Chemistry Unit 2 pp.pptx [repaired]

This is a unit of study called “Matter and Energy”…
…in which basic properties of matter and energy transformations are discussed.
Slides that utilize media are:
…Slide 52 has video on demand that can be viewed online at any time for students.
…slide 69 has sound to correspond to blocks falling in water. Slide is animated so students can visual what is happening.
A transparent overlay covers the block of wood and ice. As a teacher I point out that the wood is 25% submerged as it has a
specific gravity of 0.25 while the ice is submerged 90% as it has a specific gravity of 0.90. I then show this actual demonstration
to students using a document camera.
…slide 71 has sound to correspond to cloud of CO2 falling in down stairs because it is denser than air.
A voice explanation is also included to give a brief explanation.
… slide 202-207 demonstrate the theory and show chromatography by animated clicks on PowerPoint
I would give a verbal explanation in class to explain the slides.
… slide 227-231 concern electrolysis
slides 230 and 231 have an embedded video I that plays when clicked. Video came from PBS tv show.
A voice explanation is also included to give a brief explanation
…slide 238-261 explain properties of matter.
A slide 238 and 241 have video to give additional information
…Slide 255 are a series of jpeg images I captured from video I shot to sequence as a slide that looks like video.
Slide 256 has the actual video of class demonstration I recorded. It plays regular speed and slow motion.
Each was created using Windows Movie Maker.
…slide 261 has an external link to an animation by a textbook author (bottom right corner)
…Many slides have additional links to additional information (e.g. slide 8 (mouse over the word “matches “to get more
information from Wikipedia)

www.unit5.org/chemistry
Energy and Matter
Unit 2

Guiding Questions
Why do substances boil or freeze at different temperatures?
Why do we put salt on the roads in the winter?
Why does sweating cool us?
What is energy?
How do we measure energy?

Table of Contents
‘Matter and Energy’
(13) Introduction - Bonding
(14) Temperature vs. Heat
(11) Density
(6) Carbon Dioxide & Monoxide
(4) Archimede’s Principle
(3) Galilean Thermometer
(11) Golf Ball Lab
(15) Solid, Liquid, and Gas
(3) Heating Curve
(13) Classification of Matter
(6) Crystalline Structure
(10) Allotropes
(9) Alloys
(4) Separation Techniques
(11) Distillation
(2) Centrifugation
(3) Electrolysis
(5) Properties of Matter
(6) Energy
(11) Exothermic vs. Endothermic
(29) Calorimetry
(12) Nuclear Energy

Lecture Outline – Energy and Matter
Keys
student notes outline
textbook questions
http://www.unit5.org/chemistry/Matter.html
textbook questions
text

Chemistry of Matches
P4S3 + KClO3 P2O5 + KCl + SO2
tetraphosphorus
trisulfide
potassium
chlorate
diphosphorus
pentaoxide
potassium
chloride
sulfur
dioxide
D
The substances P4S3 and KClO3 are both
present on the tip of a strike anywhere match.
When the match is struck on a rough surface,
the two chemicals (reactants) ignite and
produce a flame.
Charles H.Corwin, Introductory Chemistry 2005, page 182
Safety matches
The products from this reaction are P2O5, KCl, and SO2,the
last of which is responsible for the characteristic sulfur smell.
Strike anywhere matches
The substances P4S3 and KClO3
are separated. The P4S3 is on
the matchbox cover.
Only when the chemicals combine
do they react and produce a flame.

block of wood: length = 2.0 m width = 0.9 m height = 0.5 m
block of wood: force = 45 N
2.205 pounds = 1 kilogram 10 Newton (9.8 N)

Force versus Pressure
Area = 0.9 m x 2.0 m
= 1.8 m2
Area = 0.5 m x 2.0 m
= 1.0 m2
Area = 0.5 m x 0.9 m
= 0.45 m2
area
force
Pressure 
2
m1.8
N45.0
Pressure 
2
m1.00
N45.0
Pressure 
2
m0.45
N45.0
Pressure 
block of wood: length = 2.0 m width = 0.9 m height = 0.5 m
25 N/m2
45 N/m2
100 N/m2
Herron, Frank, Sarquis, Sarquis, Schrader, Kulka, Chemistry, Heath Publishing,1996, page Section 6.1

Pressure
area
force
pressure 
Which shoes create the most pressure?

During a
“physical change”
a substance changes
some physical
property…
H2O

…but it is still the
same material with
the same chemical
composition.
H2O
gas
solid
liquid

Chemical Property:
The tendency of a
substance to change into
another substance.
caused by iron (Fe)
reacting with oxygen (O2)
to produce rust (Fe2O3)
Steel rusting:
4 Fe + 3 O2 2 Fe2O3

Chemical Change:
Any change involving a
rearrangement of atoms.

Chemical Reaction:
The process of a
chemical change...

During a
“chemical reaction”
new materials are
formed by a change
in the way atoms are
bonded together.

Physical and Chemical Properties
Examples of Physical Properties
Boiling point Color Slipperiness Electrical conductivity
Melting point Taste Odor Dissolves in water
Shininess (luster) Softness Ductility Viscosity (resistance to flow)
Volatility Hardness Malleability Density (mass / volume ratio)
Examples of Chemical Properties
Burns in air Reacts with certain acids Decomposes when heated
Explodes Reacts with certain metals Reacts with certain nonmetals
Tarnishes Reacts with water Is toxic
Ralph A. Burns, Fundamentals of Chemistry 1999, page 23
Chemical properties can ONLY be observed during a chemical reaction!

The formation of a
mixture
The formation of a
compound

Physical & Chemical Changes
Limestone,
CaCO3
crushing
PHYSICAL
CHANGE
Crushed limestone,
CaCO3
heating
CHEMICAL
CHANGE
Pyrex
CO2
CaO
Lime and
carbon dioxide,
CaO + CO2

Pyrex
O2
H2O
Pyrex
H2O2
Light hastens the decomposition of hydrogen peroxide, H2O2.
The dark bottle in which hydrogen peroxide is usually stored
keeps out the light, thus protecting the H2O2 from decomposition.
Sunlight
energy
H H
O O

Three Possible Types of Bonds
+ -
d+ d-
Covalent
e.g. H2
Polar Covalent
e.g. HCl
Ionic
e.g. NaCl

Metallic Bonding
Metallic bonding is the attraction between positive ions and
surrounding freely mobile electrons. Most metals contribute
more than one mobile electron per atom.
“electron sea”
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
e1-
Free electrons
+
+
+
+
+
++
+
+
+
+
+
+
+
Cations
Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 245

Shattering an Ionic Crystal; Bending a Metal
Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 248
+ +
++
+ ++
+++
+
+ +
+
+
+
+
++
+ ++
+++
+
+ +
+
+
++ + + + ++ +
+ +
+
+
+ + + ++ + +
+
+ +
++
+ ++
+++
+
+ +
+
+
+
+
++
+ ++
+++
+
+ +
+
+
++ + + + ++ +
+ +
+
+
+ + + ++ + +
+
+-
+ --
- +
+
- -+
- ++-
+ -
-
+
+ -+-- - -++
-
+
+ - + - -+ + -
+-
+ --
- +
+
- -+
- ++-
+ -
-
+
+ -+-- - -++
-
+
+ - + - -+ + -
An ionic crystal
A metal
No electrostatic forces of repulsion –
metal is deformed (malleable)
Electrostatic
forces
of repulsion
Force
Force
broken crystal

Properties of Ionic Compounds
• Crystalline solids
• Hard and brittle
• High melting points
• High boiling points
• High heats of vaporization
• High heats of fusion
• Good conductors of electricity when molten
• Poor conductors of heat and electricity when
solid
• Many are soluble in water

Chemical Bonds
Increasing ionic character
Covalent bonding
Electrons are shared
equally
Cl Cl
Polar covalent bonding
unequally
ClH
Ionic bonding
Electrons are
transferred
Cl1-Na1+
• between two identical nonmetal atoms are non-polar covalent.
• between two different nonmetal atoms are polar covalent.
• between nonmetals and reactive metals are primarily ionic.

