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Chapter 1
Lecture Presentation
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Matter and
Measurement
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Chemistry
In this science we studymatter, its properties,
and its behavior.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
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Matter
We define matter as anything that has mass and takes
up space.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
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Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
Atoms are the building blocks of matter.
Each element is made of the same kind of atom.
A compound is made of two or more different kinds of
elements.
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States of Matter
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
2012 Pearson Education, Inc.
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
2012 Pearson Education, Inc.
Matter
And
Measurement
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Classification of Matter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
2012 Pearson Education, Inc.
Matter
And
Measurement
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Sample Exercise 1.1 Distinguishing among Elements, Compounds,
and Mixtures
Solution
White gold contains gold and a white metal, such as palladium. Two samples of white gold differ in the relative
amounts of gold and palladium they contain. Both samples are uniform in composition throughout. Use Figure 1.9
to classify white gold.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Aspirin is composed of 60.0% carbon, 4.5%hydrogen, and 35.5% oxygen by mass, regardless
of its source. Use Figure 1.9 to classify aspirin.
Answer:
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Sample Exercise 1.1 Distinguishing among Elements, Compounds,and Mixtures
Because the material is uniform throughout, it ishomogeneous. Because its composition differs for
the two samples, it cannot be a compound.
Instead, it must be a homogeneous mixture.
Solution
White gold contains gold and a white metal, such as palladium. Two samples of white gold differ in the relative
amounts of gold and palladium they contain. Both samples are uniform in composition throughout. Use Figure 1.9
to classify white gold.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Aspirin is composed of 60.0% carbon, 4.5%hydrogen, and 35.5% oxygen by mass, regardless
of its source. Use Figure 1.9 to classify aspirin.
Answer:
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Sample Exercise 1.1 Distinguishing among Elements, Compounds,and Mixtures
Because the material is uniform throughout, it ishomogeneous. Because its composition differs for
the two samples, it cannot be a compound.
Instead, it must be a homogeneous mixture.
Solution
White gold contains gold and a white metal, such as palladium. Two samples of white gold differ in the relative
amounts of gold and palladium they contain. Both samples are uniform in composition throughout. Use Figure 1.9
to classify white gold.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Aspirin is composed of 60.0% carbon, 4.5%hydrogen, and 35.5% oxygen by mass, regardless
of its source. Use Figure 1.9 to classify aspirin.
Answer: It is a compound because it has constant
composition and can be separated into several
elements.
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Properties and
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Changes of Matter
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Types of Properties Physical Properties
Can be observed without changing a substanceinto another substance.
Boiling point, density, mass, volume, etc.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Chemical Properties Can only be observed when a substance is
changed into another substance.
Flammability, corrosiveness, reactivity with acid,etc.
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Types of Properties
Intensive Properties
Are independent of the amount of the substance
that is present.
Densit boilin oint color etc.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Extensive Properties
Depend upon the amount of the substance
present.
Mass, volume, energy, etc.
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Types of Changes
Physical Changes
These are changes in matter that do not change thecomposition of a substance.
Chan es of state tem erature volume etc.
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Chemical Changes
Chemical changes result in new substances.
Combustion, oxidation, decomposition, etc.
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Chemical Reactions
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In the course of a chemical reaction, the reacting
substances are converted to new substances.
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Separation of
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Mixtures
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Filtration
In filtration, solid
substances are separated
from liquids and solutions.
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Distillation
Distillation uses
differences in the
boiling points of
substances to se arate a
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
homogeneous mixtureinto its components.
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ChromatographyThis technique separates substances on the basis of
differences in solubility in a solvent.
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Units of
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Measurement
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SI Units
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Systme International dUnits
A different base unit is used for each quantity.
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Metric SystemPrefixes convert the base units into units that are
appropriate for the item being measured.
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Sample Exercise 1.2 Using SI Prefixes
Solution
What is the name of the unit that equals (a) 109 gram, (b) 106 second, (c) 103 meter?
(a) How many picometers are there in one meter?
(b) Express 6.0 103 m using a prefix to replace the
power of ten. (c) Use exponential notation to
Practice Exercise
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
express 4.22 mg in grams. (d) Use decimal notation
to express 4.22 mg in grams.
