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Chemistry 205
For all students: Pickup and fill out a copy of the course questionnaire. If you are not currently enrolled in the course, write “Not Enrolled” at the top of the form.
For all enrolled students: Please pick up a copy of each of the following Course Syllabus (If you are using the 3rd edition of the textbook, you will need to the syllabus for that edition. Course Study Guide Homework Assignment #1 & #2 (same sheet)
Fundamentals of Chemistry I: General ChemistrySection 1: Dr. Dennis Pederson
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Chinese Five-element View of Matter
Built on basis of process and change: Generating Interactions and Overcoming Interactions
Generating Interactions:Wood feeds fireFire creates earth (ash)Earth bears metalMetal collects waterWater nourishes wood
Overcoming Interactions:Wood parts earthEarth absorbs waterWater quenches fireFire melts metalMetal chops wood
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Plato’s Solids
Earth WaterFire Air
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Greek Four-element View of Matter
Each element is a combination of two properties: Fire = hot + dry Earth = cold + dry Water = cold + wet Air = hot + wet
Aristotle’s definition of an element: Let us define theElement in bodies as that into which other bodies may beanalyzed, which are present in them either potentiallyor actually ---, and which cannot itself be analyzed intoconstituents differing in kind.
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Robert Boyle wrote The Skeptical Chymist
Boyle’s definition of an element:Certain primitive and simple, or perfectly unmingled bodies;which not being made of any other bodies, or of one another,are the ingredients of which all those perfectly mixt bodiesare immediately compounded, and into which they areultimately resolved.
Aristotle’s definition of an element: Let us define theelement in bodies as that into which other bodies may beanalyzed, which are present in them either potentiallyor actually ---, and which cannot itself be analyzed intoconstituents differing in kind
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Chemistry 205 Questionnaire Is Still Needed From The Following Student
Elisha Alexander
Any Students Who Are Trying To Add Chemistry 205 Should See Me After Class
Laboratory places are available in the following sections.Section 4: M 12:00 - 2:50 One placeSection 5: M 3:00 - 5:50 One placeSection 6: M 6:00 - 8:50 One placeSection 8: T 3:00 - 5:50 One place Section 11: F 7:40 - 10:30 One placeSection 12: Th 2:00 - 4:50 One place
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01_03.JPG
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01_02.JPG
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Rules for Determining the Number of Significant Digits in a Measurement
1. A number is a significant digit if it is
a. not a zero 892 g 3 significant digits26943 m 5 significant digits
b. a zero between two nonzero digits
6032 mL 4 significant digits963022 dg 6 significant digits
c. a zero after a nonzero digit and after the decimal point
46.0 cL 3 significant digits183.70 mm 5 significant digits
2. A zero is a not significant digit if it is
a. a leading zero, after the decimal and before the first nonzero digit
0.0025 L 2 significant digits0.000105 cg 3 significant digits
b. a trailing zero, after a nonzero digit and only showing the power of ten
4500 dm 2 significant digits139000 mg 3 significant digits
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Rules for conversion of a number to scientific notation.
1. Move the decimal point enough places to the left or Right so that only one integer is to the left of the decimal.
3. If you’ve moved the decimal to the right, decrease the exponent on the power of 10 by one for each place moved.
2. If you’ve moved the decimal to the left, increase the exponent on the power of 10 by one for each place moved.
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Significant Digits4169 g
5.040 dL
Four significant digits: all integers
0.0087 cg
12.306 m
35900 mL Three significant digits: the two zeros only tell the power of tenWritten in scientific notation shows the significant digits: 3.59 x 104
Five significant digits: the zero is between integers
Two significant digits: the two zeros between the decimal and the first integer only tell the power of tenWritten in scientific notation shows the significant digits: 8.7 x 10–3
Four significant digits: a zero after the decimal and after an integer counts
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The measurement 0.0000043 m, expressed correctly using scientific notation, is
A) 4.3 x 10-7 m B) 4.3 x 10-6 m
C) 4.3 x 106 mD) 0.43 x 10-5 m
Answer: 4.3 x 10-6 m
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5.21 cm is the same distance as A) 0.0521 m B) 52.1 dm C) 5.21 mm
D) 521 m
Answer: Checking all conversions (5.21 cm)(1 m/100 cm) = 0.0521 m (5.21 cm)(10 cm/1 dm) = 0.521 dm (5.21 cm)(10 mm/1 cm) = 52.1 mm
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States of MatterGas Liquid
Low density
Little or no attractive forces between particles
Takes shape of container
Particles in rapid motion and largedistance between particles
Fills container
High density
Strong attractive forces between particles
Takes shape of container
Particles in slow motion and closetogether
Definite volume
High density
Very strong attractive forces between particles
Definite shape
Particles closeTogether and essentially no motion
Definite volume
Solid
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Physical and Chemical Properties and ChangesIdentify each of the following as a physical or chemical change or property
The melting point of aluminum is 660 oC
Baking soda dissolves in water
White phosphorus can react with air
When white phosphorus is put in air it bursts into flame
Gold can be pounded into a thin sheet
Liquid alcohol evaporates
Cooking a hamburger
Making a wire out of a piece of copper
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Solid-Liquid Transformations
endothermic
exothermic
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Liquid-Gas Transformations
endothermic
exothermic
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Phase Transition Summary
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02_03.JPG
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+++10.00 g of leadLaw of Definite Proportions1.55 g of sulfur11.55 g of lead sulfide18.00 g of lead10.00 g of lead3.00 g of sulfur11.55 g of lead sulfide1.45 g of sulfur(leftover)
1.55 g of sulfur11.55 g of lead sulfide8.00 g of lead(leftover)
++
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Dalton’s Atomic Theory
All matter is composed of tiny, indivisible particles called atoms.
