MatterHas mass and volume

Classes of Matter

Matter

MixturesSubstances

Homogeneous Heterogeneous Elements Compounds

solution

Mixture-2 or more substances are not chemically combined

-proportions can vary- Each part keeps its own properties

Heterogeneous mixture

- Can see different parts, substances are not evenly mixedSand and water Iron and sand

Toenails and scabs

Homogeneous mixture- Cannot see different parts, substances are evenly mixed

- Called a solution

Salt water NaCl (aq)

Pure substances- Cannot by physically separated

- All samples have the same properties

- A sample has the same properties throughout

- m.p., density, f.p. etc.

Element Pure substance which cannot be broken down by chemical means

Cu Cu cannot be broken down into different substances

Compound - Substance which can be separated into unlike parts using chemical means

- Made of 2 or more elements in a definite ratio

NaCl H2O CuSO4

- Smallest part is a molecule

Special symbolsFirst discovered elements were based on Latin names

Symbol Latin Name Common name

Cu

Au

Fe

Pb

Hg

K

Ag

Na

Sn

cuprum

aurum

ferrum

plumbum

hydrargyrum

kalium

argentum

natrium

stannum

copper

gold

iron

lead

mercury

potassium

silver

sodium

tin

Chemical Properties - How a substance interacts with other substances

Ex. Iron + damp air rust

Chemical Change- Production of new substance from original-Sour milk- Decomposition

Evidence of a chemical change

1. Energy is either absorbed or given offExothermic – given offEndothermic – energy is absorbed2. Gas is formed

3. Precipitate is formed

Conservation of Mass- Matter can not be created nor destroyed, it can only change form

Physical Properties- Can be observed without the production of a new substance

Ex. Color, hardness, melting point, density

Density = Mass / volume

Sample Problem: A student measures out 12.1 cm3 Hg and finds the sample weighs 164.56 g. Calculate

a.) the density

b) The mass if the sample has a volume of only 2.15 cm3

c) The volume if the sample has a mass of 94.2 g

Physical change- Results in change in size or shape of substance, but substance remains the same

Describe what you observe when ice is placed in a glass of water.

What happened when ice was added to beaker 1? 2? And 3?

Give the best explanation for what you observed.

The Levitating Golf Ball

What do you think will happen when the golf ball is added to the salt solution?

Examine the golf ball for the next few days and observe what will happen.

Solid

- Definite shape and volume

Liquids- Definite volume / no definite shape

- Particles close together, and slowly vibrating

- Lowest state of P.E.

- Lowest state of entropy (disorder)

- Particles further apart

> P.E.> entropy

- Molecules can vibrate and rotate over one another

- Regular geometric structureCrystal lattice structure

Gases - No definite shape nor volume

-Particles in random motion all over container

-Temperature and pressure changes have large effects on gases

. Changing Phases

-During a phase change, all the energy added does not cause a change in temperature

-All energy added goes to change the phase

-Potential energy stored as a change of phase

-Boiling water stays at 100oC, it does not heat up

- Highest P.E. and entropy

-The ability to do work

Types of EnergyTypes of Energy

Potential EnergyStored energy, usually due to a change in position

Kinetic energyEnergy of motion

Faster an object moves = more KEForms of EnergyForms of Energy

Thermal Heat energyChemical energyEnergy from chemical bonds

Released during chemical changes

Radiant Light energy

Nuclear Energy due to a change in an atom’s nucleus

Some PE

More PEEnergy

Law of conservation of EnergyLaw of conservation of Energy Energy can be converted from one form to another but is never created of destroyed

Measuring EnergyTemperature- Average Kinetic Energy of molecules in a substance- Average Kinetic Energy of molecules in a substance

- The higher the temp the more K.E. the substance has- Does NOT measure heat, only the relative motion of particles

Which has a higher average K.E. ?

20g of Water at 40oC or 10 g of CO2 at 30oC

Thermometer Hollow tube

Filled with a liquid that expands easily

Mercury HgVery high boiling point, very low melting point

Very little evaporation

Alcohol Smaller liquid range, Less expensive and less dangerous

Temperature scales

Need two reference points to create a scaleChemists use the boiling and the freezing point of pure waterChemists use the boiling and the freezing point of pure water

Fahrenheit oF Water boils at212oFWater freezes at 32oF

Problem - Numbers are sometimes difficult to work with

Celsius oC Easier to use in scienceWater boils at 100oC

Water freezes at 0oC

Kelvin Scale

Indicates the total kinetic energy of the sample

Changing oC and K

K

1 degree of Celsius = 1 unit of Kelvin

Kelvin is always higher by 273

10oC = ______ K283 -10oC = ______ K263

Formula on Table T of reference tables!!!

