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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- Spring '08
- Larsen
- Mole, Reaction, Chemical reaction, amu 62.07 amu
-
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