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

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?

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