Chemical Bonds
Increasing ionic character
Nonpolar covalent
equally
Cl Cl
Polar covalent
unequally
ClH
Ionic bonding
Electrons are
transferred
Cl1-Na1+
• between two identical nonmetal atoms are nonpolar covalent.
• between two different nonmetal atoms are polar covalent.
• between nonmetals and reactive metals are primarily ionic.

Covalent vs. Ionic
Covalent
Transfer
electrons
(ions formed)
+ / -
Between
Metal and
Nonmetal
Strong
Bonds
(high melting point)
Share
electrons
(polar vs. nonpolar)
Between
Two
Nonmetals
Weak
Bonds
(low melting point)
Alike Different
Electrons
are
involved
Chemical
Bonds
Ionic
Different
Topic Topic

Photoelectric
Generator
Solar Calculator
Radiant energy Evacuated
chamber
Metal
surface
Current
indicator
Positive
terminal
Voltage
source
cathode anode
Symbolic representation
of a photoelectric cell
cathode anode
evacuated glass
envelope
Photoelectric Cell

Celsius & Kelvin Temperature Scales
Boiling point
of water
Freezing point
of water
Absolute
zero
Celsius
100
Celsius
degrees
100oC
0oC
-273oC
Kelvin
100
Kelvins
373 K
273 K
0 K

Temperature is Average Kinetic Energy
Fast Slow
“HOT” “COLD”
Kinetic Energy (KE) = ½ m v2
*Vector = gives direction and magnitude

Temperature Scales
Fahrenheit
212 oF
180 oF
32 oF
Celcius
100 oC
100 oC
0 oC
Kelvin
373 K
100 K
273 K
Boiling point
of water
Freezing point
of water
Notice that 1 kelvin degree = 1 degree Celcius

Kelvin Scale
blue
white
yellow
4300 K PIAA HID Bulb
5000 K PIAA Plasma Blue
5250 K Sunlight
4150 K PIAA Xtreme White
3800 K PIAA Super White
3200 K Halogen Bulb
2600 K Incandescent Bulb

Temperature Scales
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 136

Compare Celsius to Fahrenheit
oF – 32 = 1.8 oC

Converting
70 degrees
Celsius to
Kelvin
units.
oC + 273 = K

Temperature Scales
• Temperature can be subjective and so fixed scales had to be
introduced.
• The boiling point and freezing point of water are two such points.
• Celsius scale (oC)
– The Celsius scale divides the range from freezing to boiling into 100
divisions.
– Original scale had freezing as 100 and boiling as 0.
– Today freezing is 0 oC and boiling is 100 oC.
• Fahrenheit scale (oF)
• Mercury and alcohol thermometers rely on thermal expansion

Thermal Expansion
• Most objects e-x-p-a-n-d when heated
• Large structures such as bridges must be
built to leave room for thermal expansion
• All features expand together
COLD
HOT
Cracks in sidewalk.

Equal Masses of Hot and Cold Water
Thin metal wall
Insulated box Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Water Molecules in Hot and Cold Water
Hot water Cold Water
90 oC 10 oC

Water Molecules in the same
temperature water
Water
(50 oC)
Water
(50 oC)Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Heat versus Temperature
Kinetic energy
Fractionsofparticles
lower temperature
higher temperature
TOTAL
Kinetic ENERGY
= Heat

Molecular Velocities
speed
many different molecular speeds
molecules sorted by speed
the Maxwell speed distribution
http://antoine.frostburg.edu/chem/senese/101/gases/slides/sld016.htm

Temperature vs. Heat
Measured
with a
Calorimeter
Total
Kinetic
Energy
Joules
(calories)
Measured
with a
Thermometer
Average
Kinetic
Energy
oCelcius
(or Kelvin)
Alike Different
A Property
of
Matter
Have
Kinetic
Energy
Heat
Different
Topic Topic
Temperature

Conservation of Matter
Reactants yield Products

Heavy Metal Poisoning
Exposure to mercury
made the Hatter “mad”.
Eating chips of lead paint
causes brain damage.
TREATMENT: Chelation therapy
EDTA (ethylenediamine tetra acetic acid)
Arsenic treated lumber.
‘Green-treated’ wood
will not rot outdoors
for 50 years.

Density
• Density is an
INTENSIVE property
of matter.
- does NOT depend
on quantity of
matter.
- color, melting point, boiling point, odor, density
• Contrast with
EXTENSIVE
- depends on
quantity of matter.
- mass, volume, heat content (calories)
Styrofoam Brick

Properties of Matter
http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld001.htm
Pyrex Pyrex
Extensive
Properties
Intensive
Properties
volume:
mass:
density:
temperature:
100 mL
99.9347 g
0.999 g/mL
20oC
15 mL
14.9902 g
0.999 g/mL
20oC

Styrofoam Brick
?
It appears that the brick is ~40x
more dense than the Styrofoam.

MM
V
= =D
V
D
BrickStyrofoam
Styrofoam Brick

Which liquid has the highest density?
52
3
1
4
Coussement, DeSchepper, et al. , Brain Strains Power Puzzles 2002, page 16
least dense 1 < 3 < 5 < 2 < 4 most dense

Cube
Representations
1 m3 = 1 000 000 cm3

Volume and Density
Relationship Between Volume and Density for Identical Masses of Common Substances
Cube of substance Mass Volume Density
Substance (face shown actual size) (g) (cm3) (g/cm3)
Lithium
Water
Aluminum
Lead
10 19 0.53
10 10 1.0
10 3.7 2.7
10 0.58 11.4

Density
D
M
Vensity
ass
olume
D =
M
V
M = D x V
V =
M
D

Volume
Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 41
6 cm
3 cm
2 cm
1 cm
4 cm4 cm

2 cm
2 cm
2 cm
Volume
Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 41
V = length x width x heightV = 2 cm x 2 cm x 2 cmV = 8 cm3
Volume = length x width x heightVolume = 6 cm x 2 cm x 3 cm
6 cm
3 cm
2 cm
1 cm
4 cm
Volume = 36 cm3
Volume =
Volume = 28 cm3
36 cm3
8 cm3
-

Density of Some Common Substances
Density of Some
Common Substance
Substance Density
(g / cm3)
Air 0.0013*
Lithium 0.53
Ice 0.917
Water 1.00
Aluminum 2.70
Iron 7.86
Lead 11.4
Gold 19.3
*at 0oC and 1 atm pressure

Consider Equal Volumes
The more massive object
(the gold cube) has the
_________ density.
Equal volumes…
…but unequal masses
aluminum gold
GREATER
Density =
Mass
Volume
Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 71

Consider Equal Masses
Equal masses…
…but unequal volumes.
The object with the
larger volume
(aluminum cube) has
the density.
aluminum
gold
smaller
Christopherson Scales
Made in Normal, Illinois USA

Two ways of
viewing
density
Equal volumes…
…but unequal masses
The more massive object
(the gold cube) has the
greater density.
aluminum gold
(A)
Equal masses…
…but unequal volumes.
(B)
gold
aluminum
The object with the
larger volume
(aluminum cube) has
the smaller density.

Carbon Dioxide Detector
• Where is the best location to place a
CO2 detector in your home?
Recall: Density Air = 1.29 g/L
Density CO2 = 1.96 g/L
A. Top floor of home
B. Basement (near ceiling)
C. Basement (near floor)
D. It doesn’t matter, if your batteries are dead in the detector
C. Basement (near floor) Carbon dioxide is denser than air and sinks.

