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## Module 1: Basic Concepts Matter, Mathematical Manipulation, Dimensional Analysis, Density, Specific

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Title: Module 1: Basic Concepts Matter, Mathematical Manipulation, Dimensional Analysis, Density, Specific

1
Module 1 Basic Concepts Matter, Mathematical
Manipulation, Dimensional Analysis, Density,
Specific Gravity, Temperature, and Heat Transfer
• By Alyssa Jean-Mary
• Source Modular Study Guide for First Semester
Chemistry by Anthony J. Papaps and Marta E.
Goicoechea-Pappas

2
Basic Ideas What is Chemistry?
• Chemistry is the science that studies the
composition, characterization, and transformation
of matter
• It started out as alchemy.

3
Basic Ideas What is the Scientific Method?
• The Scientific Method is a systematic approach to
research that involves collecting and analyzing
data.
• The first step is to formulate a problem.
• The second step is to make observations and
conduct experiments on that problem.
• The third step is to interpret the observations
and experiments from the second step.
• Here, a hypothesis is first formed. A hypothesis
is a tentative explanation for a set of
observations that is made after the collection of
enough information.
• After the formation of a hypothesis, a more
definite explanation is developed a law or a
theory is made. A law is a concise verbal or
mathematical statement of a relationship between
phenomena that is always the same under the same
conditions that is made after the collection of a
large amount of information. A theory is a step
further than a law. It is a unifying principle
that explains a body of facts and those laws that
are based on them.
• The fourth step is testing the interpretation,
experiments support the interpretation made is
the thrid step.

4
Basic Ideas What is Matter and how is it
classified?
5
Basic Ideas States of Matter
• The physical state that a sample of matter exists
in depends on the temperature
• There are three physical states of matter
• Solid
• Solids have a definite shape and volume, a high
density, and are virtually incompressible.
• Liquid
• Liquids have a definite volume, but assume the
shape of the container they are in. They also
have a high density like solids, but are slightly
compressible unlike solids.
• Gas
• Gases assume the volume and shape of the
container they are in (i.e. no definite volume or
shape), have a low density unlike solids and
liquids, and are very compressible.
• Solids have stronger forces of attraction and are
more ordered than liquids, and liquids, in turn,
have stronger forces of attraction and are more
ordered than gases.

6
Basic Ideas Conversion between States of Matter
• Matter can convert between the three physical
states by changes in temperature and pressure
• Gas to Liquid cool or increase pressure
Condensation
• Gas to Solid cool Deposition
• Liquid to Gas heat or decrease pressure
Vaporization (Evaporation) (Boiling)
• Liquid to Solid cool Freezing
(Solidification) (Crystallization)
• Solid to Gas heat Sublimation
• Solid to Liquid heat Melting (Fusion)

7
Basic Ideas Common Elements to Know
8
Basic Ideas Types of Solutions (i.e. -
Homogeneous Mixtures)
• Every solution has two components a solute and a
solvent.
• The solute(s) are the component(s) that are
present in the lesser amount.
• The solvent is the component that is present in
the greatest amount.
• Liquid solutions are the most common, but there
are also gas solutions and solid solutions. For
example, an alloy is a solid solution. It is a
homogeneous mixture of metals.

9
Basic Ideas Characteristics of Solutions
• Solutions
• have uniform distribution.
• have components that do not separate upon
standing.
• have components that cannot be separated by
filtration.
• only vary their compositions within certain
limits.
• are almost always transparent (i.e. you cannot
see through it).

10
Basic Ideas Energy
• Energy is the capacity to do work or transfer
heat. Two forms of energy are potential energy
(PE) and kinetic energy (KE).
• Potential energy is the energy an object
possesses because of its position or composition.
This is the kind of energy that is found in
chemicals. For example, natural gas and gasoline
contain PE.
• Kinetic energy is the energy of motion.

11
Basic Ideas Properties and Changes
• Properties
• A physical property is a property that can be
observed in the absence of any change in
composition. Some examples are color, odor,
taste, melting point, boiling point, freezing
point, density, length, and specific heat.
• A chemical property is a property that matter
exhibits as it undergoes changes in composition.
Some examples are that coal and gasoline burn in
air to form carbon dioxide and water, iron reacts
with oxygen in the air to form rust, and bleach
turns hair blonde.
• Changes
• A physical change is a change that is observed
without a change in composition. Some examples
are cutting wood, melting a solid, and boiling a
liquid.
• A chemical change is a change that is observed
only when a change in composition is occurring.
Some examples are the burning of wood, the
rusting of iron, and the dying of hair.

