3. 3
Numbers in Scientific Notation
A number written in scientific notation contains a
• coefficient.
• power of 10.
Examples:
coefficient power of ten coefficient power of ten
1.5 x 102
7.35 x 10-4
4. 4
Writing Numbers in
Scientific Notation
To write a number in scientific notation,
• move the decimal point to give a number 1-9.
• show the spaces moved as a power of 10.
Examples:
52 000. = 5.2 x 104
0.00178 = 1.78 x 10-3
4 spaces left 3 spaces right
6. 6
Comparing Numbers in Standard
and Scientific Notation
Here are some numbers written in standard format
and in scientific notation.
Number in Number in
Standard Format Scientific Notation
Diameter of the Earth
12 800 000 m 1.28 x 107
m
Mass of a typical human
68 kg 6.8 x 101
kg
Length of a pox virus
0.000 03 cm 3 x 10-5
cm
7. 7
Study Tip: Scientific Notation
In a number 10 or larger, the decimal point
• is moved to the left to give a positive power of 10
In a number less than 1, the decimal point
• is moved to the right to give a negative power of 10
8. 8
Learning Check
Select the correct scientific notation for each.
A. 0.000 008 m
1) 8 x 106
m, 2) 8 x 10-6
m, 3) 0.8 x 10-5
m
B. 72 000 g
1) 7.2 x 104
g, 2) 72 x 103
g, 3) 7.2 x 10-4
g
10. 10
Learning Check
Write each as a standard number.
A. 2.0 x 10-2
L
1) 200 L, 2) 0.0020 L, 3) 0.020 L
B. 1.8 x 105
g
1) 180 000 g, 2) 0.000 018 g, 3) 18 000 g
14. 14
. l2
. . . . l . . . . l3
. . . . l . . . . l4
. . cm
• The markings on the meterstick at the end of the
orange line are read as:
the first digit 2
plus the second digit 2.7
• The last digit is obtained by estimating.
• The end of the line may be estimated between 2.7–
2.8 as half way (0.5) or a little more (0.6), which gives
a reported length of 2.75 cm or 2.76 cm.
Reading a Meterstick
15. 15
Known & Estimated Digits
If the length is reported as 2.76 cm,
• the digits 2 and 7 are certain (known).
• the final digit, 6, is estimated (uncertain).
• all three digits (2, 7, and 6) are significant, including
the estimated digit.
16. 16
. l8
. . . . l . . . . l9
. . . . l . . . . l10
. . cm
What is the length of the orange line?
1) 9.0 cm
2) 9.04 cm
3) 9.05 cm
Learning Check
17. 17
. l8
. . . . l . . . . l9
. . . . l . . . . l10
. . cm
The length of the orange line could be reported as
2) 9.04 cm
or 3) 9.05 cm
The estimated digit may be slightly different. Both
readings are acceptable.
Solution
18. 18
. l3
. . . . l . . . . l4
. . . . l . . . . l5
. . cm
• For this measurement, the first and second known
digits are 4 and 5.
• When a measurement ends on a mark, the estimated
digit in the hundredths place is 0.
• This measurement is reported as 4.50 cm.
Zero as a Measured Number
19. 19
Significant Figures in
Measured Numbers
Significant Figures
• obtained from a measurement include all
of the known digits plus the estimated
digit.
• reported in a measurement depend on the
measuring tool.
21. 21
All nonzero numbers in a measured number are
significant.
Number of
Measurement Significant Figures
38.15 cm 4
5.6 ft 2
65.6 lb 3
122.55 m 5
Counting Significant Figures
22. 22
Sandwiched Zeros
• occur between nonzero numbers.
• are significant.
Number of
Measurement Significant Figures
50.8 mm 3
2001 min 4
0.0702 lb 3
0.405 05 m 5
Sandwiched Zeros
23. 23
Trailing Zeros
• follow nonzero numbers in numbers without
decimal points.
• are usually placeholders.
• are not significant.
Number of
Measurement Significant Figures
25 000 cm 2
200 kg 1
48 600 mL 3
25 005 000 g 5
Trailing Zeros
24. 24
Leading Zeros
• precede nonzero digits in a decimal number.
• are not significant.
Number of
Measurement Significant Figures
0.008 mm 1
0.0156 oz 3
0.0042 lb 2
0.000 262 mL 3
Leading Zeros
25. 25
State the number of significant figures in each of
the following measurements.
