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 QuestionsWhy 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

KeysKeys

Lecture Outline – Energy and Matter

Lecture Outline – Energy and Matter

student notes outline

textbook questions

http://www.unit5.org/chemistry/Matter.html

textbook questions

text

Chemistry of Matches

P4S3 + KClO3 P2O5 + KCl + SO2

tetraphosphorustrisulfide

potassiumchlorate

diphosphoruspentaoxide

potassiumchloride

sulfurdioxide

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 onthe 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

2m 1.8

N 45.0 Pressure

2m 1.00

N 45.0 Pressure

2m 0.45

N 45.0 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

forcepressure

Which shoes create the most pressure?

During a“physical change”

a substance changes some physical property…

During a“physical change”

a substance changes some physical property…

H2O

…but it is still the same material with the same chemical composition.

…but it is still the same material with the same chemical composition.

H2Ogas

solid

liquid

Chemical Property:Chemical Property:

The tendency of a substance to change into another

substance.

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:Chemical Change:

Any change involving a rearrangement of atoms.Any change involving a

rearrangement of atoms.

Chemical Reaction:Chemical Reaction:

The process of a chemical change...The process of a

chemical change...

During a“chemical reaction”new materials are

formed by a change in the way atoms are bonded together.

During a“chemical reaction”new materials are

formed by a change in the way atoms are bonded together.

Physical and Chemical PropertiesExamples 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 23Chemical properties can ONLY be observed during a chemical reaction!

The formation of a mixture

The formation of a mixture

The formation of a compound

The formation of a compoundChemical Change

Chemical Change

Physical Change

Physical Change

Physical & Chemical Changes

Limestone,CaCO3

crushing

PHYSICALCHANGE

Crushed limestone,CaCO3

heating

CHEMICALCHANGE

Pyrex

CO2

CaO

Lime andcarbon 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 storedkeeps 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 contributemore 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 sharedequally

Cl Cl

Polar covalent bonding

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A. Burns, Fundamentals of Chemistry 1999, page 229

• 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

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A. Burns, Fundamentals of Chemistry 1999, page 229

• 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

Transferelectrons

(ions formed)

+ / -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs. nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Photoelectric Generator

Solar Calculator

Radiant energy Evacuatedchamber

Metalsurface

Currentindicator

Positiveterminal

Voltagesource

cathode anode

Symbolic representationof a photoelectric cell

cathode anode

evacuated glassenvelope

Photoelectric Cell

Celsius & Kelvin Temperature Scales

Boiling pointof water

Freezing pointof water

Absolutezero

Celsius

100Celsiusdegrees

100oC

0oC

-273oC

Kelvin

100Kelvins

373 K

273 K

0 K

Temperature is Average Kinetic Energy

Fast Slow“HOT” “COLD”

Kinetic Energy (KE) = ½ m v 2

*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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 139

Converting 70 degrees Celsius to

Kelvin units.

oC + 273 = K

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 137

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

COLDHOT

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 Water90 oC 10 oC

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

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

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les 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

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

Conservation of Matter

Reactants yield Products

Heavy Metal Poisoning

Exposure to mercurymade the Hatter “mad”.

Eating chips of lead paint causes brain damage.

TREATMENT: Chelation therapy EDTA (ethylenediamine tetra acetic acid)

Arsenic treated lumber.‘Green-treated’ woodwill 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

ExtensiveProperties

IntensiveProperties

volume:mass:

density:temperature:

100 mL99.9347 g

0.999 g/mL20oC

15 mL14.9902 g

0.999 g/mL20oC

Styrofoam Brick

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

MMMM

VV= =DD

VVDD

BrickBrickStyrofoamStyrofoam

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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 119

Volume and DensityRelationship 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 cm2 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

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 MassesEqual masses……but unequal volumes.

The object with the larger volume (aluminum cube) has the density.

aluminum

gold

Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 71

smaller

Christopherson Scales

Made in Normal, Illinois USA

Two ways of viewing density

Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 71

Equal volumes…

…but unequal masses

The more massive object(the gold cube) has thegreater density.

aluminum gold

(A)

Equal masses……but unequal volumes.