12
• An extensive property is a property that depends
on the amount of material present. Some examples
are volume and mass.
• An intensive property is a property that does not
depend on the amount of material present. Some
examples are melting point, boiling point,
freezing point, color, and density.

13
Basic Ideas Laws
• Law of Conservation of Mass There is no
observable change in the quantity of matter
during an ordinary chemical reaction. For
example, 58.7g of Ni and 12.0g of C make 70.7g
NiC when reacted together.
• Law of Conservation of Energy Energy cannot be
created or destroyed, but it can be converted
from one form to another. For example, potential
energy can be converted to kinetic energy, but if
you have 10kJ of potential energy, then you are
going to get 10kJ of kinetic energy.
• Law of Definite Proportions Different samples of
any pure compound contain the same elements, each
in the same proportion by mass. For example, all
samples of water (H2O) contain 11.1 hydrogen (H)
by mass and 88.9 oxygen (O) by mass.

14
Basic Ideas Accuracy vs. Precision
• Accuracy tells us how close a measurement is to
the true value.
• Precision tells us how closely two or more
measurements agree with one another.
• For example, Say you measured the density of a
substance in two separate trials to be 1.05g/mL
and 0.998g/mL, and the density of the substance
is actually 1.00g/mL. The accuracy of the
measurement is obtained by comparing it to
1.00g/mL, the actual density, whereas the
precision of the measurement is obtained by
comparing the 1.05g/mL to the 0.998g/mL.

15
Rounding Off Numbers
• The rounding off of a number depends on the
identity of the digit after the cut-off point
(i.e. the next digit). The digit before the
cut-off point is known as the previous digit.
The rules for rounding off are
• Rule 1 If the next digit is less than 5, then
the previous digit remains the same.
• Rule 2 If the next digit is greater than 5 or
5 followed by nonzeros, then the previous digit
is increased by one.
• Rule 3 If the next digit is 5 or 5 followed by
all zeros, then the previous digit
• A. remains the same if it is even OR
• B. is increased by one if it is odd.
• Steps to Round Off a Number
• Step 1 Identify the next digit.
• Step 2 Based on the next digit, identify the
rule that should be followed for rounding off.
• Step 3 Based on the rule, either keep the
previous digit the same or increase it by one.

16
Examples of Rounding Off Numbers
• Example 1 Round off 2.637 to the second decimal
point.
• Step 1 The next digit is 7.
• Step 2 Since 7 is greater than 5, rule 2
applies.
• Step 3 Since rule 2 applies, the previous
digit, 3, is increased by one, making it 4. So,
the number rounded off is 2.64.
• Example 2 Round off 4.45120 to the second
decimal point.
• Step 1 The next digit is 1.
• Step 2 Since 1 is less than 5, rule 1 applies.
• Step 3 Since rule 1 applies, the previous
digit remains the same. So, the number rounded
off is 4.45.
• Example 3 Round off 1.67500 to the second
decimal point.
• Step 1 The next digit is 5 followed by zeros.
• Step 2 Since the next digit is a 5 followed by
zeros, rule 3 applies. And since the previous
digit, 7, is odd, part B of rule 3 applies.
• Step 3 Since part B of rule 3 applies, the
previous digit, 7, is increased by one, making
it 8. So, the number rounded off is 1.68.

17
Scientific Notation
• The general from of scientific notation is
• N x 10e
• N is a number that is between /- 1 and /- 9. e
is an exponent (i.e. a power of ten), and thus is
always a whole number.
• Some examples are 4.5 x 1011 or 8.7 x 10-6.
• Steps to convert a number into scientific
notation
• Step 1 If your number is not in scientific
• Step 2 Convert your number so that it is between
/- 1 and /- 9.
• If you need to move the decimal place to the left
to do this, then the exponent needs to be
increased by the same amount.
• If you need to move the decimal place to the
right to do this, then the exponent needs to be
decreased by the same amount.