A. 0.030 m
B. 4.050 L
C. 0.0008 g
D. 2.80 m
Learning Check
26. 26
State the number of significant figures in each of
the following measurements.
A. 0.030 m 2
B. 4.050 L 4
C. 0.0008 g 1
D. 2.80 m 3
Solution
27. 27
Significant Figures in
Scientific Notation
In scientific notation all digits in the coefficient
including zeros are significant.
Number of
Measurement Significant Figures
8 x 104
m 1
8.0 x 104
m 2
8.00 x 104
m 3
28. 28
Study Tip: Significant Figures
The significant figures in a measured number are
• all the nonzero numbers.
12.56 m 4 significant figures
• zeros between nonzero numbers.
4.05 g 3 significant figures
• zeros that follow nonzero numbers in a decimal
number.
25.800 L 5 significant figures
29. 29
A. Which answer(s) contain 3 significant figures?
1) 0.4760 2) 0.00476 3) 4.76 x 103
B. All the zeros are significant in
1) 0.00307. 2) 25.300. 3) 2.050 x 103
.
C. The number of significant figures in 5.80 x 102
is
1) one (1). 2) two (2). 3) three (3).
Learning Check
30. 30
A. Which answer(s) contain 3 significant figures?
2) 0.00476 3) 4.76 x 103
B. All the zeros are significant in
2) 25.300. 3) 2.050 x 103
.
C. The number of significant figures in 5.80 x 102
is
3) three (3).
Solution
31. 31
In which set(s) do both numbers contain the
same number of significant figures?
1) 22.0 and 22.00
2) 400.0 and 40
3) 0.000 015 and 150 000
Learning Check
32. 32
Solution
In which set(s) do both numbers contain the same
number of significant figures?
3) 0.000 015 and 150 000
Both numbers contain 2 significant figures.
33. 33
Examples of Exact Numbers
An exact number is obtained
• when objects are counted.
Counted objects
2 soccer balls
4 pizzas
• from numbers in a defined relationship.
Defined relationships
1 foot = 12 inches
1 meter = 100 cm
35. 35
Learning Check
A. Exact numbers are obtained by
1. using a measuring tool.
2. counting.
3. definition.
B. Measured numbers are obtained by
1. using a measuring tool.
2. counting.
3. definition.
36. 36
Solution
A. Exact numbers are obtained by
2. counting.
3. definition.
B. Measured numbers are obtained by
1. using a measuring tool.
37. 37
Learning Check
Classify each of the following as (1) exact or (2) measured
numbers.
A.__Gold melts at 1064 °C.
B.__1 yard = 3 feet
C.__The diameter of a red blood cell is 6 x 10-4
cm.
D.__There are 6 hats on the shelf.
E.__A can of soda contains 355 mL of soda.
38. 38
Classify each of the following as (1) exact or (2) measured
numbers.
A. 2 A measuring tool is required.
B. 1 This is a defined relationship.
C. 2 A measuring tool is used to determine
length.
D. 1 The number of hats is obtained by counting.
E. 2 The volume of soda is measured.
Solution
41. 41
Rounding Off Calculated
Answers
When the first digit dropped is 4 or less,
• the retained numbers remain the same.
45.832 rounded to 3 significant figures
drops the digits 32 = 45.8
When the first digit dropped is 5 or greater,
• the last retained digit is increased by 1.
2.4884 rounded to 2 significant figures
drops the digits 884 = 2.5 (increase by 0.1)
42. 42
Adding Significant Zeros
• Sometimes a calculated answer requires more
significant digits. Then, one or more zeros are
added.
Calculated Zeros Added to
Answer Give 3 Significant Figures
4 4.00
1.5 1.50
0.2 0.200
12 12.0
43. 43
Learning Check
Round off or add zeros to the following calculated
answers to give three significant figures.
A. 824.75 cm
B. 0.112486 g
C. 8.2 L
44. 44
Solution
Adjust the following calculated answers to give answers
with 3 significant figures.
A. 825 cm First digit dropped is greater than 5.
B. 0.112g First digit dropped is 4.
C. 8.20 L Significant zero is added.
46. 46
When multiplying or dividing
• the final answer must have the same number of
significant figures as the measurement with the
fewest significant figures.
• use rounding rules to obtain the correct number of
significant figures.
Example:
110.5 x 0.048 = 5.304 = 5.3 (rounded)
4 SF 2 SF calculator 2 SF
Multiplication and Division
47. 47
Select the answer with the correct number of
significant figures.
