2010 Exam 3 Key

2010 Exam 3 Key - Name Section Number Chemistry 1314 Honors...

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N a m e Section Number Chemistry 1314 Honors Midterm Exam #3 November 10, 2010 This exam is closed book, notes, papers, etc. You may use a calculator. If you need a physical constant or unit conversion factor that is not provided, please ask. This exam must represent your own work. Scores: 1. / 20 2. a. / 4 b. / 4 c. / 6 d. / 6 3. / 20 4. / 11 5. / 9 6. / 20 Total: / 100
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CHEM 1314 H Midterm Exam #3, 2010 Page 2 1. For each of the following reactions, (i) predict any products (if they aren’t already given); (ii) complete balanced equations; (iii) identify the reactions as precipitation, acid-base, redox, or none of the above; and (iv) for acid-base reactions identify the acid and the base, for redox reactions write the appropriate oxidation numbers above each atom and identify what is oxidized and what is reduced, and for precipitation reactions identify the precipitate. You do not have to complete the states of matter in the reactions (other than identifying a precipitate). a. C 3 H 6 O + O 2 ⎯→ b. Mg(OH) 2 + HI c. NaOH + Fe(NO 3 ) 2 d. K 2 Cr 2 O 7 + HCl + H 2 C 2 O 4 CO 2 CrCl 3 + H 2 O + KCl e. NH 4 + + C 2 H 5 O - f. Cl - + I 2 2 H 2 O + MgI 2 2 Base Acid Predicting the products (a,b,c,e,f): 1 point Balancing reaction (a,b,c,d,e): 1 point Reaction type (a,b,c,d,e): 1 point Reaction-specific info (a,b,c,d,e): 1 point 2 NaNO 3 + Fe(OH) 2 Precipitation Reaction. Precipitate is Fe(OH) 2 Ox. #s: +1 +6 -2 +1 -1 +1 +3 -2 +4 -2 +3 -1 +1 -2 +1 -1 Redox Reaction. C is oxidized; Cr is reduced. 3 CO 2 + 3 H 2 O 4 Ox. #s: -4/3 +1 -2 0 +4 -2 +1 -2 Redox Reaction. C is oxidized; O (from O 2 ) is reduced. Acid-base reaction. NH 3 + C 2 H 5 OH Acid Base Acid-base reaction. No reaction. Note: The identification of reaction type is implicit in the reaction-specific info (e.g., identifying the acid and the base implies that it is an acid-base reaction). 2 8 3 6 2 7 2
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CHEM 1314 H Midterm Exam #3, 2010 Page 3 2. The oxidation number for carbon that you determined for the first reactant in problem 1 (a) probably seemed unusual to you. In this problem we will explore this reactant and, among other things, try to explain why that unusual oxidation number is actually correct. a. There are at least nine distinct compounds that share the molecular formula C 3 H 6 O. Draw Lewis structures for two such distinct molecules, and demonstrate their reasonableness by assigning formal charges for each atom in each structure. b. Redraw below the two structures you have chosen and assign oxidation numbers to each atom in each structure. c. Explain why the oxidation number you determined for carbon in the first reactant in problem 1 (a) is consistent with the numbers you determined in problem 2 (b). d. Oxidation numbers and formal charges are both methods of assigning charges to atoms in a molecule. Explain why these two methods produce different results, and discuss how one should view the “true” charge on each atom.
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2010 Exam 3 Key - Name Section Number Chemistry 1314 Honors...

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