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Unformatted text preview: Honors General Chemistry II (CHM 2046H)
Potentially useful information, Final Exam
1 in = 2.54 cm (exactly)
e = 1.602 × 1019 C 1 lb = 453.59 g °C = 5/9 (°F – 32) NA = 6.022×1023 1 L = 1.0567 qt °F = 9/5 °C + 32 1 J = 1 kg·m2/s2 1 amu = 1.6605 ×1024 g
K = °C + 273.15 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 L ⋅ atm = 8.314
mol ⋅ K q = s·m·∆T
PA = XA P° A
A = e ln A = ln(e A ) [ A]0 − [ A]t = kt k = Ae − Ea / RT
pH = log[H3O+] J
mol ⋅ K ln P = − ∆ H vap
RT P1 = X1 Ptotal ∆H vap 1 1 P −
ln 2 = −
P R T2 T1 1 +C ∆Tb = Kb m ∆Tf = Kf m A
ln = ln A − ln B B ln A x = x ln A ln [ A]t
= − kt
[ A]0 ( a − b) 2 = a 2 − 2ab + b 2
pOH = log[OH] t1/ 2 = x= k = 1.381 × 1023 J/K 0.693
k − b ± b 2 − 4ac
2a pKa = log Ka Percent ionization = [H+]equilibrium/[HA]initial
S = k ln W Standard molar volume at STP is 22.41 L
S g = k Pg π = MRT
e ( A− B ) = eA
eB 1
1
−
= kt
[ A]t [ A]0 are the solutions for ax2 + bx + c = 0. Kw = 1.0 × 1014 at 25 °C pH = pK a + log 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) F = 96,485 C/mol ∆G = nFE
E = mc2 Avogadro constant = 6.0221421×1023 mol1;
c = 2.99792458 ×108 m s1
mass of electron = 5.48579909x104 amu
molar mass of a proton = 1.007825 amu.
molar mass of a neutron = 1.008665 amu. E = Eo – (RT/nF)ln Q Do not turn this page until you are given the instruction to begin the exam!
Raise your hand if, during testing, you need additional scratch paper. ...
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This note was uploaded on 12/13/2011 for the course CHM 2046h taught by Professor Lufaso during the Fall '11 term at UNF.
 Fall '11
 Lufaso
 Chemistry

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