Notes Chapter 3:
Matter and Energy
Matter is anything that has mass and
takes up space (volume)
In other words it is ‘stuff’, ‘things’,
whatever is NOT energy
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Notes Chapter 3
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Matter comes in three states (or phases):
State
Solid
Definition
has a fixed volume
and a fixed shape
Liquid
has a fixed volume
but takes the shape
of its container
Examples
ice cubes,
diamond, iron
bar
gasoline,
alcohol,
water, blood
has no fixed volume
Gas
or shape; it takes
the volume & shape
oxygen,
helium, air
of its container
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Notes Chapter 3
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Molecular Arrangement in the Three States
Solid (Ice)
Liquid (Water)
Gas (Steam)
Molecules are
held close
together by
strong forces;
molecules are in
a crystalline
pattern
Molecules are
loosely held
together by
weaker forces;
molecules can
flow around
each other
Molecules are
NOT held
together at all;
molecules are
independent of
each other
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Notes Chapter 3
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Physical Properties
and
Physical Changes
Physical properties of a substance include:
color, odor, mass, volume, density,
aanndd
state (solid, liquid, gas)
(In other words it is a characteristic of a
substance that can change without the
substance becoming a different substance)
Physical changes include:
cutting, breaking, expanding, contracting,
aanndd
Chemistry-1
changes in state (melting, solidifying,
evaporating, and condensing)
Notes Chapter 3
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(In other words it is a change in a substance
without the substance becoming a different
substance)
The Molecules Remain Intact (unchanged).
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Notes Chapter 3
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Distilling Saltwater is a Physical Change.
Distillation
(physical method)
+
Salt
Saltwater solution
Pure water
(homogeneous mixture)
Steam is
condensed in a
tube cooled by
water.
Cooling
water in
Saltwater
Cooling
water out
Pure
Bunsen burner
water
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Notes Chapter 3
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Salt
Pure
Water
a.k.a.
Distilled
Water
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Notes Chapter 3
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Filtration also is a
Physical Change
Mixture of Solid and
Liquid
Filter paper
traps solids
Stirring Rod
Funnel
Filtrate - liquid component of
the mixture
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Notes Chapter 3
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Chemical Properties
aanndd Chemical Changes
Chemical properties of a substance include:
ability to react with oxygen,
ability to react with water,
decomposing, aanndd burning.
In other words it is the ability to form new
substances.
Chemical changes include: wood burning, iron
rusting, food digesting, grass growing
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Notes Chapter 3
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In other words a substance changes into a
different substance (or substances)
The Molecules Change Into Different Molecules.
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Notes Chapter 3
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Elements, Compounds, aanndd Mixtures
Elements:
 they are made up of only one kind of atom
 are the simplest form of matter
 are the purest form of matter
 are represented by a chemical symbol
 cannot be broken down into other substance by
any means.
Examples: Gold (Au), Silver (Ag), Oxygen (O22)
Compounds:
 compounds are pure substances
 are made up of two or more kinds of atoms
chemically bonded into molecules.
 the molecules of a compound are always the
same.
 the kinds of atoms that make up the compound
are always in definite proportions
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Notes Chapter 3
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 can be represented by a chemical formula and
they have a chemical name
 can be broken down into other substances by
chemical changes only.
Examples: Carbon Dioxide (CO22), Water (H22O), Salt
(NaCl), Sugar (C66H1122O66)
Mixtures:
 are made up of two or more substance NNOOTT
chemically combined.
 the substances in the mixture are combined in
any proportion
 do not have a chemical formula and do not have
a chemical name
Heterogeneous mixtures:
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Notes Chapter 3
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 contain regions that have a different proportion
of substances (and so have different
properties) from other regions of the mixture.
Examples: blood, cereal & milk, cement
Homogeneous mixtures:
 are the same throughout. Every part of the
mixture contains the same proportion of
ingredients.
 Homogeneous mixtures are called
solutions
Examples: saltwater, tea, air, alloys
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Notes Chapter 3
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Heat Energy
Energy:
 the capacity to do work (using a force to move an
object).
 when the temperature of an object is raised, the
molecules have more energy and therefore move
more.
 The amount of energy (heat) needed to raise the
temperature of one gram of water by one degree
Celcius is a . . . calorie.
 the unit of energy used in the metric system is the
joule.