All atoms of a given element are alike, but differ from atoms of another element.
Compounds are formed when atoms of different elementscombine in fixed proportions.
A chemical reaction involves the rearrangement, separation,or combination of atoms. Atoms are never created or destroyedduring a chemical reaction.
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Millikan Oil-Drop Experiment
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Thompson Plum Pudding Model of the Atom
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Rutherford’s Gold-Foil Experiment
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Electron Energy Sublevels
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Order of Sublevel Filling
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Electron Configurations and the Periodic Table
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Atomic Size
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Ionization EnergyEnergy required to Remove
First Electron
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Increasing Metallic Character
Incr
easi
ng
Met
alli
c C
har
acte
r
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Summary of Results - First Examination
Grade ScaleA’s 85 - 100B’s 70 - 84C’s 50 - 69D’s 35 - 49F’s 0- 34
Examination Average: 71
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Nuclear ReactionsParticle ReactionProcess
decay He2
4
decay e-1
0
H1
1proton bombardment
positron decay e
+1
0
neutron bombardment
n0
1
U92
238 Th90
234He
2
4 +
C 6
14N
7
14+e
-1
0
e+1
0C
6
11 + B 5
11
n0
1U
92
238Np
93
239U
92
239+ + e
-1
0
H1
1+Li
3
7 Be4
8
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Positron Emission Tomography
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Nuclear Fission
n0
1U
92
235U
92
236+ +Kr
36
91Ba
56
1423 n
0
1+
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Nuclear Fusion
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Monoatomic Ions
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Polyatomic IonsOH– hydroxide CN– cyanide NH4
+ ammonium
NO3– nitrate NO2
– nitritesame charge, one less oxygen
CO32– carbonate
SO42– sulfate
HCO3– hydrogen carbonate
(bicarbonate)add H+, change charge
ClO3–
chlorate
SO32– sulfite
HSO4– hydrogen sulfate
(bisulfate) HSO3– hydrogen sulfite
(bisulfite)
ClO2–
chlorite
PO43– phosphate
ClO4–
perchlorate
one more oxygen
ClO–hypochlorite
one less oxygen
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Chemical NomenclatureFormula to Name
BaBr2 barium bromide
PCl3
Na2SO4sodium sulfate
(NH4)3PO4 ammonium phosphate
Al2(CO3)3
phosphorus trichloride
Sn(NO3)4 tin(IV) nitrate or stannic nitrate
aluminum carbonate
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Chemical NomenclatureName to Formula
potassium sulfide K1+ S2– K2S
magnesium hydrogen carbonate Mg2+ HCO31– Mg(HCO3)2
dichlorine oxide Cl2O
copper(II) nitrite Cu2+ NO21– Cu(NO2)2
zinc phosphate Zn2+ PO43– Zn3(PO4)2
tetrasulfur dinitride S4N2
Congratulations!!
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Kinetic Molecular Theory of GasesGas contains small particles, moving rapidly and randomly:
Negligible attractive forces between the particles:
Volume occupied by particles is negligible relative to total volume of the gas:
Average kinetic energy of particles is proportional to the Kelvin temperature:
Particles are in constant motion, move rapidly in straight lines until they collide with each other or the container walls:
Explains why gases diffuse quickly and fill any container they occupy
Also explains why a gas expands to fill its container
Explains low density and compressibility of gases
Increasing the temperature causes particles to move faster
Explains how gas exerts a pressure - collisions with the container walls
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Summary of Results - Second Examination
Grade ScaleA’s 85 - 100B’s 70 - 84C’s 50 - 69D’s 35 - 49F’s 0- 34
Examination Average: 64.5
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H2O + CO2 HCO3– + H+
HbH+ + O2 HbO2 + H+tissues lungs
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Vapor Pressure of Water as aFunction of Temperature
Temperature ( oC)
Pre
ssur
e (
mm
Hg)
T em p eratu re ( oC)P ressu re ( m m H g )7 6 0 m m H g (1 a tm )
760 mmHg (1 atm)
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Energy Changes in Chemical Reactions
Endothermic reaction: N2(g) + 2 O2(g) + 67 kJ 2 NO2(g)
Exothermic reaction: N2O4(g) + 4 H2(g) 2 N2(g)+ 4 H2O(g) + 977 kJ
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Activation Energy Diagram
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Collisions and Rate of Chemical Reactions
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Effect of a Catalyst
Catalyst provides an alternate pathway that has a lower activation energy.
