matter has mass and volume classes of matter matter mixtures substances...
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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
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