Quiz6KEY

Quiz6KEY - Chem 6A 2011(Sailor Name Student ID Number...

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Chem 6A 2011 (Sailor) QUIZ #6 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 q = C p m Δ T 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 #6 1. The three quantum numbers for an electron in a ruthenium atom in a certain state are n = 4, l = 2, m l = +1, and m s = + ½ . In what type of orbital is the electron located (e.g. 2s, 3p, etc.)? (10 pts) Type of orbital = 4d (10 pts if correct, all or nothing) 2. Give all possible m l values for orbitals that have each of the following: (20 pts) (a) l = 2 ANSWER: -2, -1, 0, 1, 2 (5pts all or nothing for each of these) (b) n = 1 ANSWER: 0 (c) n = 4 ANSWER: -3, -2, -1, 0, 1, 2, 3 (d) l = 3 ANSWER: -3, -2, -1, 0, 1, 2, 3 This was homework problem 7.42 in the book. 3. Photoionization occurs when a photon of light strikes an atom and removes an electron. Use the Rydberg equation to calculate the wavelength of light needed to completely remove the electron from a 2s orbital in a hydrogen atom. Set up but do
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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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Quiz6KEY - Chem 6A 2011(Sailor Name Student ID Number...

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