A. 2.19 x 4.2 =
1) 9 2) 9.2 3) 9.198
B. 4.311 ÷ 0.07 =
1) 61.59 2) 62 3) 60
C. 2.54 x 0.0028 =
0.0105 x 0.060
1) 11.3 2) 11 3) 0.041
Learning Check
48. 48
A. 2.19 x 4.2 = 2) 9.2
B. 4.311 ÷ 0.07 = 3) 60
C. 2.54 x 0.0028 = 2) 11
0.0105 x 0.060
On a calculator, enter each number, followed by
the operation key.
2.54 x 0.0028 ÷ 0.0105 ÷ 0.060 = 11.28888889
= 11 (rounded)
Solution
49. 49
When adding or subtracting
• the final answer must have the same number of
decimal places as the measurement with the fewest
decimal places.
• use rounding rules to adjust the number of digits in
the answer.
25.2 one decimal place
+ 1.34 two decimal places
26.54 calculated answer
26.5 final answer with one decimal place
Addition and Subtraction
50. 50
For each calculation, round off the calculated answer
to give a final answer with the correct number of
significant figures.
A. 235.05 + 19.6 + 2 =
1) 257 2) 256.7 3) 256.65
B. 58.925 - 18.2 =
1) 40.725 2) 40.73 3) 40.7
Learning Check
53. 53
Prefixes
A prefix
in front of a unit increases or decreases the size of that unit.
makes units larger or smaller than the initial unit by one or more
factors of 10.
indicates a numerical value.
prefix = value
1 kilometer = 1000 meters
1 kilogram = 1000 grams
55. 55
Indicate the unit that matches the description.
1. A mass that is 1000 times greater than 1 gram.
1) kilogram 2) milligram 3) megagram
2. A length that is 1/100 of 1 meter.
1) decimeter 2) centimeter 3) millimeter
3. A unit of time that is 1/1000 of a second.
1) nanosecond 2) microsecond 3) millisecond
Learning Check
56. 56
Indicate the unit that matches the description.
1. A mass that is 1000 times greater than 1 gram.
1) kilogram
2. A length that is 1/100 of 1 meter.
2) centimeter
3. A unit of time that is 1/1000 of a second.
3) millisecond
Solution
57. 57
Select the unit you would use to measure
A. your height.
1) millimeters 2) meters 3) kilometers
B. your mass.
1) milligrams 2) grams 3) kilograms
C. the distance between two cities.
1) millimeters 2) meters 3) kilometers
D. the width of an artery.
1) millimeters 2) meters 3) kilometers
Learning Check
58. 58
A. your height.
2) meters
B. your mass.
3) kilograms
C. the distance between two cities.
3) kilometers
D. the width of an artery.
1) millimeters
Solution
59. 59
An equality
states the same measurement in two different units.
can be written using the relationships between two metric
units.
Example: 1 meter is the same as 100 cm and 1000 mm.
1 m = 100 cm
1 m = 1000 mm
Metric Equalities
63. 63
Indicate the unit that completes each of the following
equalities.
A. 1000 m = ___ 1) 1 mm 2) 1 km 2) 1 dm
B. 0.001 g = ___ 1) 1 mg 2) 1 kg 2) 1 dg
C. 0.1 s = ___ 1) 1 ms 2) 1 cs 2) 1 ds
D. 0.01 m = ___ 1) 1 mm 2) 1 cm 2) 1 dm
Learning Check
64. 64
Indicate the unit that completes each of the following
equalities.
A. 2) 1000 m = 1 km
B. 1) 0.001 g = 1 mg
C. 3) 0.1 s = 1 ds
D. 2) 0.01 m = 1 cm
Solution
65. 65
Complete each of the following equalities.
A. 1 kg = ___ 1) 10 g 2) 100 g 3) 1000 g
B. 1 mm = ___ 1) 0.001 m 2) 0.01 m 3) 0.1 m
Learning Check
66. 66
Complete each of the following equalities.
A. 1 kg = 1000 g (3)
B. 1 mm = 0.001 m (1)
Solution
68. 68
Equalities
• use two different units to describe the same measured
amount.
• are written for relationships between units of the metric
system, U.S. units, or between metric and U.S. units.
