
Assigned Problems (odd numbers only)

“Questions and Problems” 3.1 to 3.41
(begins on page 61)

“Additional Questions and Problems”
3.49 to 3.69 (page 87-88)

“Challenge Questions” 3.71 and 3.75,
(page 88)





Matter is any material that has mass and occupies space
Matter is made up of small particles

Atoms

Molecules
Includes all things (living and nonliving) such as plants, soil, and rocks
Any material we use such as water, wood, clothing, etc.


Chemistry is the study of matter

The properties of different types of matter

The way matter behaves when influenced by other matter and/or energy


Nearly all changes that matter undergoes involves the release or absorption of energy
Energy is the part of the universe that has the ability to do work

Matter
Pure Substance Mixture




Matter that has a definite and constant composition
Always contains the same substance, never varies
Either elements or compounds, all of one type


A pure sample of water only contains water molecules
Pure table salt contains only salt


Elements

Substances which can not be broken down into simpler substances by chemical reactions

Fundamental substances

Compounds

Two or more elements combined chemically in a definite and constant ratio

Can be broken down into simpler substances

Most of matter is in the compound form


Compounds

Results from a chemical combination of two or more elements


Can be broken down into elements by chemical processes
Properties of the compound not related to the properties of the elements that compose it

Water is composed of hydrogen and oxygen gases
(combined in a 2:1 ratio)





Something of variable composition
Result from the physical combination of two or more substances (elements or compounds)
Made up of two or more types of substances physically mixed
Not mixed in a fixed ratio, no chemical combination between the two substances


Compounds are not mixtures

Cannot be separated by a physical process

Can be subdivided by a chemical process into two or more simpler substances

Mixtures


Unlike compounds, mixtures can be separated by a physical process
Retain the properties of their individual components


Two types of mixtures:

Homogeneous mixture:

Same uniform composition throughout

Not possible to see the two substances present

Heterogeneous mixture:

Composition is not uniform throughout the sample.

It contains visibly different parts or phases




Homogenous mixtures

A sugar solution


14 karat gold, a mixture of copper and gold
Air, a mixture of gases (oxygen, nitrogen)
Heterogeneous mixture

Oil and vinegar


Raisin cookies
Sand
Pure substance
 i.e. copper (all elements are pure substances)

Classification of Matter
Pure Substances
Elements Compounds
Chemical Methods
Matter
Mixture
Homogeneous
Mixture
Heterogeneous
Mixture
Physical Methods
Pure Substances




Many properties used to identify chemical substances
Two types

Physical Properties

Chemical Properties
Properties can be:

Directly observable

The interaction of the matter with other substances




Solid

Has a rigid, definite shape and definite volume
Liquid


Has an indefinite shape and a definite volume.
It will take the shape of the container it fills
Gas


Has an indefinite shape and an indefinite volume.
It will take the shape and completely fill the volume of the container it fills

(a) Solid (Ice)
Fig3_2
(b) Liquid (Water) (c) Gas (Steam)


Physical Properties

Characteristics of matter that can be observed or measured without changing its identity or composition

Characteristics that are directly observable

Color, odor, physical state, density, melting point, boiling point

Physical Changes

Cutting a piece of metal, melting ice


A process that alters the appearance of a substance but does not change its identity or composition

No new substance is formed

Most common are changes of state


Chemical Properties

Describes the ability of a substance to react and change into a new substance

Properties that matter exhibits as it undergoes changes in chemical composition

During a chemical change, the original substance is converted into one or more new substances with different chemical and physical properties


A change in the fundamental components of the substance:

A substance undergoes a change in chemical composition

Also called a chemical reaction

Conversion of material(s) into one or more new substances

Wood burning, iron rusting, alka seltzer tablet into water





The boiling point of ethyl alcohol is 78 °C
Physical property – describes an inherent characteristic of alcohol, its boiling point
Diamond is very hard

Physical property – describes inherent characteristic of diamond – hardness
Sugar ferments to form ethyl alcohol

Chemical property – describes behavior of sugar, ability to form a new substance
(ethyl alcohol)





Melting of snow
Physical change – a change of state but not a change in composition
Burning of gasoline

Chemical change – combines with oxygen to form new compounds
Rusting of iron

Chemical change – combines with oxygen to form a new reddish-colored substance (ferric oxide)




Iron metal is melted

Physical change – describes a state change, but the material is still iron
Iron combines with oxygen to form rust

Chemical change – describes how iron and oxygen combine to make a new substance, rust (ferric oxide)
Sugar ferments to form ethyl alcohol

Chemical change – describes how sugar forms a new substance (ethyl alcohol)



A measure of how hot or cold a substance is compared to another substance
Fahrenheit Scale, °F



Used in USA
Water’s freezing point = 32°F, boiling point = 212°F
Celsius Scale, °C

Used in science (USA) and everyday use in most of the world


Temperature unit larger than the Fahrenheit
Water’s freezing point = 0°C, boiling point = 100°C


Kelvin Scale, K

SI Unit



Used in science
Temperature unit same size as Celsius
Water’s freezing point = 273 K, boiling point
= 373 K