Answers:
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Sample Exercise 1.2 Using SI Prefixes
We can find the prefix related to each power of ten in
Table 1.5: (a) nanogram, ng, (b) microsecond, s,
(c) millimeter, mm.
Solution
What is the name of the unit that equals (a) 109 gram, (b) 106 second, (c) 103 meter?
(a) How many picometers are there in one meter?
(b) Express 6.0 103 m using a prefix to replace the
power of ten. (c) Use exponential notation to
Practice Exercise
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
express 4.22 mg in grams. (d) Use decimal notation
to express 4.22 mg in grams.
Answers:
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Volume
The most commonly usedmetric units for volume are
the liter (L) and the milliliter
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
.
A liter is a cube
1 decimeter (dm) long on
each side.
A milliliter is a cube1 centimeter (cm) long on
each side.
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Temperature
By definitiontemperature is a
measure of the average
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particles in a sample.
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Temperature In scientific measurements,
the Celsius and Kelvin
scales are most often used. The Celsius scale is based
on the properties of water.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
0C is the freezing pointof water.
100 C is the boiling point
of water.
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Temperature
The kelvin is the SI unit
of temperature.
It is based on the
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.
There are no negative
Kelvin temperatures.
K = C + 273.15
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Temperature
The Fahrenheit scale is
not used in scientificmeasurements.
=
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C = 5/9(F 32)
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Sample Exercise 1.3 Converting Units of Temperature
(a)
(b)
Solution
A weather forecaster predicts the temperature will reach 31 C. What is this temperature (a) in K, (b) in F?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5 C. What is the freezing point in (a) K,
(b) F?
Answers:
Practice Exercise
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Sample Exercise 1.3 Converting Units of Temperature
(a) Using Equation 1.1, we have K = 31 + 273 = 304 K.
(b) Using Equation 1.2, we have
Solution
A weather forecaster predicts the temperature will reach 31 C. What is this temperature (a) in K, (b) in F?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5 C. What is the freezing point in (a) K,
(b) F?
Answers:
Practice Exercise
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Sample Exercise 1.3 Converting Units of Temperature
(a) Using Equation 1.1, we have K = 31 + 273 = 304 K.
(b) Using Equation 1.2, we have
Solution
A weather forecaster predicts the temperature will reach 31 C. What is this temperature (a) in K, (b) in F?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5 C. What is the freezing point in (a) K,
(b) F?
Answers: (a) 261.7 K, (b) 11.3 F
Practice Exercise
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Derived Units
Density is a physical property of asubstance.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
,
derived from the units for mass and volume.
d =
m
V
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Sample Exercise 1.4 Determining Density and Using Density toDetermine Volume or Mass
(a) We are given mass and volume, so Equation
1.3 yields
Solution
(a) Calculate the density of mercury if 1.00 102 g occupies a volume of 7.36 cm3.(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.
(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm3) if the length of the cube is 2.00 cm?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
(b) So v ng Equat on 1.3 or vo ume an then
using the given mass and density gives
(c) We can calculate the mass from the volume of
the cube and its density. The volume of a cube
is given by its length cubed:
Solving Equation 1.3 for mass and
substituting the volume and density of the
cube, we have
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Sample Exercise 1.4 Determining Density and Using Density toDetermine Volume or Mass
(a) We are given mass and volume, so Equation
1.3 yields
Solution
(a) Calculate the density of mercury if 1.00 102 g occupies a volume of 7.36 cm3.(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.
(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm3) if the length of the cube is 2.00 cm?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
(b) So v ng Equat on 1.3 or vo ume an then
using the given mass and density gives
(c) We can calculate the mass from the volume of
the cube and its density. The volume of a cube
is given by its length cubed:
Solving Equation 1.3 for mass and
substituting the volume and density of the
cube, we have
Volume = (2.00 cm)3 = (2.00)3 cm3 = 8.00 cm3
Mass = volume density = (8.00 cm3)(19.32 g/cm3) = 155 g
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Sample Exercise 1.4 Determining Density and Using Density toDetermine Volume or Mass
Continued
(a) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm3. (b) A student needs
15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol
are needed? (c) What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?