For example,
1 m = 1000 mm
1 lb = 16 oz
2.20 lb = 1 kg
Equalities
69. 69
Exact and Measured Numbers in
Equalities
Equalities between units in
• the same system of measurement are definitions
that use exact numbers.
• different systems of measurement (metric and U.S.)
use measured numbers that have significant figures.
Exception:
The equality 1 in. = 2.54 cm has been defined as an
exact relationship. Thus, 2.54 is an exact number.
72. 72
A conversion factor is
• obtained from an equality.
Equality: 1 in. = 2.54 cm
• written as a fraction (ratio) with a numerator and
denominator.
• inverted to give two conversion factors for every
equality.
1 in. and 2.54 cm
2.54 cm 1 in.
Conversion Factors
73. 73
Write conversion factors from the equality for each
of the following.
A. liters and mL
B. hours and minutes
C. meters and kilometers
Learning Check
74. 74
Write conversion factors from the equality for each of the
following.
A. 1 L = 1000 mL 1 L and 1000 mL
1000 mL 1 L
B. 1 h = 60 min 1 h and 60 min
60 min 1 h
C. 1 km = 1000 m 1 km and 1000 m
Solution
75. 75
A conversion factor
• may be obtained from information in a word problem.
• is written for that problem only.
Example 1:
The price of one pound (1 lb) of red peppers is $2.39.
1 lb red peppers and $2.39
$2.39 1 lb red peppers
Example 2:
The cost of one gallon (1 gal) of gas is $2.89.
1 gallon of gas and $2.89
Conversion Factors in a Problem
76. 76
A percent factor
• gives the ratio of the parts to the whole.
% = parts x 100
whole
• uses the same unit in the numerator and denominator.
• uses the value 100.
• can be written as two factors.
Example: A food contains 30% (by mass) fat.
30 g fat and 100 g food
100 g food 30 g fat
Percent as a Conversion Factor
78. 78
Smaller Percents: ppm and ppb
Small percents are shown as ppm and ppb.
• Parts per million (ppm) = mg part/kg whole
Example: The EPA allows 15 ppm cadmium in food
colors
15 mg cadmium = 1 kg food color
• Parts per billion ppb = µ g part/kg whole
Example: The EPA allows10 ppb arsenic in public
water
10 µ g arsenic = 1 kg water
79. 79
Arsenic in Water
Write the conversion factors for 10 ppb arsenic
in public water from the equality
10 µ g arsenic = 1 kg water.
Conversion factors:
10 µ g arsenic and 1 kg water
1 kg water 10 µ g arsenic
80. Study Tip: Conversion Factors
An equality
• is written as a fraction (ratio).
• provides two conversion factors that are the
inverse of each other.
80
81. 81
Learning Check
Write the equality and conversion factors for each of the
following.
A. meters and centimeters
B. jewelry that contains 18% gold
C. One gallon of gas is $2.89
82. 82
Solution
A. 1 m = 100 cm
1m and 100 cm
100 cm 1m
B. 100 g jewelry = 18 g gold
18 g gold and 100 g jewelry
100 g jewelry 18 g gold
C. 1 gal gas = $2.89
1 gal and $2.89
$2.89 1 gal
83. Risk-Benefit Assessment
A measurement of toxicity is
• LD50 or “lethal dose.”
• the concentration of the substance that causes death
in 50% of the test animals.
• in milligrams per kilogram (mg/kg or ppm) of body mass.
• in micrograms per kilogram (µ g/kg or ppb) of body
mass.
83
85. Solution
The LD50 for aspirin is 1100 ppm. How many grams of
aspirin would be lethal in 50% of persons with a body
mass of 85 kg?
B. 94 g
1100 ppm = 1100 mg/kg body mass
85
85 kg
1100 mg
×
kg
1 g
1000 mg
× = 94 g
87. 87
To solve a problem,
• identify the given unit.
• identify the needed unit.
Example:
A person has a height of 2.0 meters.
What is that height in inches?
The given unit is the initial unit of height.
given unit = meters (m)
The needed unit is the unit for the answer.
needed unit = inches (in.)
Given and Needed Units
88. 88
Learning Check
An injured person loses 0.30 pints of blood. How
many milliliters of blood would that be?
Identify the given and needed units given in this
problem.
Given unit = _______
Needed unit = _______
89. 89
Solution
An injured person loses 0.30 pints of blood. How
many milliliters of blood would that be?
Identify the given and needed units given in this
problem.
