Quiz3KEY

Quiz3KEY - Chem 6A 2011 (Sailor) Name: Student ID Number:...

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Chem 6A 2011 (Sailor) QUIZ #3 Name: VERSION A KEY Student ID Number: Section Number: Some useful constants and relationships: Specific heat capacities (in J/g . K): H 2 O (l) = 4.184; Al (s) = 0.900; Cu (s) = 0.387; Steel (s) = 0.45 101.325 J = 1 L . atm 1 atm = 760 Torr 1J = 1kg . m 2 /s 2 1 eV = 1.6022 x 10 -19 J R = Ideal gas constant: 0.08206 L . atm . mol -1 . K -1 = 8.31451 J . mol -1 . K -1 Avogadro constant: 6.022 x 10 23 mole -1 Planck's constant = h = 6.6261 x 10 -34 J . s c = speed of light: 3.00 x 10 8 m/s R H = 1.097 x 10 -2 nm -1 C 2 = second radiation constant = 1.44 x 10 -2 K . m T λ max = 1 5 C 2 Emitted power (W) Surface area (m 2 ) = ( cons tan t ) T 4 e = mc 2 c = λν 1 = R H 1 n 1 2 1 n 2 2 Λ Ν Μ Ξ Π Ο E = h ν E = hc E (in Joules) = 2.18 × 10 18 Z 2 n 2 Λ Ν Μ Ξ Π Ο
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Chem 6A 2011 (Sailor) QUIZ #3 1. Write balanced net ionic equations for the following reactions and indicate the states (i.e., (s) if the reaction forms a precipitate, etc.) The first one is done for you as an example. All reactions are carried out in aqueous solution at room temperature: (5 pts each) Reactants Net ionic equation: CuCl 2 (aq) + Li 2 S (aq) Cu 2+ (aq) + S 2- (aq) CuS (s) Sr(NO 3 ) 2 (aq) + KF (aq) Sr 2+ (aq) + 2F - (aq) SrF 2(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states Pb(NO 3 ) 2(aq) + K 2 CrO 4(aq) Pb 2+ (aq) + CrO 4 2- (aq) PbCrO 4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states BaCl 2 (aq) + Na 2 SO 4(aq) Ba 2+ (aq) + SO 4 2- (aq) BaSO 4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states H 3 PO 4(aq) + CsOH (aq) H + (aq) + OH - (aq) H 2 O (aq) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states (can have H 2 O (l) or 3 in front of everything) NaI (aq) + CuCl (aq) I - (aq) + Cu + (aq) CuI (s) 3 pts correct balanced reaction, 2 pts correct states 2. Balance the following redox reactions and identify the oxidizing agent and the reducing agent in each. The first one is done for you: (5 pts each) These equations (except for the example) are all part of the Ostwald process to make nitric acid, problem 4.84 Reaction Oxidizing agent Reducing Agent 6 Li + N 2 2 Li 3 N N 2 Li 4 NH 3 + 5 O 2 4 NO + 6 H 2 O 3 pts correct balanced reaction O 2 1 pt NH 3 1 pt 2 NO + 1 O 2 2 NO 2 3 pts correct balanced reaction. OK to leave off the 1 or to use fractions O 2 1 pt NO 1 pt 3 NO 2 + 1 H 2 O 2 HNO 3 + 1 NO 3 pts correct balanced reaction. OK to leave off the 1 or to use fractions NO 2 1 pt NO
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This note was uploaded on 11/16/2011 for the course CHEM 6A taught by Professor Pomeroy during the Spring '08 term at UCSD.

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Quiz3KEY - Chem 6A 2011 (Sailor) Name: Student ID Number:...

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