1. CHAPTER 3 – MATTER
MATTER – Anything that occupies space and has mass
Desk Rock Water Air You
Matter is found in 3 phases or states:
SOLID – definite shape and volume
LIQUID – indefinite shape, definite volume
GAS – indefinite shape and volume
KINETIC MOLECULAR THEORY – Matter is composed of tiny particles that are in constant motion
2A-1 (of 18) (3 – ) (4 – 5-16) (5 – )

2. Solid Liquid Gas
Vibrating
Rolling
Straight-Line
2A-2

3. PHYSICAL CHANGE – One that does not change the identity of the matter
CHANGES IN MATTER
PHYSICAL CHANGE – One that does not change the identity of the matter
Tearing Crushing Freezing Melting Boiling
CHEMICAL CHANGE – One that does change the identity of the matter,
also called CHEMICAL REACTIONS
Burning Rusting Tarnishing Spoiling Decomposing
CHEMICAL ACTIVITY – The likelihood of matter to undergo chemical change
ACTIVE - reacts vigorously
INACTIVE - reacts slowly
INERT - does not react
2A-3

4. EVIDENCE FOR CHEMICAL CHANGE
Color change
A solid forms
Bubbles are formed
A flame is produces
Heat is absorbed or released
2A-4

5. TESTS FOR GASES
Hydrogen - makes a burning splint “pop”
Oxygen - makes a glowing splint ignite
Carbon dioxide - puts out a burning splint
2A-5

6. SUBSTANCE – A homogeneous material consisting of one kind of matter
TYPES OF MATTER
SUBSTANCE – A homogeneous material consisting of one kind of matter
Water Iron Gold Nitrogen Salt
There are 2 types of substances
ELEMENT – A substance that cannot be broken down into simpler substances by chemical means
Iron Gold Nitrogen (Found on the Periodic Table)
COMPOUND – A substance that can be broken down into simpler substances by chemical means
Water Salt
2A-6

7. Water
↓electricity
Hydrogen and Oxygen
↓electricity ↓electricity
Compound
Elements
Compound
Elements
Salt
↓electricity
Sodium and Chlorine
↓electricity ↓electricity
2A-7

8. Granite Air Seawater Wine Spanish Omelet Granite Spanish Omelet
MIXTURE – A physical combination of substances with a variable composition
Granite Air Seawater Wine Spanish Omelet
There are 2 types of mixtures
HETEROGENEOUS – A mixture that has parts with visibly different properties
Granite Spanish Omelet
HOMOGENEOUS – A mixture that has the same properties throughout
Air Seawater Wine
SOLUTION – A homogeneous mixture
2A-8

9. + PROPERTIES OF COMPOUNDS 1) Components (elements) combine chemically
2) Components (elements) are in a definite proportion
3) Components lose their characteristic properties
4) Components can only be separated by chemical means +
2 g Sodium
3 g Chlorine
5 g Sodium chloride
4 g Sodium
6 g Chlorine
10 g Sodium chloride
2A-9

10. + PROPERTIES OF MIXTURES
1) Components (elements or compounds) combine physically
2) Components (elements or compounds) can be in any proportions
3) Components retain their characteristic properties
4) Components can be separated by physical means +
2 g Tin
3 g Copper
Bronze
2 g Tin
6 g Copper
Still Bronze
2A-10

11. SEPARATION OF MIXTURES
DISTILLATION – Separating the substances in a mixture by vaporizing (boiling) one of them
FILTRATION – Separating the substances in a mixture if one is a solid and one is a liquid
2A-11

12. Matter
Mixture
Substance
Heterogeneous
Homogeneous
Compound
Element
2A-12

13. CHAPTER 10 – THE CONCEPT OF ENERGY
ENERGY – The capacity to do work
Units of energy:
CALORIE (cal) – The energy needed to raise the temperature of 1 gram of water 1 Cº
JOULE (J) – The energy needed to accelerate a 1 kg object 1 m/s2 over a 1 m distance
1 cal = J
2A-13

14. Convert 25.0 calories to joules
x J
__________
1 cal
= J
2A-14

15. SPECIFIC HEAT CAPACITY (s) – The energy needed to raise the temperature of 1 gram of a substance 1 Cº
Substance Specific Heat Capacity (s)
Water
Sand
Iron
J/gCº
0.84 J/gCº
0.45 J/gCº
2A-15

16. To calculate the energy needed to raise the temperature of a substance:
Q = smΔT
Q = energy (J)
s = specific heat capacity (J/gCº)
m = mass of the substance (g)
ΔT = temperature change (Cº)
2A-16

17. Calculate the energy needed to raise the temperature of 50
Calculate the energy needed to raise the temperature of 50.0 g of water from 22ºC to 37ºC.
Q = smΔT
= (4.184 J/gCº)(50.0 g)(15 Cº)
= 3,138 J
= 3,100 J
2A-17

18. A 2. 3 g sample of gold requires 6
A 2.3 g sample of gold requires 6.0 J of energy to raise its temperature from 25ºC to 45ºC. Find gold’s specific heat capacity.
Q = smΔT
Q = s
______
mΔT
= J
_________________
(2.3 g)(20. Cº)
= J/gCº
2A-18