CHEM 110 (BEAMER)
EXAM 1 REVIEW
Last Name
Date
First Name
Section
M
W
REVIEW: TWO-STEP AND MULTI-STEP CONVERSIONS CALCULATIONS
Note:
This is Practice Work. Unless I give this during lecture, this is not due for a grade.
Note:
I did not give you room on this sheet to do your work. You will need your own paper.
Relevant Exam 1 Information

Do not expect these exact problems to show up on the Exam. (They won’t). Copying down the
solutions quickly will not help understand the material.

There are no conversion calculations in Part 1 of the Exam.

You may only use conversion factors from the memorized list and the Appendices.

You must show your work for full credit

You must use units through each step.

The answer must be numeralan.b.ically correct.

The answer must have the exact number of significant digits correct.

The units in the answer must match the units in the work.

For this practice work, I have given several hints. Do not expect these hints on the Exam. (See the
practice exam)

The Exam will have approximately three of these questions.

Max Time. You should practice these problems so that you are working under Max Time. If you
use the Max Time for every problem, you will not finish the Exam.
Questions
1)
Convert 0.750 meters to feet. Max Time: 6 min
2)
Convert 360,000 minutes to days. Max Time: 5 min
3)
Convert 874.0 cm to feet. Max Time: 6 min
4)
Convert 15,000. mL to quarts. Max Time: 6 min
5)
A piece of plywood weighs 400.00 oz. Convert this measurement to kilograms. Max Time: 6 min
6)
Mount Everest is an approximate elevation of 8.8 km. Convert this measurement to feet.
Max Time: 6 min
7)
Convert 1.87 km/min to miles per hour. Max Time: 6 min
Page 1 of 5
CHEM 110 (BEAMER)
8)
EXAM 1 REVIEW
Chemical CX is 42.0%(v/v) vegetable oil. The density of vegetable oil is 0.91 g/mL. Determine the
mass (in grams) of vegetable oil present in a 400.0-mL sample of CX. You may use “oil” to represent
“vegetable oil.” Max Time: 9 min
Hint: It might be helpful to write the two conversion factors before beginning the problem.
9)
The density of silver is 10.49 g/cm3. A sample of density has a volume of 151.25 cm3. Calculate the
mass of the sample silver in pounds. You may not use the density equation for this question. Max
Time: 6 min
Hint: Write density as a conversion factor to get the ball rolling.
10)
Calculate the mass (in kg) of a sample of water that has a volume of 2500 mL.
Max Time: 4 min
Note: You must have the density of water memorized to do this question.
Page 2 of 5
CHEM 110 (BEAMER)
EXAM 1 REVIEW
ANSWERS/SOLUTIONS
A few of notes:

I have included “timing” to help you gauge your work output speed.
Again: My use of term “competence,” I am


I am referring to “competence with the material.”

I am not referring to your “competence as a person, a parent, a bowler, etc…”

After printing this out, you may want to get your multi-color clicky pens to cancel units.

Remember that there may be more than one way to get from the initial value to the answer.

I will write the answers in reverse order: The conversion factors will appear before the solution.

Check your final answer against this answer key. If you get the same answer, but used a different “path,” that’s okay.
Question 1 Solution
1)
(
0.750 m
)
1
(
1.0936 yd
)
1m
(
3 ft
)
1 yd
=
x2.46 ftx
Sample Point Breakdown (Question 1): Total of 13 points
(
0.750 m
)
1
(
3 pt
1.0936 yd
)
1m
(
3 ft
)
1 yd
=
x2.46 ftx
3 pt
3 pt
4 pt

1 pt: trying

1 pt: correct conversion factor

1 pt: correct conversion factor

1 pt: correct numerical answer

1 pt: correct initial value

1 pt: units written

1 pt: units written

1 pt: correct number of sigdigs

1 pt: units written

1 pt: numerator/denominator

1 pt: numerator/denominator

1 pt: correct units

no credit if initial value
correct

1 pt: units must match work
correct
is “flipped”
Page 3 of 5
CHEM 110 (BEAMER)
EXAM 1 REVIEW
Question 2 Solution
2)
(
360,000 min
)
1
(
1 hr
)
60 min
(
1 day
)
24 hr
=
2
x250 day or 2.5 × 10 dayx
Question 3 Solution
3)
(
874.0 cm
)
1
(
1 in
)
2.54 cm
(
1 ft
)
12 in
=
x28.67 ftx
=
x15.851 qtx
Question 4 Solution
4)
(
15,000. mL
)
1
Note:
(
1 L
)
1000 mL
(
1.0567 qt
)
1L
The conversion factor 1 L = 1000 mL must be shown in your work. You cannot
simply “move the decimal point.”
Question 5 Solution
5)
(
400.00 oz
)
1
(
1 lb
)
16 oz
(
1 kg
)
2.2046 lbs
=
x11.340 kgx
5280 ft
)
1 mi
=
x2.9 × 10 ftx
60 min
)
1 hr
=
x69.7 mi/hrx
Question 6 Solution
6)
(
8.8 km
)
1
(
0.62137 mi
)
1 km
(
4
Question 7 Solution
7)
(
1.87 km
)
1 min
(
1 mi
)
1.6093 km
(
Question 8 Solution
42.0 mL OIL = 100 mL CX
Conversion Factors:
8)
(
400.0 mL Cx
)
1
(
42.0 mL OIL
)
100 mL Cx
(
0.91 g OIL
)
1 mL OIL
0.91 g OIL = 1 mL OIL
=
x152.9 g vegetable oilx
Page 4 of 5
CHEM 110 (BEAMER)
EXAM 1 REVIEW
Question 9 Solution
10.49 g Ag = 1 cm3 Ag
Conversion Factor from density:
9)
(
151.25 cm3 Ag
)
1
(
10.49 g Ag
)
1 cm3 Ag
(
1 lb Ag
)
453.59 g Ag
x3.4979 lbs Agx
=
Question 10 Solution
Memorized conversion factor for density of water:
10)
(
2500 H2 O
)
1
(
1 g H2 O
)
1 mL H2 O
(
1 g H2O = 1 mL H2O
1 kg H2 O
)
1000 g H2 O
=
x25 kg H2 Ox
Page 5 of 5