103a-fall2007-exam2-key - Exam II KEY- There were 2...

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Exam II KEY- There were 2 versions so look at the answers closely. 1) (3 each, all or nothing) You had either the top set or bottom set of PES data. For the top set A) [Ar] 4s 1 3d 5 Cr B) 1s 2 2s 2 2p 4 O C) 1s 2 2s 2 2p 6 3s 2 Mg For the bottom set A) [Ar] 4s 2 4p 2 Ge B) 1s 2 2s 2 2p 3 N C) 1s 2 2s 2 2p 6 3s 1 Na 2) Consider the element that corresponds with spectrum B and answer the following questions. SHOW YOUR WORK FOR CREDIT. a) (5) The ionization energy of the 1s electrons in this element are 543.1 eV OR 409.9 eV (1eV = 1.602 x 10 -19 J). What is the longest photon wavelength required to ionize a 1s electron in this element? Report you answer in nm. Answer = Either 2.282 or 3.024 nm A B C A B C
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b) (5) Convert 543.1 eV OR 409.9 eV to kJ/mol. Answer = 52394 kJ/mol or 39544 kJ/mol c) (1) Is the common ion of this element diamagnetic or paramagnetic? Diamagnetic 3) (5) Consider the plot shown of the kinetic energies of electrons ejected from the surface of potassium metal or silver metal at different frequencies of incident light. In one sentence, explain why the line for Ag starts at a higher frequency than K? The ionization energy of Ag is higher than K so the incoming light must be off higher frequency to eject electrons. 4)
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This note was uploaded on 04/03/2008 for the course CHEM 103a taught by Professor Weso during the Spring '08 term at University of Arizona- Tucson.

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103a-fall2007-exam2-key - Exam II KEY- There were 2...

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