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Kinetics and Equilibrium
Equilibriumrate of forward reaction = rate of reverse reaction
Equilibriumconcentration of reactants andproducts no longer changing
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Equilibrium ConstantsAt equilibrium, a mathematical relationship exists between theconcentration(s) of the products and the concentration(s) of theproducts - called the equilibrium constant expression.
Example: aA + bB cC + dD
Kc = [C]c [D]d
[A]a [B]b[ ] = concentration in mol/liter
Kc = [SO3]2
[SO2]2 [O2]
Other examples:
2SO2(g) + O2(g) 2SO3(g)
CH4(g) + 2H2S(g) CS2(g) + 4H2(g) Kc = [CH4]
[H2S]2
[CS2] [H2]
4
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Equilibrium Constant and Extent of Reaction
If Kc is large (>> 1) then equilibrium mixture is mostly products.
Examples:
2H2(g) + S2(g) 2H2S(g) Kc = 1.1 x 107
N2(g) + 3H2(g) 2NH3(g) Kc = 1.6 x 102
If Kc is small (<< 1) then equilibrium mixture is mostly reactants.
PCl5(g) PCl3(g) + Cl2(g) Kc = 1.2 x 10–2
N2(g) + O2(g) 2NO(g) Kc = 2 x 10–9
Examples:
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Changing Equilibrium Conditions
Le Châtelier's PrincipleWhen a system at equilibrium is disturbed, the system willshift in the direction that will reduce that stress.
Three types of stress: (1) Concentration change (2) Temperature change (3) Volume change
2SO2(g) + O2(g) 2SO3(g) + heat
Concentration change effects:
Increase concentration: Shift in direction that uses what was added.
Decrease concentration: Shift in direction that replaces what was removed.
Add some SO2 (reactant)- shift toward products
Add some SO3 (product) - shift toward reactants
Remove some SO2 (reactant)- shift toward reactants
Remove some SO3 (product)- shift toward products
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Changing Equilibrium Conditions
Le Châtelier's Principle
2SO2(g) + O2(g) 2SO3(g) + heat
Temperature change effects:
Addition of heat (increase temperature) Shift toward reactants to use added heat
Addition of heat (increase temperature) Shift toward products to use added heat
Removal of heat (decrease temperature) Shift toward products to produce heat
N2(g) + 2 O2(g) + 67 kJ 2 NO2(g)
Removal of heat (decrease temperature) Shift toward reactants to produce heat
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Changing Equilibrium Conditions
Le Châtelier's Principle
2SO2(g) + O2(g) 2SO3(g)
Volume change effects: Requires that there is a difference in the volume of the products and the volume of the reactants.
Decrease volume Shift toward products (less moles of gas, less volume)
Increase volume Shift toward reactants (more moles of gas, more volume)
Decrease volume Shift toward reactants (less moles of gas, less volume)Increase volume Shift toward products (more moles of gas, more volume)
2 NO2(g) N2(g) + 2 O2(g)
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*Expired air is a mixture of alveolar air and inspired air.
*
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Le Châtelier's Principle Oxygen and Carbon Dioxide Transport in the Blood
Hemoglobin - oxygen binding equilibrium HbH+ + O2(g) HbO2 + H+
In the lungs, concentration of oxygen is high, equilibrium shifts right ( ) binding more oxygen.In the tissues, concentration of oxygen is low, equilibrium shifts left ( ) releasing more oxygen.
Oxygen Transport
Carbon Dioxide Transport
Carbon dioxide - bicarbonate ion equilibrium
CO2(g) + H2O H2CO3(aq) HCO3– (aq) + H+
In the tissues, concentration of CO2 is high, equilibrium shifts right ( ) forming more soluble bicarbonate ion.In the lungs, concentration of CO2 is low, equilibrium shifts left ( ) releasing CO2.
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H2O + CO2 HCO3– + H+
HbH+ + O2 HbO2 + H+tissues lungs
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08_T01.JPG
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Dissolving NaCl in H2O
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Like Dissolves Like
H2O (polar)
CH2Cl2 (nonpolar)
Add I2 (nonpolar)
Add Ni(NO3)2 (ionic)
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Equivalent (concentration unit) = one mole of positive or negative charge:
1 mole Na+ = 1 Eq,
1 mol Mg2+ = 2 Eq
Electrolyte Concentrations in Body Fluids
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Solubility as a Function of Temperature
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Colloids
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Osmosis
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QuickTime™ and aTIFF (Uncompressed) decompressor
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