K = oC + 273

What K temperature = -33 oC ?What C temperature = 260 K?

At which temperature would molecules of 1 g of water have the greatest K.E. ? 5 o C or 5 K

KelvinCelsius

0oC 273 K1oC 274 K

-1oC 272 K

Absolute Zero

Coldest possible temperatureO K or -273oC

All motion stops

Cooking with Paper Demo:

Predict what will happen to the paper cup when placed over the bunsen burner.

Explain what did happen to the cup and give your best explanation for what you observed.

Predict what will happen to the balloon placed over the hot flame and give a reason for your prediciton.

Heating Curve

Time

Temp Solid phase

Melting

No temp change

Liquid phase

Boiling

No temp change

Gasphase

PhasesKE (temp) increasesPhase changeKE does not change, PE changes

KE KEPE

KE

KEKEPE

- Heat added at a constant rate

Cooling Curve

Reverse of Heating Curve

Time

Temp

Reverse of Boiling CondensationOccurs at the same temp as boiling

Reverse of MeltingFreezing or CrystallizationOccurs at the same temp as melting

Condensation

Freezing or Crystallization

Heating curve

- Heat removed at a constant rate

Heat - The amount of energy transferred from one substance to another

Calorimeter - Used to measure temp change of a known quantity of water

Heat always flows from object with greater temp to one with lower temp

- Unit of heat energyJoule

Specific heat of water = 4.18J/g oC

Formula for Heat

q = m c T

q = Heat ( lost or gained) joules

m = mass grams

T = Change in temperature oC

C = Specific heat Amount of energy needed to raise the temperature of 1 g of a substance 1oC

Water

Copper

Aluminum

Iron

4.18 J/g oC

0.373 J/g oC

0.880 J/g oC

0.461 J/g oC

Heats up easier and cools down easierTable B

Examples

How much energy is needed to raise the temperature of 10.0 g of Cu from 16oC to 21oC?

q = m c T

q = (10 g) (0.373 J/g oC) (5oC)

q = 18.65 Joules

How much water can be heated from 2.0oC to 50.0oC using 3.0 Kjoules?

Kjoule = 1000 Joules

q = m c T

3000 Joules = (m) (4.18 J/g oC) (48oC)3000 Joules = 201 (m)

14.92 g

How many joules of energy is required to heat 22.0 g of water from 10.0oC to 30.0oC? q = m c T

q = (22.0 g)(4.18 J/goC) (20.0oC)

q = 1839.2 Joules 1840 J

If we add 150 Joules of heat to 20.0 g of water, and the original temperature of the water is 8.00oC, what is the final temperature?

Step 1 – Find the temperature change

q = m c T

150 J = (20.0 g)(4.18 J/goC) (ΔT)

150 = 84 (ΔT)1.8 oC1.79 = ΔT

Step 2 – Find the final temperature

8.00 + 1.8 = 9.8oC

Calculate the specific heat of ice if 164 J of heat is removed from 10g of ice to cool it 8.0 oC.

q = m c ∆t

164 J = 10 g (c) 8.0 oC

164 J = 80.0 g oC (c)

164 J / 80 g oC = c

2.05 J / g oC = c

Heat of Fusion

Energy added /lost is used to change the phase

Energy needed to turn 1 g of a solid into 1 g of a liquid

q = m c T does not apply during a phase change ( no T )

( melting / crystallization )

Hf of water 334 Joules/g

Example - How much energy is required to melt 10.0 g of ice at 0oC to 10 g of water at 0oC?

Q = Hf x mQ = 334 x 10.0 gQ = 3340 joules

How much ice at 0oC can we melt to water at 0oC using 1.36 Kjoules of heat?

Q = Hf x m

1360 joules = 334 x m4.07 g

Found on the first page of your reference tables!Found on the first page of your reference tables!

Table B

Heat of Fusion = Heat of Crystallization

Remove 334 joules of heat from 1 g of water to turn it into 1 g of ice

Heat of Vaporization ( boiling / condensation )

q = m Hv Hv = heat of vaporizationHv = amount of heat that must be added / removed from 1

g of a substance to boil / condense it at it’s boiling/condensation point

Ex. Calculate the Hv of alcohol if it takes 42750 J of heat to vaporize 50.0 g of it at its b.p.

