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Unformatted text preview: Honors General Chemistry II (CHM 2046H) Potentially useful information, Exam III
1 amu = 1.6605 ×1024 g e = 1.602 × 1019 C
°C = 5/9 (°F – 32)
°F = 9/5 °C + 32
K = °C + 273.15
23
22
NA = 6.022×10
1 J = 1 kg·m /s
1 Pa = 1 N/m2
1 cal = 4.184 J (exactly)
1 atm = 101.325 kPa = 760 mm Hg (exactly) = 760 torr (exactly) = 29.921 in Hg = 1.01325 bar
PV = nRT R = 0.08206 q = s·m·∆T L ⋅ atm
J
= 8.314
mol ⋅ K
mol ⋅ K
ln P = − ∆ H vap
RT P1 = X1 Ptotal Standard molar volume at STP is 22.41 L ∆H vap 1 1 P −
ln 2 = −
P R T2 T1 1 +C ∆Tb = Kb m ∆Tf = Kf m A = e ln A = ln(e A ) A
ln = ln A − ln B B ln A x = x ln A [ A]0 − [ A]t = kt ln PA = XA P° A k = Ae − Ea / RT
pH = log[H3O+] [ A]t
= − kt
[ A]0 t1/ 2 = ( a − b) 2 = a 2 − 2ab + b 2
pOH = log[OH] x= pKa = log Ka Percent ionization = [H+]equilibrium/[HA]initial
S = k ln W − b ± b 2 − 4ac
2a 0.693
k π = MRT e ( A− B ) eA
=B
e 1
1
−
= kt
[ A]t [ A]0 are the solutions for ax2 + bx + c = 0. Kw = 1.0 × 1014 at 25 °C pH = pKa + log k = 1.381 × 1023 J/K S g = k Pg Ka × Kb = Kw [conjugate _ base]
[acid ] ∆S = qrev / T ∆Suniv = ∆Ssys + ∆Ssurr ∆S°rxn = Σ nS°(products) – Σ mS°(reactants) ∆H°rxn = Σ n∆Hf°(products) – Σ m∆Hf °(reactants) ∆G = ∆H – T∆S ∆G°rxn = Σ n∆Gf°(products) – Σ m∆Gf °(reactants) ∆G = ∆G° + RTlnQ ∆G° = RTln K ∆G = wmax E = Eored(cathode) – Eored(anode) Do not turn or remove this page until you are told to begin the exam.
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 Fall '11
 Lufaso
 Chemistry

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