MT1_key_Sp09 - Chemistry 1A, Spring 2009 KEY Midterm 1,...

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Page 1 of 11 Chemistry 1A, Spring 2009 KEY Midterm 1, Version February 9, 2009 (90 min, closed book) Name:___________________ SID:_____________________ TA Name:________________ There are 22 Multiple choice questions worth 3 points each. There are 3, multi-part short answer questions. For the multiple choice section, fill in the Scantron form AND circle your answer on the exam. Put your written answers in the boxes provided. Full credit cannot be gained for answers outside the boxes provided. The lecture, homework, chemquizzes, discussion or experiment that each question is based upon is listed after the question e.g. [L3, HW 1.13, CQ 7.3] Question Points Score Multiple Choice Section 66 Question 23 12 Question 24 9 Question 25 13 Total 100
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Page 2 of 11 Thermodynamics: G ° = H ° - T S ° H ° = Σ H ° f (products) - Σ H ° f (reactants) S ° = Σ S ° (products) - Σ S ° (reactants) G ° = Σ G ° f (products) - Σ G ° f (reactants) S = k B lnW S = q rev /T E = q + w w = - P ext V for aA + bB cC + dD b a d c B A D C Q ] [ ] [ ] [ ] [ = At equilibrium, Q = K G = G ° + RTln Q G = G° + RTln(a); a = activity = γ P/P° or γ [A]/[A]° G ° = - RTln K G ° = - nF Є º Є = Є º - (RT/nF) lnQ R S T R H K ° + ° = 1 ln T = ik b,f m Π = iMRT P total = P A + P B = X A P A ° + X B P B ° Acid Base: pH = - log[H 3 O + ] pX = - log X ] [ ] [ log HA A pK pH a + = Kinetics: [A] t = [A] 0 e -kt ln[A] t = ln[A] 0 – kt t 1/2 = ln2/k 1/[A] t = 1/[A] 0 + kt k = A e (-Ea/RT) ln(k 1 /k 2 ) = E a /R ( 1/T 2 – 1/T 1 ) t 1/2 = 1/[A] 0 k t 1/2 = [A] 0 /kt Quantum: E = h ν λν = c λ deBroglie = h / p = h / mv E kin (e-) = h ν - Φ = h ν - h ν 0 = R n Z E n 2 2 x p ~ h p = mv Particle in a box (1-D Quantum): E n = h 2 n 2 /8mL 2 ; n = 1, 2, 3. .. Vibrational: E v = (v + ½) hA/2 π ; A =(k/m) ½ Rotational: E n = n(n + 1) hB; B = h/8 π 2 I; I = 2mr 2 m = m A m B /(m A + m B ) Ideal Gas: PV = nRT RT E kin 2 3 = M 3RT v rms = Constants: N 0 = 6.02214 x 10 23 mol -1 R = 2.179874 x 10 -18 J R = 3.28984 x 10 15 Hz k = 1.38066 x 10 -23 J K -1 h = 6.62608 x 10 -34 J s m e = 9.101939 x 10 -31 kg c = 2.99792 x 10 8 m s -1 T (K) = T (C) + 273.15 F = 96,485 C / mol 1 V = 1 J / C Gas Constant: R = 8.31451 J K -1 mol -1 R = 8.20578 x 10 -2 L atm K -1 mol -1 1 nm = 10 -9 m 1 kJ = 1000 J 1 atm = 760 mm Hg = 760 torr 1 bar 1 L atm 100 J
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Page 3 of 11 M ULTIPLE C HOICE The illustration to below represents a mixture of nitrogen (black) and oxygen (grey) molecules. [Disc1] The molecules shown react to form N 2 O 4 according to the equation N 2 (g) + 2O 2 (g) N 2 O 4 (g) 1) Which of the following amounts is conserved in the reaction? A) mass B) number of moles C) number of molecules D) atoms E) both A) and D) 2) The limiting reagent is A) N 2 B) O 2 C) N 2 O 4 D) there is no limiting reagent 3) The number of N 2 O 4 molecules formed is _______. A) 2 B) 4 C) 6 D) 8 4) After the reaction of 2 moles of H 2 and 2 moles of O 2 to form H 2 O, which species has the greater number of moles? [CQ 2.1] A) H 2 B) O 2 C) H 2 O D) they all have an equal number of moles after the reaction
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Page 4 of 11 5) One method for reducing the lead content in used motor oil is to add ammonium phosphate, (NH 4
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MT1_key_Sp09 - Chemistry 1A, Spring 2009 KEY Midterm 1,...

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