18
Examples of Scientific Notation
• Example 1 What is 425630.5 in scientific
notation?
• Step 1 Since the number is not in scientific
notation, it becomes 425630.5 x 100.
• Step 2 To get the number between /- 1 and /-
9, I have to move the decimal point to the left 5
places. Since I moved the decimal point to the
left, I have to increase my exponent by the same
amount, 5. So, my number in scientific notation
is 4.256305 x 105.
• Example 2 What is -0.0000586 in scientific
notation?
• Step 1 Since the number is not in scientific
notation, it becomes 0.00586 x 100.
• Step 2 To get the number between /- 1 and /-
9, I have to move the decimal point to the right
3 places. Since I moved the decimal point to the
right, I have to decrease my exponent by the same
amount, 3. So, my number in scientific notation
is 5.86 x 10-3.
• Example 3 What is 625.366 x 105 in correct
scientific notation?
• Step 1 Since the number is already in scientific
notation, nothing is added to it.
• Step 2 To get the number between /- 1 and /-
9, I have to move the decimal point to the left 2
places. Since I moved the decimal point to the
left, I have to increase my exponent by the same
amount, 2. So, my number in scientific notation
is 6.25366 x 107.

19
Significant Figures
• The amount of significant figures indicates how
accurate a measurement is.
• Exact numbers have an infinite number of
significant figures, which means that there is an
infinite number of zeros after the number. These
zeros are not shown for convenience. For example,
the amount in 1 dozen, 12, is an exact number,
but we dont write 12.00000 every time.
• Steps to identify the amount of significant
figures a number has
• Step 1 Going from left to right, locate the
first nonzero digit.
• Step 2 Again going from left to right, count the
amount of digits present in the number, starting
with the first nonzero digit.
• Note If there are zeros at the end of a number
without a decimal point, they may or may not be
significant. Thus, if the zeros are significant,
a decimal point should be at the end of the
number. For example, 300 could have 1 significant
figure, 2 significant figures, or 3 significant
figures, but 300. has 3 significant figures.
• Example 1 How many significant figures are
present in the number 564.32?
• Step 1 The first nonzero digit is 5.
• Step 2 Starting with the 5, there are 5
significant figures.
• Example 2 How many significant figures are
present in the number 0.00042?
• Step 1 The first nonzero digit is 4.
• Step 2 Starting with the 4, there are 2
significant figures.

20
Manipulating Powers of Ten
• Three ways to manipulate powers of ten
• If powers of ten are multiplied, the exponents
• If powers of ten are divided, the exponents are
subtracted.
• If powers of ten are raised to an exponent, the
exponents are multiplied.
• Example 1 What is 106 x 104?
• Answer Since it is multiplication, the exponents
are added together 6 4 10, so, it is 1010.
• Example 2 What is 1013/10-3?
• Answer Since it is division, the exponents are
subtracted from each other 13 (-3) 16, so,
it is 1016.
• Example 3 What is (105)3?
• Answer Since it is raised to an exponent, the
exponents are multiplied together 5 x 3 15,
so, it is 1015.

21
Numbers with Powers of Ten Multiplying and
Dividing
• Steps when multiplying and dividing numbers with
powers of ten
• Step 1 Place the numbers (also called
coefficients) together and the powers of ten
together.
• Step 2 Multiply or divide the numbers and add or
subtract the powers of ten, depending on which
operation is present.
• Step 3 The final answer should
• have the same amount of significant figures as
the number with the least amount of significant
figures.
• have been rounded off correctly.
• preferably be reported in scientific notation.

22
Examples of Multiplying and Dividing Numbers with
Powers of Ten
• Example 1 Multiplication What is (3.443 x 106)
x (0.43 x 104)?
• Step 1 (3.443 x 0.43) x (106 x 104)
• Step 2 1.48049 x 1010
• Step 3 3.443 has 4 significant figures and 0.43
has 2 significant figures, so the answer should
have 2 significant figures. To round off the
number to 2 significant figures, the previous
digit, 4, should become 5 since the next digit
is 8. So, it is 1.5 x 1010.
• Example 1 Division What is (4.25 x 104)/(3.21 x
102)?
• Step 1 (4.25/3.21) x (104/102)
• Step 2 1.32399 x 102
• Step 3 4.25 has 3 significant figures and 3.21
also has 3 significant figures, so the answer
should also have 3 significant figures. To round
off the number to 3 significant figures, the
previous digit, 2, should stay the same since
the next digit is 3. So, it is 1.32 x 102.