(B)

gold

aluminumThe 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 homeB. Basement (near ceiling)C. Basement (near floor)D. It doesn’t matter, if your batteries are dead in the detectorC. Basement (near floor) Carbon dioxide is denser than air and sinks.

Symptoms of CO Poisoning

Concentration of COin 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

Exchange of Blood Gases

Tank of Water

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 143

Person Submerged in Water

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 143

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 determinethe 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 MicroscopePhotograph of NaCl

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Copyright 2007 Pearson Benjamin Cummings. All rights reserved.

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 youbegin 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 = Massvolume

119 g100 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…floatAdd 3.0 mL water,stir…sink

mL 100salt gx

mL 6 mL 100

g 119

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).

ORPoster (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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 31

Ice

H2O(s) Ice

Photograph of ice model Photograph of snowflakes

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

Liquid

H2O(l) Water

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 31

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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 31

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Gas, Liquid, and Solid

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 441

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 apartParticles

Interactions between Very strong Strong Essentially noneparticles

• 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

Ene

rgy

of s

yste

m

Brown, LeMay, Bursten, Chemistry 2000, page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

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

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 487

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 fusionLv 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

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Elements

only one kindof atom; atomsare bonded itthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

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

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

element

hetero. mixture

compound

solution homo. mixture

hetero. mixture

Pure Substances

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

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

Pure Substances

Compound– composed of 2 or more elements

in a fixed ratio

– properties differ from those of individual elements

– EX: table salt (NaCl)

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

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.

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

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

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Mixtures

Variable combination of two or more pure substances.

Heterogeneous Homogeneous

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Mixtures

Solution– homogeneous– very small particles– no Tyndall effect Tyndall Effect

– particles don’t settle– EX: rubbing alcohol

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Mixtures

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

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Mixtures

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

lemonade

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Mixtures

Examples:

– mayonnaise

– muddy water

– fog

– saltwater

– Italian salad dressing

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

colloid

suspension

colloid

solution

suspension

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic /

Gen

eral

Order / Disorder

Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 43

Classification of Matter

MATTER(gas. Liquid,

solid, plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 31

Classification of Matter

uniformproperties?

fixedcomposition?

chemicallydecomposable?

no

no

no

yes

hetero-geneousmixture

solution

element

compound

http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld003.htm

Elements, Compounds, and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Elements, Compounds, and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs. Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

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 7H2

N2 O2 F2

Cl2

Br2

F2

Products made from Sulfur

Magazines and printing papersWriting and fine papersWrapping and bag papersSanitary and tissue papersAbsorbent papers

RayonCellophaneCarbon TetrachlorideRuber processing chemicals

Containers and boxesNewsprintPulp for rayon and film

PULP 3%

OTHER 3%

NONACID 12%

InsecticidesFungicidesRubber vulcanizingSoil sulfur

Specialty steels Magnessium Leather processing PhotographyDyestuffs

Bleaching Soybean extraction

Aluminum reductionPaper sizingWater treatmentPharmaceuticalsInsecticidesAntifreeze

Superphosphates Ammonium phosphate Ammonium sulfate Mixed fertilizers

AutosAppliancesTin and other containersGalvanized 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

SULFURICACID 88%

CARBONDISULFIDE 3%

GROUND &DEFINED 3%

IRON & STEEL 1%

PETROLEUM 2%

CHEMICAL 17%

OTHER INDUSTRIES 6%

RAYON & FILM 3%

TITANIUM AND O

THER

PIGM

ENTS 5%

Synthetic detergents Feed additives Anti-knock gasoline Synthetic resins Protective coating Dyestuffs Oil well acidizingPetroleum catalysts

• Rhombic sulfur– “Brimstone” (when

molten)– Polyatomic (S8)– Forms SO2

Amorphous sulfur – (without shape)

Sulfur

The sudden cooling of m-sulfur produces amorphous sulfur.

