#Chem 161-2007 final exam + solutions

#Chem 161-2007 final exam + solutions - 1 Chem 161-2007...

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1 Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 3A – STOICHIOMETRY: CHEMICAL CALCULATIONS Benzene and acetylene are different substances with the same empirical formulas. Which of the following is true? A . Their molar masses will have a simple integer ratio. B. Their molecular formulas are the same. C. Their structural formulas are similar. D. They have similar chemical properties. E. They have similar physical properties. Benzene is C 6 H 6 ; its empirical formula is C 1 H 1 or CH. Acetylene is C 2 H 2 ; its empirical formula is C 1 H 1 or CH. A. True. The molar mass of C 6 H 6 is 78; the molar mass of C 2 H 2 is 26. Their molar masses have a simple integer ratio of 3:1. Even if the student didn’t know the molecular formulas of benzene and acetylene the student would still get “A” as the answer. Let’s say that the student thought that benzene was C 6 H 6 , but that acetylene was C 4 H 4 . The molar masses would be 78 and 52, respectively, which is 3:2, a simple integer ratio. B. False. Their molecular formulas are different, C 6 H 6 and C 2 H 2 . C. H H | | C C H ― C C ― H C C | | H H H ─C ≡ C ─ H False. Their structural formulas are different. D. False. The molecules are completely different and therefore have different reactions with varied reagents, e.g., bases. E. False. The molecules are completely different and therefore have different physical properties, e.g., boiling point and melting point. A 1
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2 Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 5 - GASES Reaction stoichiometry 5.00 grams of alum, KAl(SO 4 ) 2 . xH 2 O, is heated in an oven, and the water vapor driven off is collected and found to exert a pressure of 1.33 atm in a 3.00 Liter container at 111 o C. Calculate the value of “x” in the hydrate. A. 2 B. 4 C. 5 D. 7 E . 12 KAl(SO 4 ) 2 . xH 2 O → KAl(SO 4 ) 2 + xH 2 O 5.00 g 1.33 atm 3.00L 111 + 273K Plan: V,P,TH 2 O → molH 2 O → gH 2 O → g KAl(SO 4 ) 2 → molKAl(SO 4 ) 2 → mol ratio PV = nRT n = PV/RT n = (1.33 atm x 3.00 L)/(0.08205Latmdeg -1 mol -1 x (273 + 111)) = 0.1266 mol H 2 O 0.1266 mol H 2 O x 18.02gH 2 O/mol = 2.28g H 2 O 5.00g alum – 2.28g H 2 O = 2.72g KAl(SO 4 ) 2 2.72gKAl(SO 4 ) 2 x (1/(39.098+26.982+(96.02 x 2))) = 0.01054 mol KAl(SO 4 ) 2 X = mol H 2 O/mol KAl(SO 4 ) 2 = 0.1266/0.01054 = 12.01 E 2
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3. Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 4B - CHEMICAL REACTIONS IN AQUEOUS SOLUTIONS Oxidation-Reduction Rxns Consider the following oxidation-reduction reaction: CH 4 ( g ) + NO 2 ( g ) → N 2 ( g ) + CO 2 ( g ) + H 2 O( l ) Which of the following statements is true about the reaction above? A. N 2 is the oxidizing agent. B
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This note was uploaded on 04/12/2008 for the course CHEM 161 taught by Professor Vacillian during the Fall '08 term at Rutgers.

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#Chem 161-2007 final exam + solutions - 1 Chem 161-2007...

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