Symptoms of CO Poisoning
Concentration of CO
in air (ppm)*
Hemoglobin
molecules as HbCO
Visible effects
100 for 1 hour or less 10% or less no visible symptoms
500 for 1 hour or less 20% mild to throbbing headache,
some dizziness, impaired
perception
500 for an extended period
of time
30 - 50% headache, confusion, nausea,
dizziness, muscular weakness,
fainting
1000 for 1 hour or less 50 - 80% coma, convulsions, respiratory
failure, death
*ppm is parts per million
Davis, Metcalfe, Williams, Castka, Modern Chemistry, 1999, page 760

Carbon Monoxide Poisoning
‘The Silent Killer’
Poisoning: Hb + CO  HbCO
Hemoglobin (Hb) binds with carbon monoxide (CO) in the capillaries of the lungs.
If caught in time, giving pure oxygen (O2) revives victim of CO poisoning.
Treatment causes carboxyhemoglobin (HbCO) to be converted slowly to
oxyhemoglobin (HbO2).
Treatment: O2 + HbCO  CO + HbO2
Carbon monoxide, CO, has almost 200 times the affinity to bind
with hemoglobin, Hb, in the blood as does oxygen, O2.
Davis, Metcalfe, Williams, Castka, Modern Chemistry, 1999, page 760

Tank of Water

Person Submerged in Water

Galilean Thermometer
• Density = Mass / Volume
• Mass is constant
• Volume changes with temperature
– Increase temperature  larger volume
In the Galilean thermometer, the small glass bulbs are partly
filled with a different (colored) liquid. Each is filled with a
slightly different amount, ranging from lightest at the
uppermost bulb to heaviest at the lowermost bulb. The clear
liquid in which the bulbs are submerged is not water, but
some inert hydrocarbon (probably chosen because its
density varies with temperature more than that of water
does).
Temp = 68 oC

Galilean Thermometer
In the Galilean thermometer, the small
glass bulbs are partly filled with a different
(colored) liquid. Each is filled with a slightly
different amount, ranging from lightest at
the uppermost bulb to heaviest at the
lowermost bulb. The clear liquid in which
the bulbs are submerged is not water, but
some inert hydrocarbon (probably chosen
because its density varies with temperature
more than that of water does).
The correct temperature is the lowest
floating bulb. As temperature increases,
density of the clear medium decreases
(and bulbs sink).
RECALL: Density equals mass / volume.
80o
76o
80o
76o
72o
64o
68o
64o
68o
72o
76oF 68 oF

Dissolving of Salt in Water
NaCl(s) + H2O  Na+(aq) + Cl-(aq)
Cl-
ions
Na+
ions Water molecules

Determine the minimum amount of salt needed
to make a golf ball float in 100 mL water.
Weigh out 50.0 g of NaCl
Trial Salt (g) Total Float /Sink
1 5.0 g 5.0 g Sink
2 5.0 g 10.0 g Sink
3 5.0 g 15.0 g Sink
4 5.0 g 20.0 g Sink
5 5.0 g 25.0 g Float
Add 5 g additions of salt to the water, dissolve, check to see if ball floats.
Continue with this method of successive additions until ball floats.
Re-weigh remaining salt and subtract this amount from 50.0 g to determine
the amount of salt needed.
Finally, repeat…begin 5 g less salt and add 1 g increments to narrow range.

Theorize, but Verify
Jaffe, New World of Chemistry, 1955, page 1
…We must trust in nothing but
facts. These are presented to us
by nature and cannot deceive. We
ought in every instance to submit
our reasoning to the test
of experiment. It is especially
necessary to guard against the
extravagances of imagination which
incline to step beyond the bounds
of truth.
Antoine Laurent Lavoisier, 1743 - 1794

Theory Guides, Laboratory Decides!
Density of water = 1.0 g/mL
Need to determine density of a golf ball.
mass =______ g (electronic balance)
volume = ______ mL (water displacement method) or formula?
Density of golf ball cannot be made to decrease. Therefore, you need to
increase the density of the water by dissolving salt into the water.
Limiting Factor: accurate determination of volume of golf ball
Solubility Curve of salt in water.
Water has a limit to how much salt can be dissolved.
Saturation – point at which the solution is full and cannot hold anymore solute.

Packing of NaCl Ions
Electron Microscope
Photograph of NaCl

Dissolving of NaCl
Timberlake, Chemistry 7th Edition, page 287
HH
O
Na+
+
-
- + -
+
+
-
Cl-
+ -
+
hydrated ions

100 mL
Interstitial Spaces and Particle Size
Interstitial spaces
(holes in water where substances dissolve)
Parking at school if you arrive at 7:00 AM = _____
Parking at school if you arrive at 7:45 AM = _____
More available spaces if you arrive early. Salt dissolves quicker when you
begin because there are more available spaces to 'park'.
Analogy: Compact car is easier to park than SUV.
STIR
Easy
Hard
Theory: Crush salt to make particles smaller (increase surface area)
…it will dissolve more rapidly.

100 mL of water = 100 g
Add 3.0 mL water,stir…float
You determine the density of golf ball to be 1.18 g/mL
density of water= 1.00 g/mL
Add 19 g salt to 100 g water = 119 g salt + water
Volume remains100 mL (saltwater)
Density = Mass
volume
119 g
100 mLor
Density (saltwater) = 1.19 g/mL
If golf ball doesn’t float, add 2 mL additions of salt until it floats.
Add 3.0 mL water,stir…float
Add 3.0 mL water,stir…sink
mL100
saltgx
mL6mL100
g119



Goals and Objectives:
a. Given materials and problem, formulate and test a hypothesis to
determine if a golf ball can float in salt water.
b. Collect accurate data and compare own data to other class
data. Evaluate own results.
Investigation Procedure:
a. Design an experiment to accurately determine how dense salt water
must be in order for a golf ball to float. Use metric units. Be sure to
control as many variables as possible.
b. Write down the procedure that you and your partner(s) are going to use
prior to lab day. Record any researched facts that may be useful in
knowing before conducting your experiment.
c. Carefully run your experiment, make observations and record your
measurements in a data table. Use grams and milliliters in your
measurements. Include a calculation column in your data table.
d. Critique your own procedure, discuss and compare your process with
another group, then modify your own steps as needed.
e. Repeat your experiment to check for accuracy, if time allows.

Discussion Questions for Understanding:
a. How did you determine the density of your golf ball?
b. Why does a golf ball normally sink to the bottom of a pond at the golf course?
c. What variables were difficult or impossible for you to control during this
experiment?
How much salt can be dissolved in 100 mL of water? (saturated)
effect of temperature on solubility
Surface area of salt may affect rate of dissolving (may need to crush salt finely)
d. What variables may have changed as time went on that could have affected
the outcome of your results?
e. Did you improve the accuracy of your results after conferring with another
group?
f. Describe your sources of error.
(Human error and faulty equipment are unacceptable answers)

Materials:
electronic balance 100 mL & 500 mL graduated cylinder
mortar / pestle glass stirring rod
golf ball salt (Kosher, iodized table salt, table salt)
250 mL beaker
Extension:
a. Research the manufacturing of golf balls to determine why they sink in
pond water.
b. Research to determine which body of salt water in the world would float a
golf ball the highest.
Lab Report : (10 - 12 point font two page maximum length)
Background / problem
Hypothesis (if...then)
Procedure (protocol)
Data (table, graph)
Analysis
Conclusions / Future directions (limitations)
Sample calculations - Appendix
Do not use references to yourself or others in your writing of a lab report
(except for citing past research).
OR
Poster (25 words or less) A picture is worth 1000 words!

