CHEM 1001/CHM 1020/CHM 1025C/CHM 1032C
Module Two Sample Pretest: Part M Multiple Choice 10 points
1 . The
density of a silver coin can be calculated from the following data:
The density of the coin should be
reported as using strict rules of significant figures |
|||||||||
|
a. |
10.847 g/mL |
|||||||
|
b. |
10.85 g/mL |
|||||||
|
c. |
10.8 g/mL |
|||||||
|
d. |
11 g/mL |
|||||||
|
e. |
1 x 101
g/mL |
|||||||
2. What mass of sodium chloride is
dissolved in 85.0 g of water to make a 15.0% solution? |
|||||||||
|
a. |
13.2 g |
|||||||
|
b. |
15.0 g |
|||||||
|
c. |
17.6 g |
|||||||
|
d. |
18.0 g |
|||||||
|
e. |
19.3 g |
|||||||
3. The number of significant
figures in 0.0060501000 is |
|||||||||
___ |
a |
5 |
|||||||
___ |
b |
6 |
|||||||
___ |
c |
8 |
|||||||
___ |
d |
9 |
|||||||
___ |
E |
11 |
|||||||
4. Multiply 1.05 cm times 2.2 cm
and round off the product to the proper number of significant digits.
a) 2 cm2
b) 2.0 cm2
c) 2.3 cm2
d) 2.31 cm2
e) 3 cm2
5. One cubic centimeter of helium
gas contains 26,900,000,000,000,000,000 atoms at standard conditions. Express
this number of helium atoms in scientific notation.
a) 2.69 * 1017
b) 2.69 * 1018
c) 2.69 * 1019
d) 26.9 * 1018
e) 26.9 * 1019
6. According to the metric system,
1000 g = 1 ______.
a) cg
b) kg
c) mg
d) Mg
e) µg (µ stands for Greek
letter Mu)
7. A tall glass cylinder contains
four separate liquid layers: mercury (d = 13.6 g/mL),
chloroform (d = 1.49 g/mL), acetic acid (d = 1.05 g/mL), ether (d = 0.708 g/mL). If
ice
(d = 0.99 g/mL)
is dropped into the cylinder, where does it come to rest?
a) on top of the mercury layer
b) on top of the chloroform
layer
c) on top of the acetic acid
layer
d) on top of the ether layer
e) on the bottom of the
cylinder
8. The term that refers to the close
clustering of similar laboratory measurements is |
||
|
a. |
precision. |
|
b. |
repeatability. |
|
c. |
accuracy. |
|
d. |
exactness. |
|
e. |
the number of significant
figures |
9. How many scruples are there
in 100. lbs? Some
equivalents which may be helpful are given below: 1.00
scruple = 20.0 grains |
|||||||||||||||||
|
a. |
|
|||||||||||||||
|
b. |
|
|||||||||||||||
|
c. |
|
|||||||||||||||
|
d. |
|
|||||||||||||||
|
e. |
|
10. The average velocity of
carbon dioxide molecules at 0oC is 3.62 x 104 cm/sec.
Which of the indicated calculations below would convert this value to the
velocity in miles per hour? |
|||||||||||||||||
|
a. |
|
|||||||||||||||
|
b. |
|
|||||||||||||||
|
c. |
|
|||||||||||||||
|
d. |
|
|||||||||||||||
|
e. |
|
ANSWERS: 1.this is tricky: 6.5081g/0.6mL = E
(five
significant figures divided by one significant figure number equals one
significant #answer technically)
2.b ; 3.c ;
4 c; 5.c;
6.b; 7.c; 8. a; 9.
d; 10. e