23
Numbers with Powers of Ten Adding and Subtracting
• Steps when adding and subtracting numbers with
powers of ten
• Step 1 Make all of the powers of ten the same.
• Step 2 Once all of the powers of ten are the
same, add or subtract the numbers, depending on
which operation is present, but keep the powers
of ten the same
• Step 3 The final answer should
• have the same amount of decimal places as the
number with the fewest decimal places at the time
• have been rounded off correctly.
• preferably be reported in scientific notation.

24
Examples of Adding and Subtracting Numbers with
Powers of Ten
• Example 1 Addition What is 5.34 x 1043 4.2 x
1044?
• Step 1 To make the powers the same, either
change 43 to 44 or 44 to 43. If 43 is turned into
44, since the value is increased by one, the
decimal point is moved to the left by one, making
it 0.534 x 1044.
• Step 2 (0.534 4.2) x 1044 4.734 x 1044.
• Step 3 At the time of addition, 0.534 has 3
decimal places and 4.2 has 1 decimal place, so
the answer should have 1 decimal place. To round
off the number to 1 decimal point, the previous
digit, 7, should remain the same because the
next digit is 3. So, it is 4.7 x 1044.
• Example 1 Subtraction What is 0.98 x 1056
3.12 x 1055?
• Step 1 To make the powers the same, either
change 56 to 55 or 55 to 56. If 56 is turned into
55, since the value is decreased by one, the
decimal point is moved to the right by one,
making it 9.8 x 1055.
• Step 2 (9.8 3.12) x 1055 6.68 x 1055.
• Step 3 At the time of addition, 9.8 has 1
decimal place and 3.12 has 2 decimal places, so
the answer should have 1 decimal place. To round
off the number to 1 decimal point, the previous
digit, 6, should be increased to 7, because the
next digit is 8. So, it is 6.7 x 1055.

25
Numbers with Powers of Ten Mixing
• When all operations are combined, make sure to
carry out those in parenthesis first.
• Example 1 What is
• (2.63 x 105 x (3.64 x 10-4 1.22 x 10-3))/(7.36
x 106)2?
• (3.64 x 10-4 1.22 x 10-3) (0.364 x 10-3 -
1.22 x 10-3) (0.364 1.22) x 10-3 -0.856 x
10-3
• (7.36 x 106)2 (7.36 x 106) x (7.36 x 106)
(7.36 x 7.36) x (106 x 106) 54.1696 x 1012
• (2.63 x 105 x -0.856 x 10-3) (2.63 x -0.856) x
(105 x 10-3) -2.2513 x 102
• -2.2513 x 102/54.1696 x 1012 (-2.2513/54.1696)
x (102/1012) 0.04156 x 10-10 4.16 x 10-12

26
Units of Measurement Equalities in the English
System
• Volume
• 1 pint (pt) 16 fl oz (fluid ounce)
• 1 quart (qt) 2 pt
• 1 gallon (gal) 4 qt
• Mass
• 1 pound (lb) 16 ounces (oz)
• 1 ton 2000 lb
• Length
• 1 foot (ft) 12 inches (in)
• 1 yard (yd) 3 feet (ft)
• 1 mile (mi) 5280 ft
• Time
• 1 minute (min) 60 seconds (sec)
• 1 hour (hr) 60 minutes (min)
• 1 day 24 hours (hr)

27
Units of Measurement The Metric System and the
International System of Units (SI Units)
• These two systems are decimal systems, where the
units are related to each other by powers of ten.
Prefixes are used to indicate fractions and
multiples of ten.
• The basic units for these systems are
• Volume liter (L)
• Mass gram (g)
• Length meter (m)
• Time second (s)
• The relation between length units and volume
units 1 mL 1 cm3 1 cc
• The table to the left illustrates the value of
each prefix. The number in the table always gets
placed in front of the base unit and a 1 always
gets placed in from of the prefixed base unit.
For example, 1 millimeter 10-3 meters or 1 mm
10-3 m.