Solid, Liquid, Gas
(a) Particles in solid (b) Particles in liquid (c) Particles in gas

Solid
H2O(s) Ice

Ice
H2O(s) Ice
Photograph of ice model Photograph of snowflakes
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

Liquid
H2O(l) Water
In a liquid
• molecules are in
constant motion
• there are appreciable
intermolecular forces
• molecules are close
together
• Liquids are almost
incompressible
• Liquids do not fill the
container

Gas
H2O(g) Steam

Liquids
The two key properties we need to describe are
EVAPORATION and its opposite CONDENSATION
add energy and break intermolecular bonds
EVAPORATION
release energy and form intermolecular bonds
CONDENSATION

Gas, Liquid, and Solid
Gas Liquid Solid

States of Matter
Solid Liquid Gas
Holds Shape
Fixed Volume
Shape of Container
Free Surface
Fixed Volume
Shape of Container
Volume of Container
heat heat

Some Properties of Solids, Liquids, and Gases
Property Solid Liquid Gas
Shape Has definite shape Takes the shape of Takes the shape
the container of its container
Volume Has a definite volume Has a definite volume Fills the volume of
the container
Arrangement of Fixed, very close Random, close Random, far apart
Particles
Interactions between Very strong Strong Essentially none
particles

• To evaporate, molecules must have
sufficient energy to break IM forces.
• Molecules at the surface break away
and become gas.
• Only those with enough KE escape.
• Breaking IM forces requires energy. The
process of evaporation is endothermic.
• Evaporation is a cooling process.
• It requires heat.
Evaporation

Change from gas to liquid
Achieves a dynamic equilibrium with
vaporization in a closed system.
What is a closed system?
A closed system means
matter can’t go in or out.
(put a cork in it)
What the heck is a
“dynamic equilibrium?”
Condensation

When first sealed, the molecules
gradually escape the surface of the
liquid.
As the molecules build up above the
liquid - some condense back to a
liquid.
The rate at which the molecules
evaporate and condense are equal.
Dynamic Equilibrium

As time goes by the rate of vaporization
remains constant but the rate of
condensation increases because there
are more molecules to condense.
Equilibrium is reached when:
Rate of Vaporization = Rate of Condensation
Molecules are constantly changing phase “dynamic”
The total amount of liquid and vapor remains constant
“equilibrium”
Dynamic Equilibrium

• Vaporization is an endothermic process - it
requires heat.
• Energy is required to overcome intermolecular
forces
• Responsible for cool earth
• Why we sweat
Vaporization

Energy Changes Accompanying Phase Changes
Solid
Liquid
Gas
Melting Freezing
Deposition
CondensationVaporization
Sublimation
Energyofsystem
Brown, LeMay, Bursten, Chemistry 2000, page 405

solid
liquid
gas
Heat added
Temperature(oC)
A
B
C
D
E
Heating Curve for Water
0
100
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 487

solid
liquid
gas
vaporization
condensation
melting
freezing
Heat added
Temperature(oC)
A
B
C
D
E
0
100

Latent Heat
• Take 1 kg of water from –10 oC up to 150 oC we can
plot temperature rise against absorbed heat
water
steam
(water vapor)
-10 C
0 C
100 C
ice
Lf = 80 cal/g Lv = 540 cal/g
Lf is the latent heat of fusion
Lv is the latent heat of vaporization
Q heat absorbed

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
Colloids Suspensions
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Elements
only one kind
of atom; atoms
are bonded it
the element
is diatomic or
polyatomic
Compounds
two or
more kinds
of atoms
that are
bonded
substance
with
definite
makeup
and
properties
Mixtures
two or more
substances
that are
physically
mixed
two or
more
kinds of
and
Both elements and compounds have a definite makeup and definite properties.
Packard, Jacobs, Marshall, Chemistry Pearson AGS Globe, page (Figure 2.4.1)

Matter Flowchart
Examples:
– graphite
– pepper
– sugar (sucrose)
– paint
– soda
element
hetero. mixture
compound
solution
homo. mixture
hetero. mixture

Pure Substances
Element
– composed of identical atoms
– EX: copper wire, aluminum foil

Pure Substances
Compound
– composed of 2 or more elements
in a fixed ratio
– properties differ from those of
individual elements
– EX: table salt (NaCl)

Pure Substances
Law of Definite Composition
– A given compound always contains the same,
fixed ratio of elements.
Law of Multiple Proportions
– Elements can combine in different ratios to
form different compounds.

Pure Substances
For example…
Two different compounds,
each has a definite composition.
Carbon, C Oxygen, O Carbon monoxide, CO
Carbon, C Oxygen, O Oxygen, O Carbon dioxide, CO2

Mixtures
Variable combination of two or more
pure substances.
Heterogeneous Homogeneous

Mixtures
Solution
– homogeneous
– very small particles
– no Tyndall effect Tyndall Effect
– particles don’t settle
– EX: rubbing alcohol

Mixtures
Colloid
– heterogeneous
– medium-sized particles
– Tyndall effect
– particles don’t settle
– EX: milk

Mixtures
Suspension
– heterogeneous
– large particles
– Tyndall effect
– particles settle
– EX: fresh-squeezed
lemonade

Mixtures
Examples:
– mayonnaise
– muddy water
– fog
– saltwater
– Italian salad
dressing
colloid
suspension
colloid
solution
suspension

Classification of Matter
Materials
Homogeneous
Heterogeneous
Heterogeneous
mixture
Homogeneous
mixture
Substance
Element Compound Solution Mixture
Order / Disorder
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 43

MATTER
(gas. Liquid,
solid, plasma)
PURE
SUBSTANCES MIXTURES
HETEROGENEOUS
MIXTURE
HOMOGENEOUS
MIXTURES
ELEMENTSCOMPOUNDS
Separated by
physical means into
Separated by
chemical
means into
Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 31

uniform
properties?
fixed
composition?
chemically
decomposable?
no
no
no
yes
hetero-
geneous
mixture
solution
element
compound

Elements, Compounds, and Mixtures
(a)
an element
(hydrogen)
(b)
a compound
(water)
(c)
a mixture
(hydrogen
and oxygen)
(d)
a mixture
(hydrogen
and oxygen)
hydrogen
atoms hydrogen
atoms
oxygen atoms

Mixture vs. Compound
Mixture
Fixed
Composition
Bonds
between
components
Can ONLY be
separated by
chemical means
Variable
Composition
No bonds
between
components
Can be
separated by
physical means
Alike Different
Contain
two or more
elements
Can be
separated
into
elements
Involve
substances
Compound
Different
Topic Topic

Compounds vs. Mixtures
• Compounds have properties that are
uniquely different from the elements from
which they are made.
– A formula can always be written for a compound
– e.g. NaCl  Na + Cl2
• Mixtures retain their individual properties.
– e.g. Salt water is salty and wet

Diatomic Elements, 1 and 7
H2
N2 O2 F2
Cl2
Br2
F2

Products made from Sulfur
Magazines and printing papers
Writing and fine papers
Wrapping and bag papers
Sanitary and tissue papers
Absorbent papers
Rayon
Cellophane
Carbon Tetrachloride
Ruber processing
chemicals
Containers and boxes
Newsprint
Pulp for rayon and film
PULP 3%
NONACID 12%
Insecticides
Fungicides
Rubber vulcanizing
Soil sulfur
Specialty steels
Magnessium
Leather processing
Photography
Dyestuffs
Bleaching
Soybean extraction
Aluminum reduction
Paper sizing
Water treatment
Pharmaceuticals
Insecticides
Antifreeze
Superphosphates
Ammonium phosphate
Ammonium sulfate
Mixed fertilizers
Autos
Appliances
Tin and other containers
Galvanized products
Explosives
Nonferrous metals
Synthetic rubber
Storage batteries
Textile finishing
Tire
cords
Viscose
textiles
Acetate
textiles
Blended
fabrics
Cellophane
Photographic
film
Paints and
enamels
Linoleum and
coated fabrics
Paper
Printing inks
Aviation
Gasoline
Lubricants
Other
Refinery
products
SULFURIC
ACID 88%
CARBON
DISULFIDE 3%
GROUND &
DEFINED 3%
IRON & STEEL 1%
PETROLEUM 2%
CHEMICAL 17%
Synthetic detergents
Feed additives
Anti-knock gasoline
Synthetic resins
Protective coating
Dyestuffs
Oil well acidizing
Petroleum catalysts

• Rhombic sulfur
– “Brimstone” (when
molten)
– Polyatomic (S8)
– Forms SO2
Amorphous sulfur
– (without shape)
Sulfur
The sudden cooling of m-sulfur
produces amorphous sulfur.