Answers:
Practice Exercise
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
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Uncertainty in
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Measurement
U i i M
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Uncertainty in Measurements
Different measuring devices have different uses and
different degrees of accuracy.
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Significant Figures
The term significant figures refers to digitsthat were measured.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
,
attention to significant figures so we do notoverstate the accuracy of our answers.
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Significant Figures When addition or subtraction is performed,
answers are rounded to the least significantdecimal place.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
performed, answers are rounded to thenumber of digits that corresponds to the
leastnumber of significant figures in any
of the numbers used in the calculation.
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Sample Exercise 1.5 Relating Significant Figures to the Uncertainty
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Sample Exercise 1.5 Relating Significant Figures to the Uncertaintyof a Measurement
The value 4.0 has two significant figures, whereas 4.00 has three. This difference implies that the 4.0 has
more uncertainty. A mass reported as 4.0 g indicates that the uncertainty is in the first decimal place. Thus,
the mass might be anything between 3.9 and 4.1 g, which we can represent as 4.0 0.1 g. A mass reported
as 4.00 g indicates that the uncertainty is in the second decimal place. Thus, the mass might be anythingbetween 3.99 and 4.01 g, which we can represent as 4.00 0.01 g. (Without further information, we cannot
be sure whether the difference in uncertainties of the two measurements reflects the precision or the
accuracy of the measurement.)
Solution
What difference exists between the measured values 4.0 g and 4.00 g?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
A sample that has a mass of about 25 g is placed on a balance that has a precision of 0.001 g. How manysignificant figures should be reported for this measurement?
Answer: five, as in the measurement 24.995 g, the uncertainty being in the third decimal place
Sample Exercise 1.6 Determining the Number of Significant Figures
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p g g gin a Measurement
Solution
How many significant figures are in each of the following numbers (assume that each number is a measuredquantity): (a) 4.003, (b) 6.023 1023, (c) 5000?
How many significant figures are in each of the following measurements: (a) 3.549 g, (b) 2.3 104 cm,3
Practice Exercise
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
.
Answers:
Sample Exercise 1.6 Determining the Number of Significant Figures
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p g g gin a Measurement
(a) Four; the zeros are significant figures. (b) Four; the exponential term does not add to the number of
significant figures. (c) One; we assume that the zeros are not significant when there is no decimal point
shown. If the number has more significant figures, a decimal point should be employed or the number
written in exponential notation. Thus, 5000. has four significant figures, whereas 5.00103 has three.
Solution
How many significant figures are in each of the following numbers (assume that each number is a measuredquantity): (a) 4.003, (b) 6.023 1023, (c) 5000?
How many significant figures are in each of the following measurements: (a) 3.549 g, (b) 2.3 104 cm,3
Practice Exercise
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
.
Answers: (a) four, (b) two, (c) three
Sample Exercise 1.7 Determining the Number of Significant Figures
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p g g gin a Calculated Quantity
In reporting the volume, we can show only as many significant figures as given in the dimension with the
fewest significant figures, that for the height (two significant figures):
Solution
The width, length, and height of a small box are 15.5 cm, 27.3 cm, and 5.4 cm, respectively. Calculate the volumeof the box, using the correct number of significant figures in your answer.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
It takes 10.5 s for a sprinter to run 100.00 m. Calculate her average speed in meters per second, and express
the result to the correct number of significant figures.
Answer:
Practice Exercise
Sample Exercise 1.7 Determining the Number of Significant Figures
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p g g gin a Calculated Quantity
In reporting the volume, we can show only as many significant figures as given in the dimension with the
fewest significant figures, that for the height (two significant figures):
A calculator used for this calculation shows 2285.01, which we must round off to two significant figures.
Because the resulting number is 2300, it is best reported in exponential notation, 2.3 103, to clearly
Solution
The width, length, and height of a small box are 15.5 cm, 27.3 cm, and 5.4 cm, respectively. Calculate the volumeof the box, using the correct number of significant figures in your answer.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
n cate two s gn cant gures.
It takes 10.5 s for a sprinter to run 100.00 m. Calculate her average speed in meters per second, and express
the result to the correct number of significant figures.