28
Units of Measurement Conversion between the
Metric and the English Systems
• Volume
• 1 qt 0.946 L
• Mass
• 1 lb 454 g
• Length
• 1 in 2.54 cm

29
Use of Conversion Factors in Calculations
• Obtained from equalities, which show the
relationship between two quantities that are
measuring the same quantity (i.e. mass, volume,
length, etc.)
• From the equality, two different conversion
factors can be obtained
• When using a conversion factor in a problem, you
use the one that cancels out the given units and
leaves the answer units the given units need to
be on the bottom of the conversion factor, with
the answer units on the top of the conversion
factor
• Equality
• x y,
where x and y are different units and the s
arent equal
• Two conversion factors from the equality x/
y and y / x
• If you want to convert y to x, you need to use
x / y
• y ( x / y) x
• If you want to convert x to y, you need to use
y / x
• x ( y / x) y

30
Problem Solving with Conversion Factors Factor
Label Method
• Step 1 Write down the given quantity and unit(s)
and the answer unit(s), with space between them
for the conversion factors.
• Step 2 Identify the equalities needed to change
the given unit(s) to the answer unit(s) more
than one equality may be needed.
• Step 3 Write down the two conversion factors
possible from each of the identified equalities.
• Step 4 Arrange the conversion factors between
the given quantity and unit and the answer unit,
so that the given unit cancels, leaving the
answer unit the given unit needs to be on the
bottom of the first conversion factor and the
answer unit needs to be on the top of the last
conversion factor.
• If the given quantity has two units, convert the
unit on top, and then convert the unit on the
bottom. Unlike when converting the unit on
top, which is described above, when converting
the unit on the bottom, the given unit needs to
be on the top of the first conversion factor for
the bottom and the answer unit needs to be on
the bottom of the last conversion factor for the
bottom.
• Hint A quantity with two units, x/y, can be
written as x / 1 y, since dividing by one
doesnt change the value of the number. This
will help to choose the correct conversion factor
for the problem.

31
Examples of Using the Factor Label Method 1
• Example 1 English to English Conversion How
many feet are in 3 yards?
• Step 1 3 yd.
____ ft.
• Step 2 1 yd. 3 ft. (this covers all the units
above, so no more equalities are needed)
• Step 3 Conversion factors 1 yd. / 3 ft. OR 3
ft. / 1 yd.
• Step 4 3 yd. (3 ft. / 1 yd.) 9 ft. this
conversion factor is needed to cancel yd. (given
unit is on bottom) and get ft. (answer unit is on
top)
• Example 2 Metric to Metric Conversion How many
kilograms are in 90 decigrams?
• Step 1 90 dg.
____ kg.
• Step 2 1 dg. 0.1 g. and 1 kg. 1000 g. (two
equalities are needed here since one does not
cover all of the units above )
• Step 3 Conversion factors 1 dg. / 0.1 g. OR 0.1
g. / 1 dg. and
• 1 kg. / 1000 g. OR 1000 g. / 1 kg.
• Step 4 90 dg. (0.1 g. / 1 dg.) (1 kg. / 1000
g.) 0.0090 kg. the first conversion factor is
needed to cancel dg (given unit is on bottom) and
get g (unit is on top), which leads to the second
conversion factor, which cancels g (unit is on
bottom) to get kg (answer unit is on top)

