Exercício Resolvido - Quimica Geral (9)

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Chapter 1: Matter – Its Properties and Measurement 
 17 
70. (D) We first determine the volume of steel needed. This volume, divided by the cross-
sectional area of the bar of steel, gives the length of the steel bar needed. 
3
3
 
1000 gsteel 1 cm steel
V = 1.000 kg steel × × = 129.87 cm of steel
1 kg steel 7.70 g steel
 
2 2
2
 
23
2
(2.50 in.)s
For an equilateral triangle of length s, area = = = 2.706 in.
4 4
volume 129.87 cm 1 in. 1 in.
length = = × = 2.93 in.
area 2.706 in. 2.54 cm 2.54 cm
3 3
 
 
 
71. (D) Conversion pathway approach: 
 
3
3 
3
5280 ft 12 in. 2.54 cm 1 mL 1.03 g
NaCl mass 330,000,000 mi
1 mi 1 ft 1 in. 1 cm 1 mL
 
 tons105.5
lb 2000
 ton1
g 453.6
lb 1
 watersea g 0.100
chloride sodium g 5.3
 16 
 
 Stepwise approach: 
3
3 19 3
3
19 3 22 3
3
22 3 24 3
24 3 24
3
5280 ft
 330,000,000 mi 4.9 10 ft
1 mi
12 in.
4.9 10 ft 8.4 10 in.
1 ft
2.54 cm
 8.4 10 in. 1.4 10 cm
1 in.
1 mL 1.03 g
 1.4 10 cm = 1.4 10 g
1 cm 1 mL
 
24 22
22 20
20 16
3.5 g sodium chloride
1.4 10 g 4.9 10 g NaCl
100.0 g sea water
1 lb
4.9 10 g NaCl 1.1 10 lb
453.6 g
1 ton
1.1 10 lb 5.4 10 tons
2000 lb
 
 
The answers for the stepwise and conversion pathway approaches differ slightly due to a 
cumulative rounding error that is present in the stepwise approach. 
Chapter 1: Matter – Its Properties and Measurement 
18 
 
72. (D) First, we find the volume of the wire, then its cross-sectional area, and finally its 
length. We carry an additional significant figure through the early stages of the 
calculation to help avoid rounding errors. 
3
3 2
2
2
3
2
453.6 g 1 cm
V = 1 lb × × = 50.85cm Note: area = r
1 lb 8.92 g
0.05082 in. 2.54 cm
area = 3.1416× × = 0.01309cm
2 1 in.
volume 50.85 cm 1 m
length = = × = 38.8 m
area 0.01309 cm 100 cm
 
 
73. (M) 
5
seawater
1000 g seawater2000 lb Mg 453.6 g Mg 0.001 L
V = 1.00×10 ton Mg× × × ×
1 ton Mg 1 lb Mg 1.4 g Mg 1.025 g seawater
 
 
3
7 31 m 6 10 m seawater 
1000 L
 
 
74. (D) (a) 
2
2
10 ton 1 mi 1 ft 39.37 in. 2000 lb 454 g 1000 mg
dustfall
1 mi 1 mo 5280 ft 12 in. 1 m 1 ton 1 lb 1 g
 
 
3
2 2
3.5 10 mg 1 month 1 d 5 mg
 
1m 1mo 30 d 24 h 1m 1h
 
 
(b) This problem is solved by the conversion factor method, starting with the volume 
that deposits on each square meter, 1 mm deep. 
22 2
2 3
5 1
(1.0 mm 1 m ) 1 cm 100 cm 2 g 1000 mg 1 m h
1 m 10 mm 1 m 1 cm 1 g 4.9 mg
4.1 10 h 5 10 y It would take about half a century to accumulate a depth of 1 mm.
 
75. (D) (a) 
311
22
ft 10 54.1
mi 1
ft 5280
acre 640
mi 1
feet -acre 103.54volume 
 (b) 
39
3
311 m 1036.4
cm 100
m 1
in. 1
cm 54.2
ft 1
in. 12
ft 10 1.54 volume 
 (c) gal 1015.1
L3.785
gal 1
m1
L 1000
m 10 4.36 volume 12
3
39