Exam 1-solutions

# Exam 1-solutions - Version 238 Exam 1 Mccord(50970 This...

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Version 238 – Exam 1 – Mccord – (50970) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points How many s electrons and p electrons does Mn possess? 1. 6 s , 18 p 2. 8 s , 6 p 3. 7 s , 12 p 4. 8 s , 12 p correct 5. 7 s , 18 p 6. 6 s , 12 p 7. 6 s , 6 p 8. 8 s , 18 p 9. 7 s , 13 p Explanation: The electron confguration For Mn is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 5 2 + 2 + 2 + 2 = 8 s 6 + 6 = 12 p 002 10.0 points How many electrons can possess this set oF quantum numbers: principal quantum num- ber n = 4, angular quantum number = 2? 1. 6 2. 8 3. 18 4. 2 5. 12 6. 4 7. 0 8. 10 correct 9. 16 10. 14 Explanation: Use the rules For the quantum numbers: IF n = 4 and = 2 ( i.e. , 4 d ), then m = - 2 , - 1 , 0 , +1 , +2 are permitted; there are fve di±erent orbitals and m s = ± 1 2 , each holding two electrons. 003 10.0 points Which oF the Following best describes the range oF atomic radii? 1. 0.05 to 1 ˚ A 2. 1 to 100 ˚ A 3. 5 to 10 ˚ A 4. 10 to 30 ˚ A 5. 1 to 15 ˚ A 6. 0.4 to 3 ˚ A correct Explanation: Atomic radii vary From 30 pm (He) to 300 pm (²r). 1 ˚ A= 10 - 10 m = 100 pm, so the range oF atomic radii is 0.3 to 3 ˚ A. 004 10.0 points An electron is confned to a one-dimensional box oF length L. It Falls From the second en- ergy level to the ground state, and releases a photon with a wavelength oF 322 nm. What is the length oF the box? 1. 0.541 nm. correct 2. 0.303 nm. 3. 0.151 nm. 4. 1.14 nm.

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Version 238 – Exam 1 – Mccord – (50970) 2 5. 1.61 nm. 6. 292 nm. 7. 2.92 nm. 8. 322 nm. 9. 4.32 nm. Explanation: 005 10.0 points Consider the radial distribution function (RDF) plot shown below. Which of the orbitals given as choices below would corre- spond to the RDF shown. r 4 πr 2 R 1. 5 d correct 2. 4 f 3. 4 s 4. 5 p 5. 6 d 6. 4 d Explanation: The plot has 2 radial nodes (spherical nodes). The number of spherical nodes will al- ways be equal to n - - 1. Only the 5 d orbital ±ts this case. Other orbitals that WOULD ±t this would have been the 3 s , 4 p , and 6 f . 006 10.0 points C 6 H 6 = 78.11 g/mol Cl 2 = 70.91 g/mol C 6 H 5 Cl = 112.55 g/mol HCl = 36.46 g/mol If 39.0 g of C 6 H 6 reacts with excess chlorine and produces 30.0 g of C 6 H 5 Cl in the reaction C 6 H 6 + Cl 2 C 6 H 5 Cl + HCl , what is the percent yield of C 6 H 5 Cl? 1. 69.4% 2. 53.4% correct 3. 76.9% 4. 13.2% 5. 50.0% Explanation: m C 6 H 6 = 39.0 g m C 6 H 5 Cl = 30.0 g Our ±rst step is to determine the theoretical yield of the reaction. The reaction started with 39.0 g of C 6 H 6 . We convert from grams to moles using the molar mass: ? mol C 6 H 6 = 39 . 0 g C 6 H 6 × 1 mol C 6 H 6 78 . 11 g C 6 H 6 = 0 . 4993 mol C 6 H 6 From the balanced equation we see that 1 mole C 6 H 5 Cl is produced for each mole of C 6 H 6 reacted. Therefore, if 0.4993 mol C 6 H 6 were reacted we would expect to pro- duce 0.4993 mol C 6 H 5 Cl. We use the molar mass of C
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Exam 1-solutions - Version 238 Exam 1 Mccord(50970 This...

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