Practice Problems 3
Chapter 1
CHE 151
Graham/07
Express all answers in correct scientific notation and significant figures.
1.
Convert 2.75 m to each of the following:
a.)
mm
2.75m x 1 mm =
10-3m
c.)
2.75 x 103mm
d.)
=
2.75 x 10-3km
2.75 x 109nm
m
2.75m x 1 m =
10-6m
2.75 x 106 m
Convert 3.4 mg to each of the following:
a.)
ng
b.)
3.4mg x 10-3g x 1ng = 3.4 x 106ng
1mg 10-9g
c.)
kg
3.4mg x 10-3g x 1kg = 3.4 x 10-6kg
1mg
103g
cg
3.4mg x 10-3g x 1cg = 3.4 x 10-1cg
1mg 10-9g
3.
nm
2.75m x 1 nm =
10-9m
km
2.75m x 1 km
103m
2.
b.)
d.)
Mg
3.4mg x 10-3g x 1Mg = 3.4 x 10-9Mg
1mg
106g
Convert 5.7 hours to each of the following:
a.)
s
5.7hrs x 60min x 60s = 2.1 x 104s
1hr
1min
c.)
Ms
2.1 x 104s x 1Ms = 2.1 x 10-2Ms
106s
b.)
2.1 x 104s
d.)
ms
x 1ms = 2.1 x 107ms
10-3s
ps
2.1 x 104s x 1ps = 2.1 x 1016ps
10-12s
4.
5.
Convert:
a.)
2.44 m to cm
b.)
2.44 m3 to cm3
1 gal x 3.785 dm3 x 1 L = 30.179  3.0 x 101 L (2 sf)
41 mi
1 gal
1 dm3
1 cm
10-2m
2
x
1 in
2.54 cm
2
x
1 ft
12 in
2
= 52 ft2
1 lb
0.4536 kg
x
10-2m
1 cm
x 2 54 cm
1 in
x 12 in
1 ft
=
4.50 x 102 lb
ft3
Calculate the density of a rectangular solid that has the dimensions 40.0 cm long,
10.0 cm wide, and 5.00 cm high, and has a mass of 4.50 kg.
Volume = 40.0 cm x 10.0 cm x 5.00 cm = 2.00 x 103 cm3
4.50 kg
2.00 x 103 cm3
9.
= 2.44 x 106 cm3
The density of iron is 7210 kg/m3. Calculate its density in lbs/ft3.
7210 kg x
8.
3
A laboratory bench has a surface area of 4.8 m2. What is the area in ft2?
4.8 m2 x
7.
2.44m x 1cm
10-2m
2.44 x 102 cm
An automobile is rated with a highway milage of 41 miles per gallon. How many
liters of gasoline would be required for a 526 km trip?
526 km x 1 mi x
1.609 km
6.
2.44m x 1cm =
10-2m
x
103 g
1 kg
=
2.25 g/cm3
Convert the density 5.94 kg/m3 to g/cm3.
5.94 kg x 103 g
m3
1 kg
x
10-2 m
1 cm
3
= 5.94 x 10-3 g/cm3
10.
It takes light 4 years to get from the nearest big star to the earth, and light travels
3.00 x 108 m/s. How far away is the star?
4 yrs x 365 days x 24 hr x 60 min x 60 s x 3.00 x 108 m = 3.78 x 1016 m
1 yr
1 day
1 hr
1 min
1s
11.
Under certain conditions, air has a density of 1.3 kg/m3. Calculate the mass of air
in a lecture room 10.0 m by 15.0 m by 3.0 m.
volume of room = 10.0 m x 15.0 m x 3.0 m = 4.5 x 102 m3
4.5 x 102 m3 x 1.3 kg = 5.85 x 102
m3
12.

5.9 x 102 kg
round to 2 sf
A sphere of radius 6.29 cm is made of solid iron. It has a mass of 8.201 kg.
Compute the density of iron in g/cm3 and lb/in3.
Vsphere = 4 π r3 = 4(3.14)(6.29 cm)3 = 1.04 x 103 cm3
3
3
8.201 kg
x 103 g = 7.89 g/cm3
1.04 x 103cm3 0.4536 kg
8.201 kg x 1 lb
1.04 x 103 cm3
1 kg
13.
x 2.54 cm
1 in
3
= 2.85 x 10-1 lb/in3
What would be the total surface area of a cube of iron which has a mass of 126 g?
Use the density of iron calculated in problem 12 in units of g/cm3.
Volume of cube = 126 g x 1 cm3 = 16.0 cm3
7.89 g
length of side = √16.0 = 2.52 cm
3
Surface area = (length)2 x 6 sides = (2.52)2 x 6 = 38.1 cm2
14.
What Fahrenheit temperature is the same as 50.0o Celsius?
o
F = 1.8 x oC + 32
15.
= 1.8 (50.0) + 32 = 122oF
A sheet of aluminum foil has a total area of 1.000 ft2 and a mass of 3.635 g. What
is the thickness of the foil in millimeters? (Density of aluminum = 2.699 g/cm3)
1.) convert area to metric units: 1.000 ft2 x 12 in
1 ft
2
x
2.54 cm 2 = 9.290 x 102 cm2 (area
1 in
2.) calculate volume: 3.635 g x 1 cm3 = 1.347 cm3 (volume of sheet)
2.699 g
3.) volume = l x w x h
and l x w = area; so, volume = area x h (where h = thickness)
4.) (9.290 x 102 cm2) x h = 1.347 cm3
h = 1.347 cm3
= 1.450 x 10-3cm
2
2
9.290 x 10 cm
1.450 x 10-3cm x 10-2m x 1 mm = 1.450 x 10-2 mm
1 cm
10-3m
of sheet)