Matter
Substance
Definite composition
(homogeneous)
Element
(Examples: iron, sulfur,
carbon, hydrogen,
oxygen, silver)
Mixture of
Substances
Variable composition
Compound
(Examples: water.
iron (II) sulfide, methane,
Aluminum silicate)
Homogeneous mixture
Uniform throughout,
also called a solution
(Examples: air, tap water,
gold alloy)
Heterogeneous mixture
Nonuniform
distinct phases
(Examples: soup,
concrete, granite)
Chemically
separable
Physically
separable

The Organization of Matter
MATTER
PURE
SUBSTANCES
HETEROGENEOUS
MIXTURE
HOMOGENEOUS
MIXTURES
ELEMENTS COMPOUNDS
Physical methods
Chemical
methods

Top Ten Elements
in the Universe
Percent
Element (by atoms)
1. Hydrogen 73.9
2. Helium 24.0
3. Oxygen 1.1
4. Carbon 0.46
5. Neon 0.13
6. Iron 0.11
7. Nitrogen 0.097
8. Silicon 0.065
9. Magnesium 0.058
10.Sulfur 0.044
A typical spiral galaxy
(Milky Way is a spiral galaxy)

The Composition of Air
Air
Nitrogen
OxygenHelium
Water
vapor
Neon
Carbon
dioxide
Argon

Chart Examining Some Components of Air
Nitrogen consists of molecules consisting of
two atoms of nitrogen:
Oxygen consists of molecules consisting of
two atoms of oxygen:
Water consists of molecules consisting of two
hydrogen atoms and one oxygen atom:
Argon consists of individual argon atoms:
Carbon dioxide consists of molecules consisting
of two oxygen atoms and one carbon atom:
Neon consists of individual neon atoms:
Helium consists of individual helium atoms:
N2
O2
H2O
Ar
CO2
Ne
He

Reviewing Concepts
Classifying Matter
• Why does every sample of a given substance
have the same properties?
• Explain why the composition of an element is
fixed.
• Describe the composition of a compound.
• Why can the properties of a mixture vary?
• On what basis can mixtures be classified as
solutions, suspensions, or colloids?

Unit Cells
The kind of symmetry found throughout a crystalline substance is determined
by the type of unit cell which generates the lattice structure.
Simple cubic Body-centered
cubic
Face-centered
cubic
Monoclinic HexagonalTetragonal

Simple cubic Body-centered cubic Face-centered cubic

Phosphorous (P4)
TWO ALLOTROPIC FORMS
White phosphorous
spontaneously ignites
Red phosphorous
used for matches

Sodium Chloride Crystal
= Cl-
= Na+

Packing of NaCl Ions
= Na1+
= Cl1-

Molecular Structure of Ice
Hydrogen
bonding

Allotropes of Carbon
Graphite BuckminsterfullereneDiamond

Graphite

Allotropes of
Carbon
Diamond

Diamonds in Garage
• Made gem quality
diamonds by
burning wood
• Scholarship
Molecular structure
of diamond

Molecular
structure
of
Diamond

Allotropes of
Carbon
C60 & C70
“Buckyballs”
“Buckytubes”
Buckminsterfullerene

Credit: Baughman et al., Science 297, 787 (2002)

Trojan Horse
Can use ‘camouflage’ to hide things. Be careful what’s in the Trojan!
Buckyballs can hide medicine to treat the human body.

Gold
24 karat gold 18 karat gold 14 karat gold
Gold
Copper
Silver
18/24 atoms Au24/24 atoms Au 14/24 atoms Au

Solid Brass
An alloy is a mixture of metals.
• Brass = Copper + Zinc
• Solid brass
• homogeneous mixture
• a substitutional alloy
Copper
Zinc

Brass Plated
• Brass = Copper + Zinc
• Brass plated
• heterogeneous mixture
• Only brass on outside
Copper
Zinc

Steel Alloys
• Stainless steel
• Tungsten hardened steel
• Vanadium steel
• We can engineer properties
– Add carbon to increase strength
– Too much carbon  too brittle and snaps
– Too little carbon  too ductile and iron bends
Tensilestrength
Force is added

Galvanized Nails and Screws
• Zinc coating prevents rust
– Use deck screws for any outdoor project
• Iron will rust if untreated
– Weaken and break

Nitinol Wire
• Alloy of nickel and titanium
• Remembers shape when heated
Applications:
surgery, shirts that do not need to be
ironed.

Properties of Matter
• Electrical Conductivity
• Heat Conductivity
• Density
• Melting Point
• Boiling Point
• Malleability
• Ductility

Methods of Separating Mixtures
• Magnet
• Filter
• Decant
• Evaporation
• Centrifuge
• Chromatography
• Distillation

Filtration
separates
a liquid
from a
solid
Mixture of
solid and
liquid Stirring
rod
Filtrate (liquid
component
of the mixture)
Filter paper
traps solid
Funnel

Chromatography
• Tie-dye t-shirt
• Black pen ink
• DNA testing
– Tomb of Unknown Soldiers
– Crime scene
– Paternity testing

Paper Chromatography
of Water-Soluble Dyes
orange red yellow
Initial spots of dyes
Direction of Water
(mobile phase)
movement
Filter paper
(stationary phase)
Orange mixture of
red
and
yellow
Suggested
red dye
is not
homogeneous

Separation by Chromatography
sample
mixture
a chromatographic column
stationary phase
selectively absorbs
components
mobile phase
sweeps sample
down column
detector

Ion chromatogram of orange juice
time (minutes)
detector
response
0 5 10 15 20 25
Na+
K+
Mg2+ Fe3+
Ca2+

Setup to heat a solution
Ring stand
Beaker
Wire gauze
Ring
Bunsen burner

long spout helps
vapors to condense
mixture for distillation
placed in here
Furnace
Glass retort
Glass Retort
Eyewitness Science “Chemistry” , Dr. Ann Newmark, DK Publishing, Inc., 1993, pg 13

A Distillation Apparatus
liquid with a solid
dissolved in it
thermometer
condenser
tube
distilling
flask
pure
liquid
receiving
flaskhose connected to
cold water faucetDorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 282

The solution is boiled and steam
is driven off.

Salt remains after all water is
boiled off.

No chemical change occurs
when salt water is distilled.
Saltwater solution
(homogeneous mixture)
Distillation
(physical method)
Salt
Pure water

Separation of a sand-saltwater
mixture.

Separation of Sand from Salt
1. Gently break up your salt-crusted sand with a plastic spoon.
Follow this flowchart to make a complete separation.
Salt-
crusted
sand.
Dry
sand.
Wet
sand.
Weigh the
mixture.
Decant
clear
liquid.
Evaporate
to
dryness.
Pour into
heat-resistant
container.
Fill with
water.
Stir and let
settle 1
minute.
Weigh
sand.
Calculate
weight of
salt.
Repeat
3 times?
Yes
No
2. How does this flow
chart insure a complete
separation?

Four-stroke Internal
Combustion Engine

Different Types of Fuel Combustion
2 C8H18 + 25 O2  16 CO2 + 18 H2O
__CH3OH +__O2 __CO2 +__H2O
Methanol (in racing fuel)
Gasoline (octane)

Combustion Chamber
-The combustion chamber is the area where compression and
combustion take place.
-Gasoline and air must be mixed in the correct ratio.

•Methanol can run at much higher compression ratios,
meaning that you can get more power from the engine on
each piston stroke.
•Methanol provides significant cooling when it evaporates in
the cylinder, helping to keep the high-revving, high-
compression engine from overheating.
•Methanol, unlike gasoline, can be extinguished with water if
there is a fire. This is an important safety feature.
•The ignition temperature for methanol (the temperature at
which it starts burning) is much higher than that for gasoline,
so the risk of an accidental fire is lower.
The Advantages of Methanol -
Burning Engines

•At 900 hp, it has about two to three times the horsepower of a "high-
performance" automotive engine. For example, Corvettes or Vipers
might have 350- to 400-horsepower engines.
•At 15,000 rpm, it runs at about twice the rpm of a normal automotive
engine. Compared to a normal engine, an methanol engine has larger
pistons and the pistons travel a shorter distance up and down on each
stroke.
•The motor is lighter. This lowers their inertia and is another factor in the
high rpm.
A Race Car - Basic Information

Centrifugation
• Spin sample very rapidly:
denser materials go to
bottom (outside)
• Separate blood into serum
and plasma
– Serum (clear)
– Plasma (contains red blood
cells ‘RBCs’)
• Check for anemia (lack of iron)
Blood
RBC’s
Serum
A B C
AFTER
Before

Water Molecules

The decomposition of two water
molecules.
2 H2O  O2 + 2 H2
Electric
current
Water
molecules
Diatomic Diatomic
oxygen molecule hydrogen molecules+

Electrolysis
*Must add acid catalyst
to conduct electricity
*H1+
water oxygen hydrogen
“electro” = electricity
“lysis” = to split
Water
Hydrogen
gas forms
Oxygen
gas forms
ElectrodeSource of
direct current
H2O(l) O2 (g) + 2 H2 (g)

Electrolysis of Water
Half reaction at the cathode (reduction):
4 H2O + 4 e -  2 H2 + 4 OH 1-
Half reaction at the anode (oxidation):
2 H2O  O2 + 4 H 1+ + 4 e -
hydrogen
gas
cathode
oxygen
gas
anode
D.C. power
source
water

Reviewing Concepts
Physical Properties
• List seven examples of physical
properties.
• Describe three uses of physical properties.
• Name two processes that are used to
separate mixtures.
• When you describe a liquid as thick, are
you saying that it has a high or low
viscosity?

