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.
Each problem has ONE BEST ANSWER. Bubble in your answers on the Green Scantron form. Draw
a box around the answer for the other questions. No extra time will be given to transfer answers to
the Scantron sheet. You may remove the top sheet of the exam after you have been given
instructions to begin the exam. Use correct significant figures, include units, and show all work to
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 Fall '11
 Lufaso
 Chemistry, Mole, 1 kg, Scantron, 1 Pa

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