32
Examples of Using the Factor Label Method 2
• Example 3 Metric to English Conversion How many
feet are in 5.85 centimeters?
• Step 1 5.85 cm.
____ ft.
• Step 2 2.54 cm. 1 in. and 12 in. 1 ft. (two
equalities are needed here since one does not
cover all of the units above the first equality
converts between the english and the metric
system and the second one converts within the
english system)
• Step 3 Conversion factors 2.54 cm. / 1 in. OR 1
in. / 2.54 cm. and 12 in. / 1 ft. OR 1 ft. / 12
in.
• Step 4 5.85 cm. (1 in. / 2.54 cm.) (1 ft. /
12 in.) 0.192 ft. the first conversion factor
is needed to cancel cm. (given unit is on bottom)
and get in. (unit is on top), which leads to the
second conversion factor, which cancels in. (unit
is on bottom) to get ft. (answer unit is on top)
• Example 4 Using Square Units How many m2 are in
105 in2?
• Step 1 105 in2
____ m2
• Step 2 2.54 cm. 1 in. and 1 cm. 0.01 m. (two
equalities are needed here since one does not
cover all of the units above the first equality
converts between the english and the metric
system and the second one converts within the
metric system)
• Step 3 Conversion factors 2.54 cm. / 1 in. OR 1
in. / 2.54 cm. and 1 cm. / 0.01 m. OR 0.01 m. /
1 cm.
• Step 4 105 in.2 (2.54 cm. / 1 in.)2 (0.01 m.
/ 1 cm.)2 0.0677 m.2 the first conversion
factor is needed to cancel in. (given unit is on
bottom) and get cm. (unit is on top), which leads
to the second conversion factor, which cancels
cm. (unit is on bottom) to get m. (answer unit is
on top) each conversion factor needs to be
squared so that the given units are cancelled
completely to give the answer units

33
Examples of Using the Factor Label Method 3
• Example 5 Using Cubed Units How many m3 are in
105 in3?
• Step 1 105 in3
____ m3
• Step 2 2.54 cm. 1 in. and 1 cm. 0.01 m. (two
equalities are needed here since one does not
cover all of the units above the first equality
converts between the english and the metric
system and the second one converts within the
metric system)
• Step 3 Conversion factors 2.54 cm. / 1 in. OR 1
in. / 2.54 cm. and 1 cm. / 0.01 m. OR 0.01 m. /
1 cm.
• Step 4 105 in.3 (2.54 cm. / 1 in.)3 (0.01 m.
/ 1 cm.)3 0.00172 m.3 the first conversion
factor is needed to cancel in. (given unit is on
bottom) and get cm. (unit is on top), which leads
to the second conversion factor, which cancels
cm. (unit is on bottom) to get m. (answer unit is
on top) each conversion factor needs to be
cubed so that the given units are cancelled
completely to give the answer units
• Example 6 Using Double Units How many ft./sec.
are in 731 in./min.?
• Step 1 731 in./1 min.
____ ft./sec.
• Step 2 For the top units, 12 in. 1 ft. (only
one equality is needed) for the bottom units,
1 min. 60 sec. (only one equality is needed)
• Step 3 For the top units, 12 in. / 1 ft. OR 1
ft. / 12 in. for the bottom units, 1 min. / 60
sec. OR 60 sec. / 1 min.
• Step 4 731 in./1 min. (1 ft. / 12 in.) (1
min. / 60 sec.) 1.02 ft./sec. the first
conversion factor (for the top units) is needed
to cancel in. (given unit is on bottom) and get
ft. (answer unit is on top), and the second one
(for the bottom units) is needed to cancel min.
(given unit is on top) and get sec. (answer unit
is on bottom)

34
Density (d) and Specific Gravity (sp.gr.)
• Density mass/volume OR d m/V
• The common units of density are
• g/mL for liquids and solids
• g/L for gases
• The density of a substance shows the relationship
between its mass and its volume. For example, if
the density of a substance is 3.45 g/mL, the
relationship between the mass and volume can be
written as the equality 3.45 g 1 mL.
• Specific Gravity (sp.gr.) (density of
substance)/(density of water at 4?C) (density
of substance in g/mL)/(1.00g/mL)
• Since the density of a substance is divided by
1.00 to obtain the specific gravity, the density
and specific gravity are numerically equivalent
(i.e., they are the same number since dividing
any number by 1.00 gives the same number).
Although they are numerically equivalent, they
are not equal. Since both the density of a
substance and the density of water have the same
units, they are cancelled out when the specific
gravity is obtained, making the specific gravity
a unitless quantity, unlike density, which always
has units.