Given unit = pints
Needed unit = milliliters
90. 90
• Write the given and needed units.
• Write a plan to convert the given unit to the needed unit.
• Write equalities and conversion factors that connect the
units.
• Use conversion factors to cancel the given unit and
provide the needed unit.
Unit 1 x Unit 2 = Unit 2
Unit 1
Given x Conversion = Needed
unit factor unit
Problem Setup
91. 91
Study Tip: Problem Solving
Using GPS
The steps in the
Guide to Problem
Solving (GPS) are
useful in setting up
a problem with
conversion factors.
93. 93
A rattlesnake is 2.44 m long. How many cm long is
the snake?
1) 2440 cm
2) 244 cm
3) 24.4 cm
Learning Check
94. 94
A rattlesnake is 2.44 m long. How many cm long
is the snake?
2) 244 cm
Given Conversion Needed
unit factor unit
2.44 m x 100 cm = 244 cm
1 m
Solution
95. 95
• Often, two or more conversion factors are required
to obtain the unit needed for the answer.
Unit 1 Unit 2 Unit 3
• Additional conversion factors are placed in the
setup problem to cancel each preceding unit.
Given unit x factor 1 x factor 2 = needed unit
Unit 1 x Unit 2 x Unit 3 = Unit 3
Unit 1 Unit 2
Using Two or More Factors
96. 96
How many minutes are in 1.4 days?
Given unit: 1.4 days
Factor 1 Factor 2
Plan: days h min
Set Up Problem:
1.4 days x 24 h x 60 min = 2.0 x 103
min
1 day 1 h (rounded)
2 SF Exact Exact = 2 SF
Example: Problem Solving
97. 97
• Be sure to check the unit cancellation in the setup.
• The units in the conversion factors must cancel to
give the correct unit for the answer.
What is wrong with the following setup?
1.4 day x 1 day x 1 h
24 h 60 min
= day2
/min is not the unit needed
Units don’t cancel properly.
Study Tip: Check Unit
Cancellation
98. 98
A bucket contains 4.65 L of water. Write the setup
for the problem and calculate the gallons of water in
the bucket.
Plan: L qt gallon
Equalities: 1.06 qt = 1 L
1 gal = 4 qt
Set Up Problem:
4.65 L x x 1.06 qt x 1 gal = 1.23 gal
1 L 4 qt
Learning Check
99. 99
Given: 4.65 L Needed: gallons
Plan: L qt gallon
Equalities: 1.06 qt = 1 L; 1 gal = 4 qt
Set Up Problem:
4.65 L x x 1.06 qt x 1 gal = 1.23 gal
1 L 4 qt
3 SF 3 SF exact 3 SF
Solution
100. 100
If a ski pole is 3.0 feet in length, how long is the
ski pole in mm?
Learning Check
101. 101
Equalities:
1 ft = 12 in. 1 in. = 2.54 cm 1 cm = 10 mm
Set Up Problem:
3.0 ft x 12 in. x 2.54 cm x 10 mm =
1 ft 1 in. 1 cm
Calculator answer = 914.4 mm
Final answer = 910 mm
(2 SF rounded)
Check Factors in Setup: Units cancel properly
Check Final Unit: mm
Solution
102. 102
If your pace on a treadmill is 65 meters per minute,
how many minutes will it take for you to walk a
distance of 7500 feet?
Learning Check
103. 103
Solution
Given: 7500 ft 65 m/min Need: min
Plan: ft in. cm m min
Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm
1 min = 65 m (walking pace)
Set Up Problem:
7500 ft x 12 in. x 2.54 cm x 1m x 1 min
1 ft 1 in. 100 cm 65 m
= 35 min final answer (2 SF)
Given: 7500 ft 65 m/min Need: min
Plan: ft in. cm m
Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm
1 min = 65 m (walking pace)
Set Up Problem:
7500 ft x 12 in. x 2.54 cm x 1m x 1 min
1 ft 1 in. 100 cm 65 m
= 35 min final answer (2 SF)
105. 105
How many lb of sugar are in 120 g of candy if the
candy is 25% (by mass) sugar?
Learning Check
106. 106
Solution
How many lb of sugar are in 120 g of candy if the
candy is 25%(by mass) sugar?
percent factor
120 g candy x 1 lb candy x 25 lb sugar
454 g candy 100 lb candy
= 0.066 lb of sugar
108. 108
Density
• compares the mass of an object to its volume.