How much energy is needed to completely boil 50.0 g of water at 100oC to 50.0 g of steam at 100oC?

q =m x Hv

q = 2260 x 50.0

q = 113,000 J

How much water at 100oC can we boil to steam at 100oC using 11.63 kJ of heat?

q = m x Hv

11,630 = 2260 x m

5.15 g

For water, Hv = 2260 J/g

To boil, add 2260 JTo boil, add 2260 J To condense, remove 2260 JTo condense, remove 2260 J

Heat of Vaporization = Heat of Condensation

Table B

m=

Pressure –

- The force exerted on a unit of area

Atmospheric Pressure - The force exerted by the “ocean” of air above the earth

Force due to the air being pulled to Earth by gravity

STD Pressure =Average pressure at sea level

Table A 1 atm = 760 mmHg = 760 torr = 101.3 kPa

Vacuum Pump Demo

When the pump is running what happens to the air pressure in the jar? Why?

What happens to the temperature in the jar?

A beaker of warm water is placed under the bell jar, what will happen when the vacuum is turned on?

Describe what you observed and give an explanation for this observation.

Vapor pressurePressure exerted by molecules in the gas phase

- Boiling occurs when a liquids v.p = atmospheric pressure

- As temp goes up, vapor pressure goes up

At the point when the water began to boil, how did the vapor pressure of water compare to the atmospheric pressure in the jar?

A pressure cooker is designed to cook foods faster by getting water to boil at temperatures above 100 C. How do you think this works.

High in the mountains there is less air than at lower altitudes. Why does it take so much longer to cook in the mountains than at sea level?

Table H Pressure exerted by the vapor of 4 different liquids at different temperatures

What is the vapor pressure of 100.0 ml of H2O at 75oC?

38 KPa

Using table H, predict whether ethanol or water will boil first when placed under the jar. Explain in terms of vapor pressure.

Which has stronger intermolecular forces (attraction for itself), water or ethanol? Explain.

Explain what happens to the marshmallow in terms of internal pressure and atmospheric pressure.

bubble

1) When atm pressure > v.p. bubbles collapse

2) Temp inc causing V.P. to increase – atm pressure is constant3) At B.P. the V.P. = Atm pressure bubbles get larger and erupt

Boiling

Boiling Temperature

Boiling point of a substance changes with atmospheric pressure

More pressure = higher temperature needed to get molecules to escape as a gas

Ex. Pressure cookers, cooking in the mountains

Temp at which the vapor pressure is equal to the atmospheric pressure

Food Cooks faster Food cooks slowerExamples

If the atmospheric pressure is 80.0 KPa, what will the boiling point of water be?

What is the boiling points of ethanoic acid and propanone at standard pressure?

Water’s VP line crosses the 80 KPa line at 94oC

At 101.3 KPa, they are 117oC and 56oC

EvaporationOccurs at temperatures below the boiling pointMolecules at the surface enter the gas phase

Gas = vapor

Volatile Liquids - Vaporize easily due to weak I.M.F’s

Ex. Alcohol, acetone, gasoline

If the force between molecules is weak, more molecules can enter the gas phase at lower temperatures.

Who did it, iodine phase change.

Mr. Stone found the following note on his desk. Using chemistry and the properties of iodine he was able to find the culprit!!!

What phase change took place.

What caused the fingerprint to occur.

Sublimation

Going from solid directly to the gas phase

Skips the liquid phase

2 important examples

CO2(s) CO2(g)

I2(s) I2(g)

Dry ice

Gas Pressure

Caused as gas molecules collide with the walls of their container

More collisions = more pressure

GasesKinetic Molecular Theory

1) Gases are in constant, random, straight line motion2) Gases collide with each other and with the walls of their container with no loss of energy

3) Gas particles are separated by large distances

- Volumes of molecules are negligible

4) Gas particles do not attract each other- Because mass of particles is so small

Gas Laws Show how pressure, volume, and temperature, are related to gases.

Boyle’s LawBoyle’s Law – relates Pressure and Volume

As Pressure Volume

(temperature is constant)

- Inverse relationship

P1V1 = P2V2

Ex 1. A gas at 1.5 atm is held at a volume of 545 ml. The volume of the container increases to 1065 ml with NO CHANGE IN TEMP. Calculate the new pressure.

Given:

P1 =

V1 =

V2 =

1.5 atm

545 ml

1065 ml

P2= ?

T1 =

T2 =

P1 V1 = P2 V2

T1 T2

1.5 atm ( 545 ml) = 1065 ml (?)? = .77 atm

(Pressure must be held constant)Charle’s Law Charle’s Law - relates temperature and volume

- Direct relationship - As temp increases, volume increases

V1 = V2

T1 T2

P1 V1 = P2 V2

T1 T2

Blowing up Balloon Demo:

Predict what will happen to the balloon when it is placed over the top of the flask.