Answer: 9.52 m/s (three significant figures)
Practice Exercise
Sample Exercise 1.8 Determining the Number of Significant Figures
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in a Calculated Quantity
To calculate the density, we must know both the mass and the volume of the gas. The mass of the gas is just
the difference in the masses of the full and empty container:
(837.6 836.2) g = 1.4 gIn subtracting numbers, we determine the number of significant figures in our result by counting decimal
places in each quantity. In this case each quantity has one decimal place. Thus, the mass of the gas, 1.4 g,
has one decimal place.
Solution
A gas at 25 C fills a container whose volume is 1.05 103 cm3. The container plus gas has a mass of 837.6 g. The
container, when emptied of all gas, has a mass of 836.2 g. What is the density of the gas at 25 C?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
s ng t e vo ume g ven n t e quest on, . cm , an t e e n t on o ens ty, we ave
Sample Exercise 1.8 Determining the Number of Significant Figures
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in a Calculated Quantity
To calculate the density, we must know both the mass and the volume of the gas. The mass of the gas is just
the difference in the masses of the full and empty container:
(837.6 836.2) g = 1.4 gIn subtracting numbers, we determine the number of significant figures in our result by counting decimal
places in each quantity. In this case each quantity has one decimal place. Thus, the mass of the gas, 1.4 g,
has one decimal place.
Solution
A gas at 25 C fills a container whose volume is 1.05 103 cm3. The container plus gas has a mass of 837.6 g. The
container, when emptied of all gas, has a mass of 836.2 g. What is the density of the gas at 25 C?
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
s ng t e vo ume g ven n t e quest on, . cm , an t e e n t on o ens ty, we ave
In dividing numbers, we determine the number of significant figures in our result by counting the number of
significant figures in each quantity. There are two significant figures in our answer, corresponding to the
smaller number of significant figures in the two numbers that form the ratio. Notice that in this example,following the rules for determining significant figures gives an answer containing only two significant
figures, even though each of the measured quantities contained at least three significant figures.
Sample Exercise 1.8 Determining the Number of Significant Figuresi C l l d Q i
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in a Calculated QuantityContinued
To how many significant figures should the mass of the container be measured (with and without the gas) in
Sample Exercise 1.8 for the density to be calculated to three significant figures?
Answer:
Practice Exercise
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Accuracy versus Precision
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Accuracy versus Precision
Accuracy refers to the proximity of a
measurement to the true value of a
quantity.
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rec s on re ers o e prox m y o
several measurements to each other.
Dimensional Analysis
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Dimensional Analysis
We use dimensional analysis to convert one
quantity to another.
Most commonly, dimensional analysis utilizesconversion factors (e.g., 1 in. = 2.54 cm)
1 in. 2.54 cm
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
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2.54 cm
1 in.
or
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Dimensional AnalysisUse the form of the conversion factor that
puts the sought-for unit in the numerator:
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
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Given unit = desired unitdesired unit
given unit
Conversion factor
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Dimensional Analysis For example, to convert 8.00 m to inches,
convert m to cm
convert cm to in.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
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8.00 m 100 cm
1 m
1 in.
2.54 cm= 315 in.
Sample Exercise 1.9 Converting Units
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Because we want to change from pounds to grams, we look for a relationship between
these units of mass. From the back inside cover we have 1 lb = 453.6 g. To cancel pounds
and leave grams, we write the conversion factor with grams in the numerator and pounds
in the denominator:
Solution
If a woman has a mass of 115 lb, what is her mass in grams? (Use the relationships between units given on the back
inside cover of the text.)
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0-mi
automobile race. 1 km = 0.62137 mi
Answer:
Practice Exercise
Sample Exercise 1.9 Converting Units
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Because we want to change from pounds to grams, we look for a relationship between
these units of mass. From the back inside cover we have 1 lb = 453.6 g. To cancel pounds
and leave grams, we write the conversion factor with grams in the numerator and pounds
in the denominator:
Solution
If a woman has a mass of 115 lb, what is her mass in grams? (Use the relationships between units given on the back
inside cover of the text.)
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
The answer can be given to only three significant figures, the number of significant
figures in 115 lb. The process we have used is diagrammed in the margin.