Amorphous(Glass)Crystalline

The Haber Process

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples: iron, sulfur,

carbon, hydrogen,oxygen, silver)

Element(Examples: iron, sulfur,

carbon, hydrogen,oxygen, silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples: water.

iron (II) sulfide, methane,Aluminum silicate)

Compound(Examples: water.

iron (II) sulfide, methane,Aluminum silicate)

Homogeneous mixtureUniform throughout,also called a solution

(Examples: air, tap water,gold alloy)

Homogeneous mixtureUniform throughout,also called a solution

(Examples: air, tap water,gold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples: soup, concrete, granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples: soup, concrete, granite)

Chemicallyseparable

Physicallyseparable

                                      

                                                                                                  

              

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 41

Top Ten Elements in the Universe

Top Ten Elements in the Universe

Percent

Element (by atoms) 1. Hydrogen 73.92. Helium 24.03. Oxygen 1.14. Carbon

0.465. Neon 0.136. Iron 0.117. Nitrogen

0.0978. Silicon 0.0659. Magnesium 0.05810.Sulfur 0.044

A typical spiral galaxy(Milky Way is a spiral galaxy)

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 26

The Composition of Air

AirAir

NitrogenNitrogen

OxygenOxygenHeliumHelium

Watervapor

Watervapor

NeonNeon

Carbondioxide

Carbondioxide ArgonArgon

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 34

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 twohydrogen atoms and one oxygen atom:

Argon consists of individual argon atoms:

Carbon dioxide consists of molecules consistingof 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

HeZumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 35

Reviewing ConceptsClassifying 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-centeredcubic

Face-centeredcubic

Monoclinic HexagonalTetragonal

Simple cubic Body-centered cubic Face-centered cubic

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 457

Sodium Chloride Crystal

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 455

= Cl-

= Na+

Packing of NaCl Ions

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 456

= Na1+= Cl1-

Packing of NaCl Ions

Electron MicroscopePhotograph of NaCl

Molecular Structure of Ice

Hydrogen bonding

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 455

Dry Ice – Carbon Dioxide

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 27

Allotropes of Carbon

Graphite

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

Graphite

Allotropes of Carbon

Diamond

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 455

Diamond

Diamonds in Garage

• Made gem quality diamonds by burning wood

• Scholarship

Molecular structure of diamond

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 476

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

Tens

ile s

tren

gth

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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 40

Mixture ofsolid andliquid Stirring

rod

Filtrate (liquidcomponentof the mixture)

Filter papertraps solid

Funnel

Chromatography

• Tie-dye t-shirt

• Black pen ink

• DNA testing– Tomb of Unknown Soldiers– Crime scene – Paternity testing

Paper Chromatography

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld006.htm

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld006.htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

Setup to heat a solution

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 42

Ring stand

Beaker

Wire gauze

Ring

Bunsen burner

long spout helpsvapors to condense

mixture for distillation placed in here

Furnace

Glass retortGlass Retort

Eyewitness Science “Chemistry” , Dr. Ann Newmark, DK Publishing, Inc., 1993, pg 13

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

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.

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 39

Salt remains after all water is boiled off.

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 39

No chemical change occurs when salt water is distilled.

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 40

Saltwater solution(homogeneous mixture)

Distillation(physical method)

Salt

Pure water

Separation of a sand-saltwater mixture.

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 40

Separation of Sand from Salt1. Gently break up your salt-crusted sand with a plastic spoon.

Follow this flowchart to make a complete separation.

Salt-crusted

sand.

Dry

sand.

Wetsand.

Weigh themixture.

Decant clearliquid.

Evaporateto

dryness.

Pour intoheat-resistant

container.

Fill with water.

Stir and letsettle 1 minute.

Weighsand.

Calculateweight of

salt.

Repeat3 times?

Yes

No

2. How does this flowchart insure a completeseparation?

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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 8

The decomposition of two water molecules.

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

*Must add acid catalyst to conduct electricity

*H1+

water oxygen hydrogen

“electro” = electricity “lysis” = to split

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect 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 -

hydrogengas

cathode

oxygengas

anode

D.C. powersource

water

Reviewing ConceptsPhysical 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 ConceptsPhysical Properties

• Explain why sharpening a pencil is an example of a physical change.

• What allows a mixture to be separated by distillation?

