Matter & Change Chapter 2. A. Describing Matter Understanding matter begins with observation Matter is anything that has mass and takes up space Chemistry.

Matter & ChangeMatter & ChangeChapter 2

A. Describing MatterA. Describing Matter

Understanding matter begins with observation

Matter is anything that has mass and takes up space

Chemistry – the study of matter and the changes it undergoes

B. Four States of MatterB. Four States of Matter

Solidsparticles vibrate but can’t move

aroundfixed shape fixed volumeVirtually incompressible

B. Four States of MatterB. Four States of Matter

Liquidsparticles can move

around but are still close together

variable shapefixed volumeVirtually incompressible

B. Four States of MatterB. Four States of Matter

Gasesparticles can separate and

move throughout containervariable shapevariable volumeEasily compressedVapor =

gaseous state of a substance that is a liquid or solid at room temperature

B. Four States of MatterB. Four States of Matter

Plasmaparticles collide with enough energy

to break into charged particles (+/-)gas-like, variable

shape & volumestars, fluorescent

light bulbs, TV tubes

II. Properties & Changes in MatterExtensive vs. Intensive

Physical vs. Chemical

A. Physical PropertiesA. Physical Properties

Physical Propertycan be observed without changing the

identity of the substance

B. Physical PropertiesB. Physical Properties

Physical properties can be described as one of 2 types:

Extensive Propertydepends on the amount of matter

present (example: length)

Intensive Propertydepends on the identity of substance, not the amount (example: scent)

C. Extensive vs. IntensiveC. Extensive vs. Intensive

Examples:boiling point

volume

mass

density

conductivity

intensive

extensive

extensive

intensive

intensive

D. Physical ChangesD. Physical Changes

Physical Changechanges the form of a substance without

changing its identity

properties remain the same

Examples: cutting a sheet of paper, breaking a crystal, all phase changes

D. Phase Changes – PhysicalD. Phase Changes – Physical

Evaporation =

Condensation =

Melting =

Freezing =

Sublimation =

Liquid -> Gas

Gas -> Liquid

Solid -> Liquid

Liquid -> Solid

Solid -> Gas

E. Chemical PropertiesE. Chemical Properties

Chemical Propertydescribes the ability of a substance to

undergo changes in identity

F. Physical vs. Chemical PropertiesF. Physical vs. Chemical Properties

Examples:melting point

flammable

density

magnetic

tarnishes in air

physical

chemical

physical

physical

chemical

G. Chemical ChangesG. Chemical Changes

Process that involves one or more substances changing into a new substanceCommonly referred to as a chemical

reactionNew substances have different

compositions and properties from original substances

G. Chemical ChangesG. Chemical Changes

Signs of a Chemical Changechange in color or odor

formation of a gas

formation of a precipitate (solid)

change in light or heat

H. Physical vs. Chemical ChangesH. Physical vs. Chemical Changes

Examples:rusting iron

dissolving in water

burning a log

melting ice

grinding spices

chemical

physical

chemical

physical

physical

What Type of Change?What Type of Change?

What Type of Change?What Type of Change?

I. Law of Conservation of MassI. Law of Conservation of Mass

Although chemical changes occur, mass is neither created nor destroyed in a chemical reaction

Mass of reactants equals mass of products

massreactants = massproducts

A + B C

III. Classification of Matter (pp. 80-87)Matter FlowchartPure SubstancesMixtures

A. Matter FlowchartA. Matter Flowchart

MATTER

Can it be physically separated?

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed?

noyesIs the composition uniform?

noyes

A. Matter FlowchartA. Matter Flowchart

Examples:graphite

pepper

sugar (sucrose)

paint

soda

element

hetero. mixture

compound

hetero. mixture

solution

B. Pure SubstancesB. Pure Substances

Elementcomposed of identical atomsEX: copper wire, aluminum foil

B. Pure SubstancesB. Pure Substances

Compoundcomposed of 2 or more

elements in a fixed ratio

properties differ from those of individual elements

EX: table salt (NaCl)

C. MixturesC. Mixtures

Variable combination of 2 or more pure substances.

Heterogeneous Homogeneous

C. MixturesC. Mixtures

Solutionhomogeneousvery small particlesparticles don’t settleEX: rubbing alcohol

C. MixturesC. Mixtures

Heterogeneousmedium-sized to

large-sized particles

particles may or may not settle

EX: milk, fresh-squeezed

lemonade

C. MixturesC. Mixtures

Examples: tea

muddy water

fog

saltwater

Italian salad dressing

Answers: Solution

Heterogeneous

Heterogeneous

Solution

Heterogeneous

III. Density FunFormulaSimple CalculationsDimensional Analysis

A. Derived UnitsA. Derived Units

Combination of base unitsVolume – length length length

1 cm3 = 1 mL 1 dm3 = 1 LDensity – mass per unit volume

(g/cm3)

D = MV D

M

V

Broken Heart

B. Density CalculationsB. Density Calculations

An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass.

GIVEN:

V = 825 cm3

D = 13.6 g/cm3

M = ?

WORK:

M = DV

M = (13.6 g/cm3)(825cm3)

M = 11,220 g

M = 11,200 gDM

V

B. Density CalculationsB. Density Calculations

A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid?

GIVEN:

D = 0.87 g/mL

V = ?

M = 25 g

DM

V

WORK:

V = M D

V = 25 g

0.87 g/mL

V = 28.7 mL = 29 mL

B. Density CalculationsB. Density Calculations

You have a sample with a mass of 620 g and a volume of 753 cm3. Find its density.

GIVEN:

M = 620 g

V = 753 cm3

D = ?

DM

V

WORK:

D = M V

D = 620 g

753 cm3

D = 0.82 g/cm3

C. Density Calculations with DAC. Density Calculations with DA

Used when units do not agreeConversions must be made before

using formula

D = MV

D = g

cm3

You have 3.10 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.

lb cm3

3.10 lb 1 kg

2.2 lb= 73.0 cm3

1000 g

1 kg

1 cm3

19.3 g

C. Density Calculations with DAC. Density Calculations with DA

You have 0.500 L of water. Find its mass in ounces if the density of water is 1.00 g/cm3.

L oz0.500 L 1000 mL

1L

= 17.6 oz

1.00g

1 cm3

C. Density Calculations with C. Density Calculations with DADA

1 cm3

1 mL1 kg

1000 g16 oz

1lb

2.2 lbs

1kg