35
To Obtain a Substances Density
• Since density is mass divided by volume, you need
to obtain both a substances mass and its volume
to obtain its density.
• To obtain the mass, measure the substance on a
balance.
• To obtain the volume
• If it is a liquid, measure it with a graduated
cylinder.
• If it is a solid, one of two methods can be used
• One is by water displacement. Here, a certain
volume of water is placed in a graduated
cylinder, and the value is recorded (Vwater). The
solid is then dropped into the graduated cylinder
with the water. The new volume of water is then
recorded (Vwatersolid). To obtain the volume of
the solid (Vsolid), the initial volume of the
water is subtracted from the new volume of water
(Vsolid Vwatersolid - Vwater).
• The other is through mathematical equations,
using one of the following equations
• V(cubic solid) l x w x h (length (l) width
(w) height (h))
• V(rectangular solid) l x w x h
• V(sphere) (4/3) pr3 (radius (r) diameter
(d)/2)
• V(cylinder) (pd2h)/4

36
Steps to Solving a Problem
• Step 1 Identify the information that is given in
the problem.
• Step 2 Identify what information the problem is
looking for.
• Step 3 Identify the equation that is needed to
obtain the information you are looking for, while
using the information you were given.
• Step 4 Put the given information into the
equation found in Step 3 and solve for the
information you are looking for, making sure that
all of the given information is expressed in the
proper units.

37
Examples of Calculations of Density 1
• Example 1 What is the density of a liquid that
has a mass of 5.67 grams and a volume of 10.3 mL?
• Step 1 5.67g m 10.3mL V
• Step 2 density (d)
• Step 3 d m/V
• Step 4 d 5.67 g / 10.3 mL 0.550 g/mL
• Example 2 What is the mass of a liquid if it
has a density of 1.50 g/mL and a volume of 14 mL?
• Step 1 1.50 g/mL d 14mL V
• Step 2 mass (m)
• Step 3 d m/V
• Step 4 d m / V - m d V 1.50 g/mL 14
mL 21 g
• Example 3 What is the volume of a liquid if it
has a mass of 14.6 grams and a density of 3.45
g/mL?
• Step 1 14.6g m 3.45 g/mL d
• Step 2 volume (V)
• Step 3 d m/V
• Step 4 d m / V - V m / d 14.6 g / 3.45
g/mL 4.23 mL

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Examples of Calculations of Density 2
• Example 4 If a 24g solid was placed in a
graduated cylinder containing 25.0mL of water and
the water raised to 34.2mL, what is the density
of the substance?
• Step 1 24g m 25.0mL Vwater 34.2mL
Vwatersolid
• Step 2 density (d)
• Step 3 d m/V, where V Vsolid Vwatersolid
- Vwater
• Step 4 V Vwatersolid Vwater 34.2mL-25.0mL
9.2mL d m/V 24g/9.2mL 2.6g/mL
• Example 5 What is the density of a wooden block
that has a mass of 75g and measures 2.5cm by
3.5cm by 6.1cm?
• Step 1 75g m 2.5cm length (l) 3.5cm
width (w) 6.1cm height (h)
• Step 2 density (d)
• Step 3 d m/V, where V V(rectangular solid)
l x w x h
• Step 4 V 2.5cm x 3.5cm x 6.1cm 53.375cm3
53.375mL d m/V 75g/53.375mL 1.4g/mL

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Examples of Calculations of Specific Gravity
• Example 1 What is the specific gravity of a
substance with a mass of 5.021 g and a density of
1.21 g/mL?
• Step 1 5.021 g m 1.21 g/mL d
• Step 2 specific gravity (sp.gr.)
• Step 3 sp.gr. (density of substance in
g/mL)/(1.00g/mL)
• Step 4 sp.gr. 1.21 g/mL / 1.00 g/mL 1.21
• Example 2 What is the specific gravity of a
substance with a mass of 5.021 g and a volume of
7.99 mL?
• Step 1 5.021g m 7.99mL V
• Step 2 specific gravity (sp.gr.)
• Step 3 sp.gr. (density of substance in
g/mL)/(1.00g/mL), where d m/V
• Step 4 d m/V 5.021 g / 7.99 mL 0.628 g/mL
sp.gr. (density of substance in
g/mL)/(1.00g/mL) 0.628 g/mL / 1.00 g/mL 0.628

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Temperature Conversions
• There are three temperature units degrees
Fahrenheit (F), degrees Celsius (C), and Kelvin
(K). (Note K does not have a sign.)
• To convert C to F F (1.8 C) 32
• To convert F to C C (F 32) / 1.8
• To convert C to K K C 273
• To convert K to C C K 273
• Note For the Celsius scale and the Kelvin scale,
there is a 100 difference between the freezing
point and the boiling point of water, but on the
Fahrenheit scale, the difference is 180.