Reviewing Concepts
Physical Properties
• Explain why sharpening a pencil is an
example of a physical change.
• What allows a mixture to be separated by
distillation?

Reviewing Concepts
Chemical Properties
• Under what conditions can chemical
properties be observed?
• List three common types of evidence for a
chemical change.
• How do chemical changes differ from
physical changes?

Reviewing Concepts
Chemical Properties
• Explain why the rusting of an iron bar
decreases the strength of the bar.
• A pat of butter melts and then burns in a
hot frying pan. Which of these changes is
physical and which is chemical?

2 H2 O2
2 H2O+ + E

The Zeppelin LZ 129 Hindenburg catching fire on May
6, 1937 at Lakehurst Naval Air Station in New Jersey.

S.S. Hindenburg
35 people died when the Hindenburg exploded.
May 1937 at Lakehurst, New Jersey
• German zeppelin
luxury liner
• Exploded on
maiden voyage
• Filled with
hydrogen gas

Hydrogen is the most effective buoyant gas,
but is it highly flammable. The disastrous fire
in the Hindenburg, a hydrogen-filled dirigible,
in 1937 led to the replacement of hydrogen
by nonflammable helium.

Erosion Takes a Powder
TreatedUntreated

Sodium Polyacrylate
• Absorbent Material
– Absorbs 700x volume of water
• Magicians
– Pour water in hat and it “disappears”
• Diapers
• Farmers
– Anti-erosion powder
• Add to Soils
– hold moisture between watering

Specific Heats
of Some Substances
Specific Heat
Substance (cal/ g oC) (J/g oC)
Water 1.00 4.18
Alcohol 0.58 2.4
Wood 0.42 1.8
Aluminum 0.22 0.90
Sand 0.19 0.79
Iron 0.11 0.46
Copper 0.093 0.39
Silver 0.057 0.24
Gold 0.031 0.13

(a) Radiant energy (b) Thermal energy
(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion, depending on mass and velocity.
Potential energy
Energy in Energy out
kinetic energy kinetic energy

School Bus or Bullet?
Which has more kinetic energy;
a slow moving school bus or a fast moving bullet?
Recall: KE = ½ m v2
KE = ½ m v2 KE = ½ m v2
BUS BULLET
KE(bus) = ½ (10,000 lbs) (0.5 mph)2 KE(bullet) = ½ (0.002 lbs) (240 mph)2
Either may have more KE, it depends on the mass of the bus and the velocity
of the bullet.
Which is a more important factor: mass or velocity? Why? (Velocity)2

Kinetic Energy and Reaction
Rate
Kinetic energy
lower temperature
higher temperature
minimum energy
for reaction

Hot vs. Cold Tea
Kinetic energy
Many molecules have an
intermediate kinetic energy
Few molecules have a
very high kinetic energy
Low temperature
(iced tea)
High temperature
(hot tea)
Percentofmolecules

Decomposition of Nitrogen Triiodide

Decomposition of
Nitrogen Triiodide
2 NI3(s) N2(g) + 3 I2(g)
NI3 I2
N2

Exothermic Reaction
Reactants  Products + Energy
10 energy = 8 energy + 2 energy
Reactants
Products
-DH
Energy
Energy of reactants
Energy of products
Reaction Progress

Endothermic Reaction
Energy + Reactants  Products
+DH Endothermic
Reaction progress
Energy
Reactants
ProductsActivation
Energy

Effect of Catalyst on Reaction Rate
reactants
products
Energy
activation energy
for catalyzed reaction
Reaction Progress
No catalyst
Catalyst lowers the activation energy for the reaction.What is a catalyst? What does it do during a chemical reaction?

Energy Sources in the United States
Wood Coal Petroleum / natural gas Hydro and nuclear
1850 1900 1940 1980 1990
100
80
60
40
20
0
Percent
9
91
21
71
5 3
10
50
40
20
70
10
26
58
16

1850
100
80
60
40
20
0
Percent
9
91
1900
21
71
5 3
1940
10
50
40
1980
20
70
10
1990
26
58
16

1850
100
80
60
40
20
0
Percent
9
91
1900
21
71
5 3
1940
10
50
40
1980
20
70
10
1990
26
58
16
2005
50
21
26

Energy Conversion
Timberlake, Chemistry 7th Edition, page 202
fan
electrical energy to
mechanical energy
light bulb
light energy to
thermal and radiant energy
coffee maker
thermal energy
pencil sharpener
mechanical energy

Conservation of Energy
in a Chemical Reaction
Surroundings
System
Surroundings
System
Energy
Before
reaction
After
reaction
In this example, the energy of the reactants and products increases,
while the energy of the surroundings decreases.
In every case, however, the total energy does not change.
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Reactant + Energy Product

Conservation of Energy
in a Chemical Reaction
Surroundings
System
Surroundings
System
Energy
Before
reaction
After
reaction
In this example, the energy of the reactants and products decreases,
while the energy of the surroundings increases.
In every case, however, the total energy does not change.
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Exothermic Reaction
Reactant Product + Energy

Direction of Heat Flow
Surroundings
ENDOthermic
qsys > 0
EXOthermic
qsys < 0
System
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207
System
H2O(s) + heat  H2O(l)
melting
H2O(l)  H2O(s) + heat
freezing

Caloric Values
Food joules/grams calories/gram Calories/gram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
1000 calories = 1 Calorie
"science" "food"
1calories = 4.184 joules

Units of energy
Most common units of energy
1. S unit of energy is the joule (J), defined as 1
(kilogram•meter2)/second2, energy is also
expressed in kilojoules (1 kJ = 103J).
2. Non-S unit of energy is the calorie where 1
calorie (cal) is the amount of energy needed
to raise the temperature of 1 g of water by 1°C.
One cal = 4.184 J or 1J = 0.2390 cal.
Units of energy are the same, regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass
beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for
the boiler, a calorimeter, thermometers, samples (typically samples of copper,
aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples,
and a balance.
Experimental Determination of Specific Heat of a Metal

A Coffee Cup
Calorimeter
Thermometer
Styrofoam
cover
Styrofoam
cups
Stirrer
Thermometer
Glass stirrer
Cork stopper
Two Styrofoam ®
cups nested
together containing
reactants in solution

Heating Curves
Melting - PE 
Solid - KE 
Liquid - KE 
Boiling - PE 
Gas - KE 

Heating Curves
Temperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE 
Solid - KE 
Liquid - KE 
Boiling - PE 
Gas - KE 

Heating Curves
• Temperature Change
– change in KE (molecular motion)
– depends on heat capacity
• Heat Capacity
– energy required to raise the temp of 1 gram of a
substance by 1°C
– “Volcano” clip -
– water has a very high heat capacity

Heating Curves
• Phase Change
– change in PE (molecular arrangement)
– temp remains constant
• Heat of Fusion (DHfus)
– energy required to melt 1 gram of a substance at its
m.p.

Heating Curves
• Heat of Vaporization (DHvap)
– energy required to boil 1 gram of a substance at its
b.p.
– usually larger than DHfus…why?
• EX: sweating,
steam burns, the
drinking bird

Phase Diagrams
• Show the phases of a substance at different
temps and pressures.

