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Exam%201A%20F06%20&%20Key

Exam%201A%20F06%20&%20Key - CHEM 4:011 PRINCIPLES OF...

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CHEM 4:011 – PRINCIPLES OF CHEMISTRY I EXAM #1 (30 QUESTIONS, 1½ HOURS) 14-September-2006 FORM A 1. According to the given reaction, 3 moles of Al 2 O 3 , plus 6 moles of C, plus 9 moles of Cl 2 will result in a theoretical yield of how many moles of AlCl 3 ? (You must first balance the equation.) Al 2 O 3 ( s ) + C( s ) + Cl 2 ( g ) AlCl 3 ( s ) + CO( g ) (A) 1 mole AlCl 3 (B) 3 moles AlCl 3 (C) 4 moles AlCl 3 (D) 6 moles AlCl 3 (E) 9 moles AlCl 3 2. Calculate the oxidation number of chlorine in perchloric acid, HClO 4 , a strong oxidizing agent. (A) 1 (B) +1 (C) +4 (D) +5 (E) +7 3. In the U.S. antifreeze is principally ethylene glycol, HOCH 2 CH 2 OH (a chemical so toxic that a teaspoonful will kill a cat). Congress is trying to mandate the use of a bitter additive to discourage children and pets from drinking antifreeze. What is the average molecular mass of ethylene glycol?
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2 4. In class, we determined the molecular formula for the caffeine molecule (structure shown at right). Determine the mass percent of nitrogen in caffeine. 5. If the density of nickel metal is 8.902 g/cm 3 , what is its density in kg/m 3 ? 6. Consider the two calculations (11.7 10.9) × 7.30 (i.e., subtraction first) and (11.7 × 7.30) (10.9 × 7.30) (i.e., subtraction last). Assuming these numbers are not exact, what are the correct results? (A) (11.7 10.9) × 7.30 = 6 and (11.7 × 7.30) (10.9 × 7.30) = 6 (B) (11.7 10.9) × 7.30 = 6 and (11.7 × 7.30) (10.9 × 7.30) = 5.8 (C) (11.7 10.9) × 7.30 = 5.8 and (11.7 × 7.30) (10.9 × 7.30) = 5.8 (D) (11.7 10.9) × 7.30 = 5.8 and (11.7 × 7.30) (10.9 × 7.30) = 5.84 (E) (11.7 10.9) × 7.30 = 5.84 and (11.7 × 7.30) (10.9 × 7.30) = 5.84
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