 You can convert between joules and calories using the
following equation (conversion factor)
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Notes Chapter 3
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1 cal = 4.184 Joules
In other words, it takes 4.184 J of energy
to raise the temperature of 1.0 g of water 1°C.
Specific Heat Capacity
What is it?:
You know it takes 4.184 J of energy to heat up one gram
of W
WA
AT
TE
ER
R by 1°C.
How much energy does it take to heat up one gram of
A
AL
LU
UM
MIIN
NU
UM
M by 1°C?
How about one gram of C
CA
AR
RB
BO
ON
N?
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Notes Chapter 3
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The amount of energy it takes to heat up one gram of a
substance by 1°C is that substance’s
S
SP
PE
EC
CIIF
FIIC
CH
HE
EA
AT
T.
A high specific heat means it takes a large amount of
energy to raise the temperature of that substance;
whereas a low specific heat means it takes a small amount
of energy to raise the temperature of that substance.
Chemists describe this difference by saying that substances
have different H
HE
EA
AT
TC
CA
AP
PA
AC
CIIT
TIIE
ES
S.
The Specific Heat Capacity of Water is 4.184 J/g-°C
Finding the Specific Heat of a Substance
Energy (heat) required
(Q)
Chemistry-1
=
Specific Heat
Capacity (s)
Mass (m) in
grams of sample
x
Notes Chapter 3
Change in
x
temperature (T)
in °C
Page 16 of 22
Q
Q = s xx m x T
ss ==
Q
Q
m
m xx 
TT
There are two symbols for specific heat: s and Cp.
We will use them interchangeably so you can learn both of
them.
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Notes Chapter 3
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Let’s try a sample problem
A 35.2 g sample of a metal requires 1,251 J of energy to heat
the sample by 25.0°C. Calculate the specific heat of the
metal.
ss ==
Q
Q
m
m xx 
TT
ss ==
ss ==
ss ==
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11,,225511 JJ
3355..22 gg xx 2255..00ooCC
11,,225511 // 3355..22 // 2255..00
11..4422
JJ
gg -- °°CC
JJ
gg -- °°CC
Notes Chapter 3
Page 18 of 22
Let’s try a little harder problem
A 25.0 g sample of pure iron at 85°C is dropped into 75.0 g
of water at 25°C. What is the final temperature of the water
– iron mixture?
We know the specific heat of water is: 4.184 J/g-°C
From Table 3.2 on page 73 we know
the specific heat of iron is: 0.45 J/g-°C
Remember, T is the change in temperature
T = higher temp. - lower temp.
Iron is cooling, so it’s initial temp. is higher
For Iron: T = (Tii – Tff)
Water is heating, so it’s final temp. is higher
For Water: T = (Tff – Tii)
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Notes Chapter 3
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First let’s find the amount of heat lost by the iron.
Q
Q = s xx m xx 
TT
Q
Q = 0.45 J/g-°C xx 25.0 g x (85° - Tff)
Q
Q = 11.25 x (85 - Tff) J
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Notes Chapter 3
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Next, let’s find the amount of heat gained by the water.
Q
Q = s xx m xx 
TT
Q
Q = 4.184 J/g-°C xx 75.0 g x (Tff - 25°)
Q
Q = 313.8 x (Tff - 25°) J
Since the amount of heat lost by the iron is equal to the
amount of heat gained by the water, we can set these two
‘heats’ (Q’s) equal to each other.
11.25 x (85 - Tff)
=
313.8 x (Tff - 25)
First, distribute the 11.25 and the 313.8 by multiplying them with what’s in the parenthesis. You will end up with this:
956.25 - 11.25Tff
=
313.8Tff - 7,845
Next, solve for Tf. There are a number of
ways to do this. One way would be to add
7,845 to both sides of the equation, then
add 11.25Tf to both sides of the equation.
You’ll end up with the following:
8,801
27°C
Chemistry-1
=
325Tff
=
Tff
Notes Chapter 3
Page 21 of 22
Oh, I see!
This is just basic algebra.
As long as I have my
calculator I can do these!
Oh, sure. . .
You better do a few more
practice problems kiddo.
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Notes Chapter 3
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