Calculating Energy Changes -
Temperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid

Heat Transfer
Al Al
m = 20 g
T = 40oC
SYSTEM
Surroundings
m = 20 g
T = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block “A” Block “B”
Final
Temperature
Assume NO heat energy is “lost” to the surroundings from the system.
      C30
g)20g(20
C20g20C40g20 o
oo



What will be the final temperature
of the system ?
a) 60oC b) 30oC c) 20oC d) ?

Heat Transfer
Al
Al
m = 20 g
T = 40oC
SYSTEM
Surroundings
m = 10 g
T = 20oC
20 g (40oC) 20 g (20oC) 30.0oC
Final
Temperature
20 g (40oC) 10 g (20oC) 33.3oC
      C3.33
g)10g(20
C20g10C40g20 o
oo



What will be the final temperature
of the system ?
a) 60oC b) 30oC c) 20oC d) ?
?

Heat Transfer
Al
Al
m = 20 g
T = 20oC
SYSTEM
Surroundings
m = 10 g
T = 40oC
20 g (40oC) 20 g (20oC) 30.0oC
Final
Temperature
20 g (40oC) 10 g (20oC) 33.3oC
      C7.26
g)10g(20
C40g10C20g20 o
oo



20 g (20oC) 10 g (40oC) 26.7oC

Heat Transfer
m = 75 g
T = 25oC
SYSTEM
Surroundings
m = 30 g
T = 100oC
20 g (40oC) 20 g (20oC) 30.0oC
Final
Temperature
20 g (40oC) 10 g (20oC) 33.3oC
      C46
g)30g(75
C100g30C25g75 o
oo



20 g (20oC) 10 g (40oC) 26.7oC
AgH2O
Real Final Temperature = 26.6oC
Why?
We’ve been assuming ALL materials
transfer heat equally well.

Specific Heat
• Water and silver do not transfer heat equally well.
Water has a specific heat Cp = 4.184 J/goC
Silver has a specific heat Cp = 0.235 J/goC
• What does that mean?
It requires 4.184 Joules of energy to heat 1 gram of water 1oC
and only 0.235 Joules of energy to heat 1 gram of silver 1oC.
• Law of Conservation of Energy…
In our situation (silver is “hot” and water is “cold”)…
this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy.
• Lets look at the math!

“loses” heat
Calorimetry
   
     
        
C26.6x
320.8x8550
7845313.8xx05.7705
algebra.thesolveandunitsDrop
C25-xg75CgJ184.4C100-xg30CgJ235.0
equation.intovaluesSubstitute
TTmCTTmC
TmCTmC
qq
o
oooo
ifpinitialfinalp
pp
OHAg 2





DD

m = 75 g
T = 25oC
SYSTEM
Surroundings
m = 30 g
T = 100oC
AgH2O
Tfinal = 26.6oC

Calorimetry
   
     
        
C26.6x
8550320.8x
7845313.8xx05.7705
algebra.thesolveandunitsDrop
C25-xg75CgJ184.4C100-xg30CgJ235.0
equation.intovaluesSubstitute
TTmCTTmC
TmCTmC
qq
o
oooo
ifpinitialfinalp
pp
OHAg 2





DD

m = 75 g
T = 25oC
SYSTEM
Surroundings
m = 30 g
T = 100oC
AgH2O

1 Calorie = 1000 calories
“food” = “science”
Candy bar
300 Calories = 300,000 calories
English
Metric = _______Joules
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 calorie = 4.184 Joules

Cp(ice) = 2.077 J/g oC
It takes 2.077 Joules to raise 1 gram ice 1oC.
X Joules to raise 10 gram ice 1oC.
(10 g)(2.077 J/g oC) = 20.77 Joules
X Joules to raise 10 gram ice 10oC.
(10oC)(10 g)(2.077 J/g oC) = 207.7 Joules
Heat = (specific heat) (mass) (change in temperature)
q = Cp . m . DT
Temperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x DHfus
DH = mol x DHvap

Heat = (specific heat) (mass) (change in temperature)
q = Cp . m . DT
TmCq p(ice) D
 initialfinalp(ice) TTmCq 
 C)30(C20-g10
Cg
J2.077
q oo
o








Given Ti = -30oC
Tf = -20oC
q = 207.7 Joules
Temperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x DHfus
DH = mol x DHvap

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC).
When thermal equilibrium is reached, the system has a temperature of 42oC.
Find the mass of the iron.
Calorimetry Problems 2
question #5
Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g LOSE heat = GAIN heat-
- [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)]
Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22)
205.9 X = 22091
X = 107.3 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial
temperature of 15oC. If gold has a specific heat of 0.129 J/goC, what is the final
temperature of the mixture? Assume that the gold experiences no change in state
of matter.
question #8
Au
T = 785oC
mass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)]Drop Units:
- [(12.5) (Tf - 785oC)] = (1.35x 103) (Tf - 15oC)]
-12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104
3 x 104 = 1.36 x 103 Tf
Tf = 22.1oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature
of the system.
question #9
T = 13oC
mass = 59 g
- [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [(4.184 J/goC) (59 g) (Tf - 13oC)] = (4.184 J/goC) (87 g) (Tf - 72oC)]Drop Units:
- [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)]
-246.8 Tf + 3208 = 364 Tf - 26208
29416 = 610.8 Tf
Tf = 48.2oC
T = 72oC
mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC.
Find the system's final temperature.
question #10
ice
T = -11oC
mass = 38 g
T = 56oC
mass = 214 g
- [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) + (Cf) (mass) + (Cp,H2O) (mass) (DT)
- [(4.184 J/goC)(214 g)(Tf - 56oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf - 0oC)
- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)]
- 895 Tf + 50141 = 868 + 12654 + 159 Tf
- 895 Tf + 50141 = 13522 + 159 Tf
Tf = 34.7oC
36619 = 1054 Tf
Temperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x DHfus
DH = mol x DHvap
A
B
C
D
warm icemelt ice
warm water
water cools

25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.
question #11
- [(Cp,H2O) (mass) (DT)] + (Cv,H2O) (mass) + (Cp,H2O) (mass) (DT) = [(Cp,H2O) (mass) (DT)]
- [ - 816.8 - 56400 + 104.5Tf - 10450] = 997Tf - 7972
- [qA + qB + qC] = qD
qA = [(Cp,H2O) (mass) (DT)]
qA = [(2.042 J/goC) (25 g) (100o - 116oC)]
qA = - 816.8 J
qB = (Cv,H2O) (mass)
qA = (2256 J/g) (25 g)
qA = - 56400 J
qC = [(Cp,H2O) (mass) (DT)]
qC = [(4.184 J/goC) (25 g) (Tf - 100oC)]
qA = 104.5Tf - 10450
qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)
qD = - 997Tf - 7972
- [qA + qB + qC] = qD
816.8 + 56400 - 104.5Tf + 10450 = 997Tf - 7972
67667 - 104.5Tf = 997Tf - 7979
75646 = 1102Tf
1102 1102
Tf = 68.6oC
Temperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
DH = mol x DHfus
DH = mol x DHvap
A
B
C
D
(1000 g = 1 kg)
238.4 g

A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water at 25oC.
If the system's final temperature is 46oC, what was the initial temperature of the lead?
question #12
Pb
T = ? oC
mass = 322 g
Ti = 25oC
mass = 264 g
- [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)]Drop Units:
- [(44.44) (46oC - Ti)] = (1104.6) (21oC)]
- 2044 + 44.44 Ti = 23197
44.44 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC

A sample of ice at –12oC is placed into 68 g of water at 85oC. If the final temperature
of the system is 24oC, what was the mass of the ice?
question #13
H2O
T = -12oC
mass = ? g
Ti = 85oC
mass = 68 g
GAIN heat = - LOSE heat
[ qA + qB + qC ] = - [(Cp,H2O) (mass) (DT)]
458.2 m = - 17339
m = 37.8 g
ice
Tf = 24oC
qA = [(Cp,H2O) (mass) (DT)]
qC = [(Cp,H2O) (mass) (DT)]
qB = (Cf,H2O) (mass)
qA = [(2.077 J/goC) (mass) (12oC)]
qB = (333 J/g) (mass)
qC = [(4.184 J/goC) (mass) (24oC)]
[ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)]
24.9 m
333 m
100.3 m
458.2 mqTotal = qA + qB + qC
458.2 458.2