By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0-mi
automobile race. 1 km = 0.62137 mi
Answer: 804.7 km
Practice Exercise
Sample Exercise 1.10 Converting Units Using Two or MoreConversion Factors
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Conversion Factors
To go from the given units, m/s, to the desired units, mi/hr, we must convert meters to miles and seconds to
hours. From our knowledge of SI prefixes we know that 1 km = 103 m. From the relationships given on the
back inside cover of the book, we find that 1 mi = 1.6093 km. Thus, we can convert m to km and then
convert km to mi. From our knowledge of time we know that 60s = 1 min and 60 min = 1 hr. Thus, we can
convert s to min and then convert min to hr. The overall process is
SolutionThe average speed of a nitrogen molecule in air at 25
C is 515 m/s. Convert this speed to miles per hour.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Applying first the conversions for distance and then those for time, we can set up one long equation inwhich unwanted units are canceled:
Sample Exercise 1.10 Converting Units Using Two or MoreConversion Factors
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To go from the given units, m/s, to the desired units, mi/hr, we must convert meters to miles and seconds to
hours. From our knowledge of SI prefixes we know that 1 km = 103 m. From the relationships given on the
back inside cover of the book, we find that 1 mi = 1.6093 km. Thus, we can convert m to km and then
convert km to mi. From our knowledge of time we know that 60s = 1 min and 60 min = 1 hr. Thus, we can
convert s to min and then convert min to hr. The overall process is
SolutionThe average speed of a nitrogen molecule in air at 25
C is 515 m/s. Convert this speed to miles per hour.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
Applying first the conversions for distance and then those for time, we can set up one long equation inwhich unwanted units are canceled:
Sample Exercise 1.11 Converting Volume UnitsEarths oceans contain approximately 1 36 109 km3 of water Calculate the volume in liters
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From the back inside cover, we find 1 L = 103 m3, but there is no relationship listed involving km3. From
our knowledge of SI prefixes, however, we know 1 km = 103 m and we can use this relationship between
lengths to write the desired conversion factor between volumes:
Thus, converting from km3 to m3 to L, we have
Solution
Earth s oceans contain approximately 1.36 109 km3 of water. Calculate the volume in liters.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
If the volume of an object is reported as 5.0 ft3, what is the volume in cubic meters?
Answer:
Practice Exercise
Sample Exercise 1.11 Converting Volume UnitsEarths oceans contain approximately 1 36 109 km3 of water Calculate the volume in liters
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From the back inside cover, we find 1 L = 103 m3, but there is no relationship listed involving km3. From
our knowledge of SI prefixes, however, we know 1 km = 103 m and we can use this relationship between
lengths to write the desired conversion factor between volumes:
Thus, converting from km3 to m3 to L, we have
Solution
Earth s oceans contain approximately 1.36 109 km3 of water. Calculate the volume in liters.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
If the volume of an object is reported as 5.0 ft3, what is the volume in cubic meters?
Answer: 0.14 m3
Practice Exercise
Sample Exercise 1.12 Conversions Involving DensityWhat is the mass in grams of 1 00 gal of water? The density of water is 1 00 g/mL
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Before we begin solving this exercise, we note the following:
1. We are given 1.00 gal of water (the known, or given, quantity) and asked to calculate its mass in
grams (the unknown).
2. We have the following conversion factors either given, commonly known, or available on the back
inside cover of the text:
Solution
What is the mass in grams of 1.00 gal of water? The density of water is 1.00 g/mL.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
desired result, whereas the last conversion factor must be inverted in order to cancel gallons:
Sample Exercise 1.12 Conversions Involving DensityWhat is the mass in grams of 1 00 gal of water? The density of water is 1 00 g/mL
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Before we begin solving this exercise, we note the following:
1. We are given 1.00 gal of water (the known, or given, quantity) and asked to calculate its mass in
grams (the unknown).
2. We have the following conversion factors either given, commonly known, or available on the back
inside cover of the text:
Solution
What is the mass in grams of 1.00 gal of water? The density of water is 1.00 g/mL.
2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition
Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward
desired result, whereas the last conversion factor must be inverted in order to cancel gallons:
The unit of our final answer is appropriate, and weve taken care of our significant figures. We can further
check our calculation by estimating. We can round 1.057 off to 1. Then focusing on the numbers that do not
equal 1 gives 4 1000 = 4000 g, in agreement with the detailed calculation.
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