General Chemistry Principles, Patterns, and Applications, 2011

Notice that grams cancel to leave us with an answer that is in the correct units. Always check to make sure
that your answer has the correct units.
S K I L L BUILDER ES2
Solve to find the indicated variable.
a. 16 .4 g 41 .2 g = x 18. 3 g
b. 2 .65 m 4 .02 m = 3 .28 m y
c. 3 .27 × 10 − 3 g x = 5 .0 × 10 − 1 g 3.2 g
d. Solve for V 1: P1P2=V2V1
e. Solve for T 1: P1V1T1=P2V2T2
Solution
a. Multiply both sides of the equality by 18.3 g to remove this measurement from the
denominator:
(18 .3 g) 16 .4 g 41 .2 g = (18 .3 g) x 18. 3 g 7 .28 g = x
b. Multiply both sides of the equality by 1/3.28 m, solve the left side of the equation, and then
invert to solve for y:
( 1 3.28 m ) 2 .65 m 4 .02 m = ( 1 3. 28 m ) 3. 28 m y = 1 y y = ( 4.02 ) ( 3.28) 2.65 = 4.98
m
c. Multiply both sides of the equality by 1/3.27 × 10 −3 g, solve the right side of the equation, and
then invert to find x:
( 1 3 .27 × 10 − 3 g ) 3 .27 × 10 − 3 g x = ( 1 3 .27 × 10 − 3 g ) 5 .0 × 10 − 1 g 3.2 g = 1 x
x = ( 3.2 g)(3 .27 × 10 − 3 g) 5 .0 × 10 − 1 g= 2.1 × 10 − 2 g
d. Multiply both sides of the equality by 1/V 2, and then invert both sides to obtain V 1:
( 1 V 2 ) P 1 P 2 = ( 1 V 2 ) V 2 V 1 P 2 V 2 P 1 = V 1
e. Multiply both sides of the equality by 1/P 1V 1 and then invert both sides to obtain T 1:
( 1 P 1 V 1 ) P 1 V 1 T 1 = ( 1 P 1 V 1 ) P 2 V 2 T 2 1 T 1 = P 2 V 2 T 2 P 1 V1 T 1
= T 2 P 1 V 1 P 2 V 2
Percentages
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