Explain what you observed using Charle’s Law.

Practice

If the volume of a gas at 10.0oC is 100.0 ml, then what is the volume at -2.0oC?

V1 =V2

T1 T2

100.0 ml10.0o

C

= V2

-2.0oC

10.0 x V2 -200

V2 = -20 ml

=

T1 = 10oCV1 = 100.0 mL

V2 = ?

T2 = -2.0oC

We can’t have negative volumes, so we can’t use a We can’t have negative volumes, so we can’t use a temperature scale with negativestemperature scale with negatives

With gas laws, ALWAYS USE KELVINALWAYS USE KELVIN

V1 =V2

T1 T2

100.0 ml

283 K= V2

271 K

283 x V2 27100

V2 = 95.7597 mL

=

T1 = 10oC

V1 = 100.0 mL

V2 = ?

T2 = -2.0oC

283 K

271 K95 mL

Because temp went down the volume had to go ___________

What if all three variables change?

P1 V1 = P2 V2

T1 T2Combined Gas Law

Example - A gas is collected at 273 K and 2.00 atm to a volume of 50.0 ml. What is the new pressure of a gas if the temperature drops to 200.0 K and the volume increases to 75.0 ml?

P1V1 = P2V2

T1 T2

2.00 x 50 ml

273 K= P2 x 75.0 ml

200.0 K

20,475 x P2 20,000

P2 = 0.97680

=

T1 = 273 K

V1 = 50 mL

V2 = 75.0 ml

T2 = 200.0 K

0.977 atm

P1 = 2.00 atm

P2 = ?

If any of the variables are CONSTANT, you can cross off the variable.

Always use Kelvin for temperature

Write out your variables!

ExamplesIf the pressure of 15.0 ml of gas changes from 10.0 KPa to 25.0 KPa and the temperature remains constant, what is the new volume of the gas?

P1V1 = P2V2

T1 T2

10.0 x 15.0 ml= 25.0 x V2

150 25 x V2

V2 = 6

=

V1 = 15.0 mL

V2 = ? ml

6.00 ml

P1 = 10.0 KPa

P2 = 25.0 KPa

X X

Egg in a Bottle Demo

Mr Stone will be lighting a piece of cotton that’s been soaked in isopropyl alcohol, placing this in the flask and then putting a hard boiled egg on top. Predict what will happen.

Using the combined gas law explain how Mr. Stone was able to get the egg into the bottle.

Make a hypothesis as how to remove the egg from the jar without breaking the egg up.

Brainstorm another way to the egg into the jar without using the flame.

Examine the apparatus set up in the front of the room. What do you predict will happen when the heat is removed and the cap is replaced?

Explain what you observed and why it happened.

On a large scale this happened when a cleaning crew steam cleaned the inside of the tanker below and then quickly replaced the top!!! Ouch!

In order to compare gas, we need a standard set of conditions

Standard Temperature

0oC 273 K

Standard Pressure

760 torrs 101.3 KPa 1.00 atm

What is the volume of a gas at STP if its volume at 2.2 atm and 210 K is 1.0 L?

STP

What is the temperature of a gas with a volume of 16.00 ml if the pressure remains constant and at STP it has a volume of 22.0 ml?

Ideal Gas Law - Assumes all gases act the same

1. A gas is composed of individual particles in continuous, random, straight line motion. Collisions with each other and sides of container cause pressure.

2. Gas particles are far apart so that total volume of each particle is very small in comparison to the total volume of the gas

3. Gas particles display no attraction or repulsion for one another

Gases that conform to these rules are ideal gases

No real gas is ideal under all conditions of temperature and pressureReal gases behave like ideal gases under conditions of high

temperature and low pressure

Hydrogen and Helium act most like ideal gases

Law is not perfect because gas particles do….1. have volume

2. exert some attraction for each other

1) When 20 g of water cools from 30 C to 20 C, how much heat is given off?

2) A 25 g sample of water is cooled from 363 K to 352 K. How much heat was released?

3) A sample of water is heated from 10 C to 15 C by the addition of 30 J of heat. What is the mass of the water?

4) The temperature of 50 g of water was raised to 50 C by the addition of 1000 J of heat. What was the initial temp. of the water?

5) Which change in temperature of a 1 g sample of water would cause the greatest increase in the average kinetic

energy of itsmolecules?

(1) 1 C to 10 C (2) 10 C to 1 C (3) 50 C to 60 C (4) 60 C to 50 C