• is the mass of a substance divided by its
volume.
Density Expression
Density = mass = g or g = g/cm3
volume mL cm3
Note: 1 mL = 1 cm3
Density
110. 110
Osmium is a very dense metal. What is its density
in g/cm3
if 50.0 g of osmium has a volume of 2.22
cm3
?
1) 2.25 g/cm3
2) 22.5 g/cm3
3) 111 g/cm3
Learning Check
111. 111
Given: mass = 50.0 g volume = 2.22 cm3
Plan: Place the mass and volume of the osmium
metal
in the density expression.
2) D = mass = 50.0 g
volume 2.22 cm3
calculator answer = 22.522522 g/cm3
final answer = 22.5 g/cm3
Solution
114. 114
What is the density (g/cm3
) of 48.0 g of a metal if the
level of water in a graduated cylinder rises from 25.0
mL to 33.0 mL after the metal is added?
1) 0.17 g/cm3
2) 6.0 g/cm3
3) 380 g/cm3
Learning Check
object
33.0 mL25.0
mL
115. 115
Solution
Given: 48.0 g Volume of water = 25.0 mL
Volume of water + metal = 33.0 mL
Need: Density (g/mL)
Plan: Calculate the volume difference in cm3
and place
in density expression.
33.0 mL - 25.0 mL = 8.0 mL
8.0 mL x 1 cm3
= 8.0 cm3
1 mL
Set Up Problem:
Density = 48.0 g = 6.0 g = 6.0 g/cm3
8.0 cm3
1cm3
117. 117
Which diagram correctly represents the liquid layers
in the cylinder? Karo (K) syrup (1.4 g/mL); vegetable
(V) oil (0.91 g/mL); water (W) (1.0 g/mL)
1 2 3
K
K
W
W
W
V
V
V
K
Learning Check
119. 119
The density of octane, a component of gasoline, is
0.702 g/mL. What is the mass, in kg, of 875 mL of
octane?
1) 0.614 kg
2) 614 kg
3) 1.25 kg
Learning Check
120. 120
Density can be written as an equality.
• For a substance with a density of 3.8 g/mL, the
equality is
3.8 g = 1 mL
• From this equality, two conversion factors can be
written for density.
Conversion 3.8 g and 1 mL
factors 1 mL 3.8 g
Study Tip: Density as a
Conversion Factor
121. 121
Solution
1) 0.614 kg
Given: D = 0.702 g/mL V= 875 mL
Plan: mL → g → kg
Equalities: density 0.702 g = 1 mL
and 1 kg = 1000 g
Setup: 875 mL x 0.702 g x 1 kg = 0.614 kg
1 mL 1000 g
density metric
factor factor
122. 122
If olive oil has a density of 0.92 g/mL, how many
liters of olive oil are in 285 g of olive oil?
1) 0.26 L
2) 0.31 L
3) 310 L
Learning Check
123. 123
Solution
2) 0.31 L
Given: D = 0.92 g/mL mass = 285 g
Need: volume in liters
Plan: g → mL → L
Equalities: 1 mL = 0.92 g and 1 L = 1000 mL
Set Up Problem:
285 g x 1 mL x 1 L = 0.31 L
0.92 g 1000 mL
density metric
factor factor
inverted
124. 124
A group of students collected 125 empty aluminum
cans to take to the recycling center. If 21 cans
make 1.0 lb aluminum, how many liters of
aluminum (D=2.70 g/cm3
) are obtained from the
cans?
1) 1.0 L 2) 2.0 L 3) 4.0 L
Learning Check
125. 125
Solution
1) 1.0 L
125 cans x 1.0 lb x 454 g x 1 cm3
x 1 mL x 1 L
21 cans 1 lb 2.70 g 1 cm3
1000 mL
density
factor
inverted
= 1.0 L
126. 126
Which of the following samples of metals will displace
the greatest volume of water?
1 2 3
25 g of aluminum
2.70 g/mL
45 g of gold
19.3 g/mL
75 g of lead
11.3 g/mL
Learning Check
127. 127
Solution
25 g of aluminum
2.70 g/mL
1)
Plan: Calculate the volume for each metal and select
the metal sample with the greatest volume.
1) 25 g x 1 mL = 9.3 mL aluminum
2.70 g
2) 45 g x 1 mL = 2.3 mL gold
19.3 g
3) 75 g x 1 mL = 6.6 mL lead
11.3 g
density
factors
