Page 35 - Theory and Problems of BEGINNING CHEMISTRY

24 MATHEMATICAL METHODS IN CHEMISTRY [CHAP. 2

EXAMPLE 2.38. Change 171 K and 422 K to degrees Celsius.
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Ans. 171 K − 273 =−102 C
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422 K − 273 = 149 C
Note that a change in temperature in kelvins is the same as the equivalent change in temperature in degrees
Celsius.

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EXAMPLE 2.39. Convert 0 C and 30 C to kelvins. Calculate the change in temperature from 0 Cto30 C on both
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temperature scales.
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Ans. 0 C + 273 = 273 K
30 C + 273 = 303 K
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The temperature difference on the two scales is
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30 C 303 K
−0 C −273 K
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30 C 30 K
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Solved Problems
FACTOR-LABEL METHOD
2.1. (a) Write the reciprocal for the following factor label: 4.0 mi/h. (b) Which of these—the reciprocal or
the original factor label—is multiplied to change miles to hours?
Ans. (a) 1.0 h/4.0 mi (b) The reciprocal:
1.0h

10.0mi = 2.5h
4.0mi
2.2. (a) Write the reciprocal for the following factor label: 5.00 dollars/pound. (b) Which of these—the
reciprocal or the original factor label—is multiplied to change pounds to dollars?
Ans. (a) 1 pound/5.00 dollars (b) The original:

5.00 dollars

4.6 pounds = 23 dollars
1 pound
2.3. Calculate the number of hours one must work at 12.00 dollars/h to earn 150.00 dollars.

1h
Ans. 150.00 dollars = 12.5h
12.00 dollars
2.4. Calculate the number of cents in 4.58 dollars. (a) Do the calculation by converting first the dollars to
dimes and then the dimes to cents. (b) Repeat with a direct calculation.
10 dimes

Ans. (a) 4.58 dollars = 45.8 dimes
1 dollar
10 cents

45.8 dimes = 458 cents
1 dime
10 dimes 10 cents

(b) 4.58 dollars = 458 cents
1 dollar 1 dime

100 cents
or 4.58 dollars = 458 cents
1 dollar
2.5. Percentages can be used as factors. The percentage of something is the number of parts of that thing per
100 parts total. Whatever unit(s) is (are) used for the item in question is also used for the total. If a certain
ring is 64% gold, write six factors that can be used to solve problems with this information.