Reviewing ConceptsChemical 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 ConceptsChemical 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 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

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

Top so

il runoff

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 4.18Alcohol 0.58 2.4Wood 0.42 1.8Aluminum 0.22 0.90Sand 0.19 0.79Iron 0.11 0.46Copper 0.093 0.39Silver 0.057 0.24Gold 0.031 0.13

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

(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 v 2

KE = ½ m v 2 KE = ½ m v

2

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

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs. Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

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?

An Energy Diagram

activatedcomplex

activationenergyEa

reactants

products

course of reaction

ener

gy

Animation by Raymond ChangAll rights reserved

Energy Sources in the United States

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 307

Wood Coal Petroleum / natural gas Hydro and nuclear

1850 1900 1940 1980 1990

100

80

60

40

20

0

Per

cent

9

91

21

71

5 310

50

40

20

70

10

26

58

16

Energy Sources in the United States

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 307

Wood Coal Petroleum / natural gas Hydro and nuclear

1850

100

80

60

40

20

0

Per

cent

9

91

1900

21

71

5 3

1940

10

50

40

1980

20

70

10

1990

26

58

16

Energy Sources in the United States

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 307

Wood Coal Petroleum / natural gas Hydro and nuclear

1850

100

80

60

40

20

0

Per

cent

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

fanelectrical energy tomechanical energy

light bulbelectrical energy to

light energy tothermal and radiant energy

coffee makerelectrical energy to

thermal energy

pencil sharpenerelectrical energy tomechanical energy

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

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

Endothermic Reaction

Reactant + Energy Product

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

System

En

erg

y

Beforereaction

Afterreaction

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

ENDOthermicqsys > 0

EXOthermicqsys < 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

Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51

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

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

Thermometer

Styrofoamcover

Styrofoamcups

Stirrer

Thermometer

Glass stirrer

Cork stopper

Two Styrofoam ®cups nestedtogether containingreactants in solution

A Bomb Calorimeter

Heating Curves

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

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

Heating CurvesTe

mp

erat

ure

(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

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

Heating Curves

• Phase Change– change in PE (molecular arrangement)– temp remains constant

• Heat of Fusion (Hfus)– energy required to melt 1 gram of a substance at its

m.p.

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

Heating Curves

• Heat of Vaporization (Hvap)– energy required to boil 1 gram of a substance at its

b.p.– usually larger than Hfus…why?

• EX: sweating, steam burns, the drinking bird

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

Phase Diagrams

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

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

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

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

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 Water90 oC 10 oC

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Water Molecules in the same temperature water

Water(50 oC)

Water(50 oC)

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 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) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system ?

a) 60oC b) 30oC c) 20oC d) ?

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 30.0oC

Block “A” Block “B”Final

Temperature

Assume NO heat energy is “lost” to the surroundings from the system.

20 g (40oC) 10 g (20oC) 33.3oC

C3.33

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system ?

a) 60oC b) 30oC c) 20oC d) ?

?

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 30.0oC

Block “A” Block “B”Final

Temperature

Assume NO heat energy is “lost” to the surroundings from the system.

20 g (40oC) 10 g (20oC) 33.3oC

C7.26

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 26.7oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 30.0oC

Block “A” Block “B”Final

Temperature

20 g (40oC) 10 g (20oC) 33.3oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 26.7oC

AgH2O

Real Final Temperature = 26.6oC

Why?

We’ve been assuming ALL materialstransfer 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

C26.6 x

320.8x 8550

7845 313.8x x 05.7 705

algebra. the solve and units Drop

C25 -x g 75CgJ 184.4 C100 -x g 30CgJ 235.0

equation. into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 26.6oC

Calorimetry

C26.6 x

8550 320.8x

7845 313.8x x 05.7 705

algebra. the solve and units Drop

C25 -x g 75CgJ 184.4 C100 -x g 30CgJ 235.0

equation. into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

“food” = “science”

Candy bar300 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 . DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

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 = (specific heat) (mass) (change in temperature)

q = Cp . m . DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2.077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 207.7 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

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

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

FeT = 500oCmass = ? grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(Cp,Fe) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(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.

Calorimetry Problems 2 question #8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(Cp,Au) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(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.