Energy + Reactants  Products
+DH Endothermic
Reaction progress
Energy
Reactants
Products
Activation
Energy

Catalytic Converter
N O
N O
N O
N O
N O
N
O
N
O
N
O
N O
N
O
N
O
O O
N
N
O
O
O
O
N NN
N
One of the reactions that takes place in the catalytic converter
is the decomposition of nitrogen (II) oxide (NO) to nitrogen and oxygen gas.
N O

O
Catalytic Converter
C O
N O
C
O
O
CO
N
N
One of the reactions that takes place in the catalytic converter is the decomposition of
carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas.
C
O
N
N
N
OO
O
C
OCO
2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram
H2O(g)
H2O(l)
H2(g) + ½ O2(g)
-44 kJ
Exothermic
+44 kJ
Endothermic
DH = +242 kJ
Endothermic
242 kJ
Exothermic
286 kJ
Endothermic
DH = -286 kJ
Exothermic
Energy
H2(g) + 1/2O2(g)  H2O(g) + 242 kJ DH = -242 kJ
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

Hess’s Law
Calculate the enthalpy of formation of carbon dioxide from its elements.
C(g) + 2O(g)  CO2(g)
Use the following data:
2O(g)  O2(g) DH = - 250 kJ
C(s)  C(g) DH = +720 kJ
CO2(g)  C(s) + O2(g) DH = +390 kJ
Smith, Smoot, Himes, pg 141
2O(g)  O2(g) DH = - 250 kJ
C(g) + 2O(g)  CO2(g) DH = -1360 kJ
C(g)  C(s) DH = - 720 kJ
C(s) + O2(g)  CO2(g) DH = - 390 kJ

In football, as in Hess's law, only the initial and final conditions matter.
A team that gains 10 yards on a pass play but has a five-yard penalty,
has the same net gain as the team that gained only 5 yards.
initial position
of ball
final position
of ball
10 yard pass
5 yard penalty
5 yard net gain

Fission vs. Fusion
Fuse small atoms
2H2 He
NO
Radioactive
waste
Very High
Temperatures
~5,000,000 oC
(SUN)
Split
large atoms
U-235
Radioactive
waste
(long half-life)
Nuclear
Power
Plants
Alike Different
Create
Large Amounts
of Energy
E = mc2
Transmutation
of Elements
Occurs
Change
Nucleus
of
Atoms
Fusion
Different
Topic Topic
Fission

• Use fear and
selective facts
to promote an
agenda
• Eating animals?
• Radiation = Bad
Look who is funding research;
it may bias the results.

Nuclear Fission
First stage: 1 fission Second stage: 2 fission Third stage: 4 fission

Nuclear Power Plants
map: Nuclear Energy Institute

Fermi Approximations
FERMI APPROXIMATIONS
An educated guess – based on a series
of calculations of known facts – to arrive
at a reasonable answer to a question.
How many piano tuners are
there in New York City?
ANSWER:
Enrico Fermi
400 piano tuners

Nuclear Fusion
Sun
+ +
Four
hydrogen
nuclei
(protons)
Two beta
particles
(electrons)
One
helium
nucleus
Hee2H4 4
2
0
1-
1
1
 + Energy

Conservation of Mass
…mass is converted into energy
Hydrogen (H2) H = 1.008 amu
Helium (He) He = 4.004 amu
FUSION
2 H2  1 He + ENERGY
1.008 amu
x 4
4.0032 amu = 4.004 amu + 0.028 amu
This relationship was discovered by Albert Einstein
E = mc2
Energy= (mass) (speed of light)2

Time Travel?
…Albert Einstein also discovered the Geometry of Space Near a Black Hole
Einstein’s theory of general relativity maybe interpreted in
terms of curvature of space in the presence of a gravitational
field. Here we see how this curvature varies near a black hole.

Tokamak Reactor
• Fusion reactor
• 10,000,000 o Celcius
• Russian for torroidial
(doughnut shaped)
ring
• Magnetic field
contains plasma

Cold Fusion?
• Fraud?
• Experiments must
be repeatable to
be valid

0 1 2 3 4
Number of half-lives
Radioisotoperemaining(%)
100
50
25
12.5
Half-life of Radiation
Initial amount
of radioisotope
t1/2
t1/2
t1/2
After 1 half-life
After 2 half-lives
After 3 half-lives

Triple Point Plot
solid
liquid
gas
melting
freezing
sublimation
deposition
vaporization
condensation
Temperature (oC)
Pressure(atm)
0.6
2.6

Liquid VaporSolid
Normal
melting
point
Normal
boiling
point
101.3
0.61
0.016 1000
Temperature (oC)
Pressure(KPa)
Triple point
Triple Point
Critical
pressure
Critical point
Critical
temperature
373.99
22,058

Objectives - Matter
• Explain why mass is used as a measure of the quantity
of matter.
• Describe the characteristics of elements, compounds,
and mixtures.
• Solve density problems by applying an understanding of
the concepts of density.
• Distinguish between physical and chemical properties
and physical and chemical changes.
• Demonstrate an understanding of the law of
conservation of mass by applying it to a chemical
reaction.

Objectives - Energy
• Identify various forms of energy.
• Describe changes in energy that take place
during a chemical reaction.
• Distinguish between heat and temperature.
• Solve calorimetry problems.
• Describe the interactions that occur between
electrostatic charges.

Law of Conservation of Energy
Eafter = Ebefore
2 H2 + O2  2 H2O + energy
+  + WOOF!

ENERGY
CO2 + H2OC2H2 + O2
PEreactants
PEproducts
KEstopper
heat, light, sound
Eafter = Ebefore
2 H2 + O2  2 H2O + energy
+  + WOOF!

ENERGY
C2H2 + O2
C2H2 + O2
PEreactants
PEproducts
KEstopper
heat, light, sound
Eafter = Ebefore
2C2H2 + 5O2  4 CO2 + 2H2O + energy
Energy Changes

First experimental image showing
internal atomic structures
© 2005 University of Augsburg, Experimental Physics VI, http://www.physik.uni-augs

Heating CurvesTemperature(oC)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
Melting - PE 
Solid - KE 
Liquid - KE 
Boiling - PE 
Gas - KE 

Heating Curves
 Temperature Change
• change in KE (molecular motion)
• depends on heat capacity
 Heat Capacity
• energy required to raise the temp of 1
gram of a substance by 1°C
• “Volcano” clip - water has a very high
heat capacity

Heating Curves
 Phase Change
• change in PE (molecular arrangement)
• temp remains constant
 Heat of Fusion (DHfus)
• energy required to melt 1 gram of a
substance at its m.p.

Heating Curves
 Heat of Vaporization (DHvap)
• energy required to boil 1 gram of a
substance at its b.p.
• usually larger than DHfus…why?
 EX: sweating,
steam burns,
the drinking bird

Phase Diagrams
 Show the phases of a substance at
different temps and pressures.

Resources - Matter and Energy
Objectives - matter and energy
Objectives - measurement
Objectives - phases of matter
Worksheet - vocabulary
Worksheet II - percentage composition
Worksheet - properties
Worksheet - density problems
Activity - density blocks & Part 2
Lab - golf ball lab
Worksheet - classifying matter
Outline (general)
Activity - chromatography
Outline - causes of change - calorimetry
Worksheet - calorimetry problems 1
Worksheet - calorimetry problems 2
Worksheet - heat energy problems
Worksheet - conversion factors
Worksheet - atoms, mass, and the mole
activity - mole pattern
Article - buried in ice (questions) Lab - beverage density (PowerPoint)
Textbook - questions
Article - buckyball (pics) & (video) questions
Episode 5 - A Matter of State
General Chemistry PP

Chemistry Unit 2 pp.pptx [repaired]

More Related Content

What's hot

Viewers also liked

Similar to Chemistry Unit 2 pp.pptx [repaired]

Recently uploaded

Chemistry Unit 2 pp.pptx [repaired]