Absolute zero is the lowest temperature theoretically possible
No negative temperatures


Units are different sizes

Fahrenheit scale: 180 degree intervals between freezing and boiling

Celsius scale: 100 degree intervals between freezing and boiling
180  F
100  C

9  F
5  C

1.8  F
1  C

212ºF
180
Fahrenheit degrees
1.8

F
1

C
32ºF
Fig2_9
100ºC
0ºC
Boiling point
100
Celsius degrees
Freezing point
1.8

F
1

C


To convert from °C to °F

Different values for the freezing points
32 °F
0 °C add 32 to the °F value

Different size of the degree intervals in each scale
T
 F

1.8  F
1  C
 32


Temperature units are the same size

Differ only in the value assigned to their reference points

K = °C + 273

25 °C is room temperature, what is the equivalent temperature on the Kelvin scale?


A cake is baked at 350 °F. What is this in Centigrade/Celsius? In Kelvin?
T

F

1  C
1.8  F

1.8  F
1  C
T
 F
 32
 32
   
1  C
1.8  F
 350  32    
T
 C
 176.6667  C
176.7
 273 =
449.7
K


Capacity to do work or supply heat

Electrical, radiant, mechanical, thermal, chemical, nuclear

Two forms of Energy

Potential: Stored energy

Kinetic: Motion energy

All physical changes and chemical changes involve energy changes


Potential energy:




Determined by an objects position

Chemical energy is potential energy stored in the bonds contained within a molecule. It is released in a chemical reaction
Kinetic energy

Energy that matter acquires due to motion

Converted from the potential energy
All physical changes and chemical changes involve energy changes
These changes convert energy from one form to another



The joule (J) is the SI unit of heat energy
The calorie (cal) is an older unit used for measuring heat energy ( not an SI unit )

The amount of energy needed to raise the temperature of one gram of water by 1 °C
4.184 J = 1 cal 1 kcal = 1000 cal

The Cal is the unit of heat energy in nutrition
1 Cal = 1000 cal = 1 kcal


Heat energy is the form of energy most often released or required for chemical and physical changes

Every substance must absorb a different amount of heat to reach a certain temperature

Different substances respond differently when heat is applied


If 4.184 J of heat is applied to:

1 g of water, its temperature is raised by 1 °C

1 g of gold, its temperature is raised by 32 °C

Some substances requires large amounts of heat to change their temperatures, and others require a small amount

The precise amount of heat that is required to cause a substance to have a rise in temperature is called a substance’s “specific heat”


The amount of heat energy (q) needed to raise 1 gram of a substance by 1 °C

Specific to the substance

The higher the specific heat value, the less its temperature will change when it absorbs heat

SH values given in table 3.7, page 76

Only for heating/cooling not for changes in state
SH

SH
 grams
( q
 Δ
T


Δt g
J


J (or
C g cal)
 or

C g cal


C

Specific Heat Expression with
Calories and Joules

1 cal is the energy needed to heat 1 g of water 1 °C

1 cal is 4.184 J

Make a conversion factor from the statements
SH water

1g
1 cal

1

C

4.184
1g

1

J
C


The rearrangement of the SH equation gives the expression called the “heat equation”
SH
 heat mass ( g
( q
)

)
Δ
T heat ( q )

SH
 mass ( g )
 Δ
T q(J)



SH g

J

C


 m(g)
 ΔT( 
C)



 q = heat
SH = specific heat (different for each substance) m = mass (g)
∆T = change in temperature (°C)
 answer in joules


Energy (heat) required to change the temperature of a substance depends on:

The amount of substance being heated (g)

The temperature change (initial T and final T in °C)

The identity of the substance


Heat (q)

SH
 mass (g)
 Δt
2× 2×

The amount the temperature of an object increases depends on the amount of heat added
( q )


If you double the added heat energy (q), the temperature will increase twice as much.
When a substance absorbs energy, q is positive, temperature increases

When a substance loses energy, q is negative, temperature decreases



Use same problem solving steps as before (Chapter 2)

State the given and needed units

Write the unit plan to convert the given unit to the final unit

State the equalities and the conversion factors

Set up the problem to cancel the units
Pepsi One™ contains 1 Calorie per can.
How many joules is this?
1 Cal = 1000 cal 4.184 J = 1 cal
1 Cal 1000 cal
1 Cal
4.184 J
1 cal
 4184 J

Calculating Mass Using Specific Heat

The 4184 J from the Pepsi One™ will heat how many grams of water from 0 °C to boiling?
m

SH q
 
T
4184 J 1

C

1 g
4.184
J 100

C

10 g

10 mL

Calculating Mass Using Specific Heat

How many grams of water would reach boiling if the water started out at room temperature (25 °C)?
m

SH q
 
T
4184 J
1

C

1 g
4.184
J

100 75
 
C

13.33 g  13.33 mL

Calculating The Temperature Change
Using Specific Heat Values

If 50.0 J of heat is applied to 10.0 g of iron, by how much will the temperature of the iron increase?
50.0 J g   C
0.45 J 10.0 g
 11.11  C
Q  SH  m   T  T 
Q
SH  m

 end