Calorimetry Problems 2 question #9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(Cp,H2O) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(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.

Calorimetry Problems 2 question #10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(Cp,H2O) (mass) (T)] = (Cp,H2O) (mass) (T) + (Cf) (mass) + (Cp,H2O) (mass) (T)

- [(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

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

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

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.

Calorimetry Problems 2 question #11

- [(Cp,H2O) (mass) (T)] + (Cv,H2O) (mass) + (Cp,H2O) (mass) (T) = [(Cp,H2O) (mass) (T)]

- [ - 816.8 - 56400 + 104.5Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(Cp,H2O) (mass) (T)]

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) (T)]

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

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

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

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?

Calorimetry Problems 2 question #12

PbT = ? oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(Cp,Pb) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(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?

Calorimetry Problems 2 question #13

H2OT = -12oCmass = ? g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(Cp,H2O) (mass) (T)]

458.2 m = - 17339

m = 37.8 g

iceTf = 24oC

qA = [(Cp,H2O) (mass) (T)]

qC = [(Cp,H2O) (mass) (T)]

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

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Catalytic Converter

Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 454

N O

N O

N O

N O

N O

NO

N

O

NO

N O

N

O

N

O

O ON

NO

O

OO

N NNN

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

CO

OCO

NN

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.

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + ½ O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 1/2O2(g) H2O(g) + 242 kJ DH = -242 kJKotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

Hess’s LawCalculate 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 kJC(s) C(g) DH = +720 kJCO2(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 kJC(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 positionof ball

final positionof ball

10 yard pass

5 yard penalty

5 yard net gain

Fission vs. Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5,000,000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof 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.

Shielding Radiation

Nuclear Fission

Nuclear Fission

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

Nuclear Fission

Nuclear Power Plants

map: Nuclear Energy Institute

Fermi Approximations

FERMI APPROXIMATIONS An educated guess – based on a seriesof calculations of known facts – to arriveat 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

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Mass…mass is converted into energy

Hydrogen (H2) H = 1.008 amuHelium (He) He = 4.004 amu

FUSION

2 H2 1 He + ENERGY

1.008 amux 44.0032 amu = 4.004 amu + 0.028 amu

This relationship was discovered by Albert EinsteinE = 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 interms of curvature of space in the presence of a gravitationalfield. Here we see how this curvature varies near a black hole.

Time Travel?

…Albert Einstein also discovered the Geometry of Space Near a Black Hole

Einstein’s theory of general relativity maybe interpreted interms of curvature of space in the presence of a gravitationalfield. Here we see how this curvature varies near a black hole.

Time Travel?

…Albert Einstein also discovered the Geometry of Space Near a Black Hole

Einstein’s theory of general relativity maybe interpreted interms of curvature of space in the presence of a gravitationalfield. 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 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(%

)

100

50

25

12.5

Half-life of Radiation

Initial amountof radioisotope

t1/2

t1/2

t1/2

After 1 half-life

After 2 half-lives

After 3 half-lives

Triple Point Plot

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 488

solidliquid

gas

melting

freezing

sublimation

deposition

vaporization

condensation

Temperature (oC)

Pre

ssur

e (a

tm)

0.6

2.6

Liquid VaporSolid

Normalmelting point

Normalboilingpoint

101.3

0.61

0.016 1000Temperature (oC)

Pre

ssu

re (

KP

a)

Triple point

Triple Point

Criticalpressure

Critical point

Critical temperature

373.99

22,058

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

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 EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF!

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat, light, sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF!

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + 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 CurvesHeating CurvesTe

mp

erat

ure

(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 CurvesHeating 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 capacityCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Heating CurvesHeating Curves

Phase Change• change in PE (molecular arrangement)• temp remains constant

Heat of Fusion (Hfus)

• energy required to melt 1 gram of a substance at its m.p.

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

Heating CurvesHeating Curves

Heat of Vaporization (Hvap)• energy required to boil 1 gram of a

substance at its b.p.• usually larger than Hfus…why?

EX: sweating, steam burns, the drinking bird

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

Phase DiagramsPhase Diagrams

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

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

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