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Examples of Temperature Conversions
• Example 1 How many degrees Fahrenheit are in 73
degrees Celsius?
• Answer F (1.8 C) 32 (1.8 73 C) 32
163.4 F
• Example 2 How many degrees Celsius are in 200
degrees Fahrenheit?
• Answer C (F 32) / 1.8 (200 F 32) /
1.8 93.3 C
• Example 3 How many Kelvin are there in 650
degrees Celsius?
• Answer K C 273 650 C 273 923 K
• Example 4 How many degrees Celsius are there in
543 Kelvin?
• Answer C K 273 543 K 273 270 C
• Example 5 How many degrees Fahrenheit are there
in 321 Kelvin?
• C ? K 273 321 K -273 48 C
• F (1.8 C) 32 (1.8 48 C) 32 118.4
F

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Heat Transfer
• During chemical changes and physical changes,
heat is transferred
• If there is an evolution of heat (i.e. heat is
lost or released), it is an exothermic process.
• If there is an absorption of heat (i.e. heat is
gained or absorbed), it is an endothermic
process.
• The units to express the amount of heat
transferred are calories (cal) or joules (J),
where 1 cal 4.184J.
• Specific heat is the amount of heat necessary to
raise the temperature of 1 g of substance by 1C.
Specific heat is a physical intensive property,
like density, melting point, and boiling point.
Each substance has its own specific heat.
• Using the specific heat of a substance, the heat
that is absorbed or released in a given process
can be obtained by
• q m x s x ?T
• where q is the heat energy (cal, kcal, J, or
kJ), m is mass (g), s is specific heat
(cal/gC), and ?T is the change in temperature
(?T T2 T1)(C).

43
Examples of Calculations of Heat Transfer
• Example 1 How much heat (in cal) is needed to
heat 54g of a substance (specific heat 0.212
cal/gC) from 72C to 98C?
• Step 1 Given 54g m 0.212 cal/gC s 72C
T1 98C T2
• Step 2 Looking for heat (in cal) q
• Step 3 Equation q m x S x ?T, where ?T T2
T1
• Step 4 ?T T2 T1 98C - 72C 26C q
(54g) x (0.212 cal/gC) x (26C) 297.648 cal
3.0 cal.
• Example 2 How much of a substance is present if
56J of heat is added when the substance (specific
heat 0.532 cal/gC) is heated from 23C to
54C?
• Step 1 Given 56J q 0.532 cal/gC s 23C
T1 54C T2
• Step 2 Looking for mass m
• Step 3 Equation q m x S x ?T, where ?T T2
T1
• Step 4 ?T T2 T1 54C - 23C 31C q m
x S x ?T ? 56J (m) x (0.532 cal/gC) x
(31C), but since there is J on one side and cal
on the other side, the J have to be converted to
cal before solving for m 56J x (1 cal /4.184J)
13.384 cal 13.384 cal (m) x (0.532
cal/gC) x (31C), so m 0.81156 g 0.81 g.
• Example 3 If 86cal of heat is added to 10g of a
substance (specific heat 0.411 cal/gC), and
the final temperature was measured to be 152C,
what is the initial temperature of the substance?
• Step 1 Given 86cal q 10g m 0.411 cal/gC
s 152C T2
• Step 2 Looking for initial temperature T1
• Step 3 Equation q m x S x ?T, where ?T T2
T1
• Step 4 86cal (10g) x (0.411 cal/gC) x (?T),
so ?T 20.92C ?T T2 T1 ? 20.92C 152C
T1, so T1 131.08C 1.3 x 102C

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Some Specific Heats
• The larger the specific heat of the substance,
the more heat it needs to raise its temperature.
• The specific heat of water is 1.00 cal/gC,
which is one of the largest specific heats. Since
approx. 60 of our body weight is water, this
large specific heat allows our body to maintain a
constant body temperature (around 37C) much
easier. So, our body can absorb or release a
large amount of energy with little change in body
temperature.
• Examples of other specific heats
• wood 0.421 cal/gC
• gold 0.0306 cal/gC
• graphite 0.172 cal/gC

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THE END