Page 38 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 2] MATHEMATICAL METHODS IN CHEMISTRY 27

2.22. (a) Calculate the number of centimeters in 4.58 km. Do the calculation by converting first the kilometers
to meters and then the meters to centimeters. (b) Repeat with a direct calculation. (c) Compare this
problem with Problem 2.4.
1000 m

Ans. (a)4.58 km = 4580 m
1km

100 cm
4580 m = 458 000 cm
1m

1000 m 100 cm
(b)4.58 km = 458 000 cm
1km 1m
(c) The methods are the same.
2.23. Change (a) 2.00 mg to kilograms. (b) 2.00 mm to kilometers. (c) 2.00 cm to kilometers.
0.001 g 1kg

Ans. (a)2.00 mg = 2.00 × 10 −6 kg
1mg 1000 g

0.001 m 1km
(b)2.00 mm = 2.00 × 10 −6 km
1mm 1000 m

0.01 m 1km
(c)2.00 cm = 2.00 × 10 −5 km
1cm 1000 m
Note the identical processes in all three parts and the similar answers in the first two.

2.24. Draw a square 2 cm × 2 cm. In the upper left corner of that square, draw another square 1 cm × 1 cm.
How many of the small squares will fit into the large one?
2
2
Ans. Four small squares will fit; that is, 4 cm = (2cm) . (See Fig. 2-3).
1 cm
1 cm
2 cm

2 cm
Fig. 2-3.

2.25. (a) What is the ratio in areas of a square 1 cm on each side and a square 1 mm on each side?
(b) What is the ratio in volumes of a cube 1 cm on each side and a cube 1 mm on each side?
Ans. (a) The ratio of centimeters to millimeters is 10 : 1; the ratio of their squares is therefore 100 : 1.
(b) The ratio of the cubes is 1000 : 1.

2.26. Explain in terms of units why area × distance yields volume.
2
Ans. m × m = m 3
2.27. Give the volume corresponding to each of the following numbers of liters in terms of a dimension
involving length cubed: (a) 2000 L, (b)2L,(c) 0.002 L, and (d)2 × 10 −6 L.
Ans. (a)2 m 3 (b)2dm 3 (c)2cm 3 (d)2mm 3

3
3
3
3
2.28. Give the volume in liters of each of the following: (a)3m ,(b)3dm ,(c)3cm , and (d)3mm .
Ans. (a) 3000 L (b)3L (c) 0.003 L (d)3 × 10 −6 L