGE. ...
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This note was uploaded on 12/07/2011 for the course CHEM 100 taught by Professor Feebeck during the Fall '10 term at Purdue University-West Lafayette.

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AP Equations - 2000 3WBP The College Board Advanced...

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