Chemistry Chapter 2 Section 2: Practice Problems 11-22
pages 41–46
11. Express each number in scientific notation.
a. 700
7  102
b. 38,000
3.8104
c. 4,500,000
4.5106
d. 685,000,000,000
6.851011
e. 0.0054
5.410–3
f. 0.00000687
6.8710–6
g. 0.000000076
7.610–8
h. 0.0000000008
810–10
12. Challenge Express each quantity in regular notation along with its
appropriate unit.
a. 3.60105 s
360,000 s
b. 5.41025 g/cm3
0.000054 g/cm3
c. 5.060103 km
5060 km (cannot express in regular notation with the correct
number of significant figures)
d. 8.91010 Hz
89,000,000,000 Hz
13. Solve each problem and express the answer in scientific notation.
a. (510–5)  (210–5)
7105
b. (7  108)  (4  108)
3108
c. (9102) – (7102)
2102
d. (410212) + (110212)
51012
14. Challenge Express each answer in scientific notation in the units
indicated.
a. (1.26104 kg) + (2.5106 g) in kg
1.26104 kg  0.25104 kg  1.51104 kg
b. (7.06 g) + (1.210–4 kg) in kg
7.06103 kg + 0.12103 kg
3
 7.1810 kg
c. (4.39105 kg)  (2.8107 g) in kg
4.39105 kg – 0.28105 kg = 4.11105 kg
d. (5.36101 kg)  (7.40102 kg) in g
5.36101 kg – 0.740101 kg
= 4.62101 kg = 4.62102 g
15. Solve each problem, and express the answer in scientific notation.
a. (4102)(1108)
(41)1028 = 41010
b. (210–4)(3102)
(23)10–4 2  610–2
c. (6102)  (2101)
+
(6 2)1021  3101
d. (8104)  (4101)
(8 4)104–1  2103
16. Challenge Calculate the areas and densities. Report the answers in
correct units.
a. area of a rectangle with sides measuring
3  101 cm and 3 10–2 cm
area  (3101 cm)(3102 cm)
1 (–2)
1
2
 (33)10   910 cm
b. area of a rectangle with sides measuring
1  103 cm and 5  101 cm
area  (1103 cm)(5101 cm)
3 (–1)
2
2
 (15)10   510 cm
c. density of a substance having a mass of
9105 g and a volume of 310–1 cm3
density  mass/volume
5
–1
3
 (910 g)/(310 cm )
5– (–1)
 (9  3)10
6
3
 310 g/cm
d. density of a substance having a mass of
410–3 g and a volume of 210–2 cm3
density  mass/volume
–3
–2
3
 (410 g)/(210 cm )
–3– (–2)
 (4  2)10
–1
3
 2  10 g/cm
17. Write two conversion factors for each of the following.
a. a 16% (by mass) salt solution
16 g salt
100 g solution
;
100 g solution
16 g salt
b. a density of 1.25 g/mL
1.25 g 1 mL
;
1 mL 1.25 g
c. a speed of 25 m/s
25 m 1 s
;
1 s 25 m
18. Challenge What conversion factors are needed to convert:
a. nanometers to meters?
nanometers to
109 m
meters: 1 nm
b. density given in g/cm3 to a value in kg/m3?
g/cm 3 to kg/m 3 :
1 kg
106 cm 3
and
1000 g
1 m3
Use Table 2 to solve each of the following.
19. a. Convert 360 s toms.
360 s ×
1000 ms
= 360,000 ms
1s
b. Convert 4800 g to kg.
4800 g ×
1 kg
= 4.8 kg
1000 g
c. Convert 5600 dm to m.
5600 dm ×
1m
= 560 m
10 dm
d. Convert 72 g to mg.
72 g ×
1000 mg
= 72,000 mg
1g
e. Convert 2.45102 ms to s.
2.45 × 102 ms ×
1s
= 0.245 s
1000 ms
f. Convert 5 m to km.
5 μm ×
10-9 km
= 5 × 10-9 km
1 μm
g. Convert 6.800103 cm to km.
6800 cm ×
1m
1 km
×
= 0.068 km
100 cm 1000 m
h. Convert 2.5101 kg to Mg.
2.5 × 101 kg ×
1 mg
= 0.025 mg
1000 kg
20. Challenge Write the conversion factors needed to determine the
number of seconds in one year.
 365 d   24 h   60 min   60 s 
1 yr 




 1 yr   1 d   1 h   1 min 
21. The speedometer below displays a car’s speed in miles per hour.
What is the car’s speed in
km/h? (1 km  0.62 mile)
 65 mi

 1h

  1 km

  0.62 mi


 = 10 × 102 km/h


(Note: because significant figures and rounding haven’t been covered
yet, students might generate a calculator answer of about 104.8 km/h.)
22. How many seconds are in 24 h?
24 h ×
60 min
60 s
×
= 86,400 s
1h
1 min