1307Spring2008Ex3Atest

- Equations E = h c = Palmer-Rydberg Equation 1 1 2 2 n1 n 2 Energy Equation from Bohr model of H E =-2.18 x 10-18 J(1/n2 E = Efinal Einitial = R

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Equations: E = h· ν c = λ · ν Palmer-Rydberg Equation: ) ( 2 2 2 1 n 1 n 1 R 1 = λ Energy Equation from Bohr model of H: E = (-2.18 x 10 -18 J) (1/n 2 ) E = E final - E initial Constants: Plank constant (h): 6.626 x 10 -34 J·s Speed of light (c): 3.00 x 10 8 m/s Rydberg constant (R): R = 1.097 x 10 7 m -1 The space below is designated for scratching purpose.
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MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1. A green laser photon with the wavelength 532 nm has the frequency ____ Hz. A. 5.64 x 10 14 B. 4.55 x 10 13 C. 6.73 x 10 15 D. 7.82 x 10 16 E. 3.46 x 10 12 2. Which is the smallest? A. Mg 2+ B. Na + C. Al 3+ D. F - E. O 2- 3. Which is isoelectronic with Mg 2+ ? A. Ar B. S 2- C. Ca 2+ D. K + E. O 2- 4. What is the electron configuration of argon? A. [Ne]2s 2 2p 6 B. [Ne]3s 2 3p 6 C. [He]2s 2 2p 6 D. [Ne]3p 6 E. [Ne]2s 2 3p 6 5. What group has the general valence electron configuration ns 2 np 4 ? A. 8A (18) B. 7A (17) C. 5A (15) D. 6A (16) E. 4A (14) 6. Which atom is diamagnetic? A. neon B. carbon C. nitrogen D. oxygen E. lithium
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7. Which is the most polar bond? A.
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This test prep was uploaded on 04/07/2008 for the course CHEM 1307 taught by Professor Li during the Spring '08 term at Texas Tech.

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- Equations E = h c = Palmer-Rydberg Equation 1 1 2 2 n1 n 2 Energy Equation from Bohr model of H E =-2.18 x 10-18 J(1/n2 E = Efinal Einitial = R

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