test 1-chem - Version 171 Exam 1 McCord(53130 This...

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Version 171 – Exam 1 – McCord – (53130) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH301tt This exam is only for McCord’s Tues/Thur CH301 class. c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10 - 34 J · s m e = 9 . 11 × 10 - 31 kg N A = 6 . 022 × 10 23 mol - 1 ν = R parenleftbigg 1 n 2 y - 1 n 2 x parenrightbigg where R = 3 . 29 × 10 15 s - 1 ψ n ( x ) = parenleftbigg 2 L parenrightbigg 1 2 sin parenleftBig nπx L parenrightBig E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , · · · 001 10.0 points Dr. Schrodinger would predict that the tran- sition from the first energy level to the second energy level would require a photon with the longest wavelength for which of the following elements? 1. K 2. Na 3. Li 4. Rb correct Explanation: Based on Schrodinger’s analysis, the energy levels become closer together as the size of the “box” increases. 002 10.0 points In the spectrum of atomic hydrogen, a violet- blue line is observed at 434 nm. What are the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line? 1. n = 3, n = 2 2. n = 6, n = 3 3. n = 4, n = 3 4. n = 5, n = 3 5. n = 5, n = 2 correct 6. n = 6, n = 2 7. n = 4, n = 2 Explanation: λ = 434 nm = 4 . 34 × 10 - 7 m Because the line is in the visible part of the spectrum, it belongs to the Balmer series for which the ending n is 2. For the starting value of n , ν = c λ = 3 × 10 8 m / s 4 . 34 × 10 - 7 m = 6 . 909 × 10 14 s - 1 Using the Ryberg formula, ν = (3 . 29 × 10 15 s - 1 ) parenleftbigg 1 n 2 2 - 1 n 2 1 parenrightbigg ν 3 . 29 × 10 15 s - 1 = parenleftbigg 1 n 2 1 - 1 n 2 2 parenrightbigg 1 n 2 2 = 1 n 2 1 - ν 3 . 29 × 10 15 s - 1 = 1 4 - 6 . 909 × 10 14 s - 1 3 . 29 × 10 15 s - 1 = 0 . 04 n 2 2 = 25 n 2 = 5 003 10.0 points
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Version 171 – Exam 1 – McCord – (53130) 2 If a particle is confined to a one-dimensional box of length 300 pm, for Ψ 3 the particle is most likely to be found at 1. 50, 150, and 250 pm, respectively. cor- rect 2. 0 pm. 3. 100 and 200 pm, respectively. 4. 17.3 pm. 5. 300 pm. Explanation: 004 10.0 points What are the correct quantum numbers ( n , ℓ, m ) for the seventh electron of Cl? 1. 3, 1, 0 2. 2, 1, 2 3. 2, 2, 1 4. 3, 2, 1 5. 3, 0, 0 6. 2, 0, 1 7. 3, 1, 1 8. 2, 1, 1 correct Explanation: The electron configuration for Cl is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 5 The seventh electron is in a 2 p orbital, so n = 2, = 1, m = - 1, 0, or +1. 2, 1, 0 and 2, 1, - 1 are not given as choices. 005 10.0 points How many electrons are in the third principal level ( n = 3) of a chromium (Cr) atom? 1. 13 correct 2. 4 3. 24 4. 8 5. 2 6. 12 7. 6 Explanation: There are 2 electrons in the 3 s orbital, 6 electrons in the 3 p orbitals and 5 electrons in the 3 d orbitals. Remember that Cr is an exception. One electron from the 4 s orbital gets elevated to a 3 d orbital giving it a half filled 3 d orbital. 006 10.0 points Which of the following would be expected to have the highest first ionization energy?
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