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Chemistry 205
Fundamentals of Chemistry I: General Chemistry
Section 1: Dr. Dennis Pederson
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)
Chinese Five-element View of Matter
Built on basis of process and change:
Generating Interactions and
Overcoming Interactions
Generating Interactions:
Wood feeds fire
Fire creates earth (ash)
Earth bears metal
Metal collects water
Water nourishes wood
Overcoming Interactions:
Wood parts earth
Earth absorbs water
Water quenches fire
Fire melts metal
Metal chops wood
Plato’s Solids
Fire
Earth
Air
Water
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 the
Element in bodies as that into which other bodies may be
analyzed, which are present in them either potentially
or actually ---, and which cannot itself be analyzed into
constituents differing in kind.
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 bodies
are immediately compounded, and into which they are
ultimately resolved.
Aristotle’s definition of an element: Let us define the
element in bodies as that into which other bodies may be
analyzed, which are present in them either potentially
or actually ---, and which cannot itself be analyzed into
constituents differing in kind
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 place
Section 5: M 3:00 - 5:50 One place
Section 6: M 6:00 - 8:50 One place
Section 8: T 3:00 - 5:50 One place
Section 11: F 7:40 - 10:30 One place
Section 12: Th 2:00 - 4:50 One place
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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 digits
26943 m 5 significant digits
b. a zero between two
nonzero digits
6032 mL 4 significant digits
963022 dg 6 significant digits
c.
46.0 cL 3 significant digits
183.70 mm 5 significant digits
a zero after a nonzero digit
and after the decimal point
2. A zero is a not significant digit if it is
a.
a leading zero, after the decimal
and before the first nonzero digit
b. a trailing zero, after a nonzero
digit and only showing the power
of ten
0.0025 L 2 significant digits
0.000105 cg 3 significant digits
4500 dm 2 significant digits
139000 mg 3 significant digits
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.
2. If you’ve moved the decimal to the left, increase the
exponent on the power of 10 by one for each place moved.
3. If you’ve moved the decimal to the right, decrease the
exponent on the power of 10 by one for each place moved.
Significant Digits
4169 g
35900 mL
12.306 m
Four significant digits: all integers
Three significant digits: the two zeros
only tell the power of ten
Written in scientific notation shows the
significant digits: 3.59 x 104
Five significant digits: the zero is
between integers
0.0087 cg
Two significant digits: the two zeros
between the decimal and the first integer
only tell the power of ten
Written in scientific notation shows the
significant digits: 8.7 x 10–3
5.040 dL
Four significant digits: a zero after the
decimal and after an integer counts
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 m
D) 0.43 x 10-5 m
Answer: 4.3 x 10-6 m
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
States of Matter
Gas
Liquid
Solid
Low density
High density
High density
Takes shape
of container
Takes shape
of container
Definite shape
Fills container
Definite volume
Definite volume
Particles in rapid
motion and large
distance between
particles
Particles in slow
motion and close
together
Particles close
Together and
essentially no
motion
Little or no
attractive forces
between particles
Strong attractive
forces between
particles
Very strong
attractive forces
between particles
Physical and Chemical Properties and Changes
Identify 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
Solid-Liquid Transformations
endothermic
exothermic
Liquid-Gas Transformations
endothermic
exothermic
Phase Transition Summary
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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 elements
combine in fixed proportions.
A chemical reaction involves the rearrangement, separation,
or combination of atoms. Atoms are never created or destroyed
during a chemical reaction.
Millikan Oil-Drop Experiment
Thompson Plum Pudding Model of the Atom
Rutherford’s Gold-Foil Experiment
Electron Energy Sublevels
Order of Sublevel Filling
Electron Configurations and the
Periodic Table
Atomic Size
Ionization Energy
Energy required to Remove
First Electron
Increasing Metallic Character
Increasing Metallic Character
Summary of Results - First Examination
Grade Scale
A’s 85 - 100
B’s 70 - 84
C’s 50 - 69
D’s 35 - 49
F’s 0- 34
Examination Average: 71
Nuclear Reactions
- decay
- decay
positron
decay
neutron
bombardment
proton
bombardment
Reaction
Particle
4
238
He
2
0
e
6
11
e
+1
6
1
238
0
92
n
2
0
C
0
7
1
3
+
e +
+1
1
239
0
92
n
U +
e
-1
C
1
H
He
U
92
14
-1
0
4
+
Process
Li
+
1
H
1
234
Th
90
14
N
7
11
5
B
239
U
Np +
93
8
Be
4
0
e
-1
Positron Emission Tomography
Nuclear Fission
235
1
U + 0n
92
236
U
92
91
Kr +
36
142
1
Ba + 3 0n
56
Nuclear Fusion
Monoatomic Ions
Polyatomic Ions
OH– hydroxide
NO3– nitrate
CN– cyanide
NO2– nitrite
CO32– carbonate
NH4+ ammonium
same charge,
one less oxygen
HCO3– hydrogen carbonate
(bicarbonate)
add H+, change charge
SO42– sulfate
HSO4–
SO32– sulfite
hydrogen sulfate
(bisulfate)
ClO3– chlorate
one more oxygen
ClO4– perchlorate
PO43– phosphate
HSO3
–
hydrogen sulfite
(bisulfite)
ClO2– chlorite
one less oxygen
ClO– hypochlorite
Chemical Nomenclature
Formula to Name
BaBr2
barium bromide
Na2SO4
sodium sulfate
(NH4)3PO4
ammonium phosphate
PCl3
phosphorus trichloride
Sn(NO3)4
tin(IV) nitrate or stannic nitrate
Al2(CO3)3
aluminum carbonate
Chemical Nomenclature
Name to Formula
potassium sulfide
K1+ S2–
K2S
Mg2+ HCO31–
magnesium hydrogen carbonate
dichlorine oxide
Mg(HCO3)2
Cl2O
copper(II) nitrite
Cu2+
NO21–
Cu(NO2)2
zinc phosphate
Zn2+
PO43–
Zn3(PO4)2
tetrasulfur dinitride
S4N2
Congratulations!!
Kinetic Molecular Theory of Gases
Gas contains small particles, moving rapidly and randomly:
Explains why gases diffuse quickly and fill any container they occupy
Negligible attractive forces between the particles:
Also explains why a gas expands to fill its container
Volume occupied by particles is negligible relative to total volume
of the gas:
Explains low density and compressibility of gases
Average kinetic energy of particles is proportional to the
Kelvin temperature:
Increasing the temperature causes particles to move faster
Particles are in constant motion, move rapidly in straight lines
until they collide with each other or the container walls:
Explains how gas exerts a pressure - collisions with the container walls
Summary of Results - Second Examination
Grade Scale
A’s 85 - 100
B’s 70 - 84
C’s 50 - 69
D’s 35 - 49
F’s 0- 34
Examination Average: 64.5
tissues
lungs
HbH+ + O2
HbO2 + H+
H2O + CO2
HCO3– + H+
Pressure ( mmHg)
Vapor Pressure of Water as a
Function of Temperature
760 mmHg (1 atm)
Temperature ( oC)
Energy Changes in Chemical Reactions
Endothermic reaction: N2(g) + 2 O2(g) + 67 kJ
Exothermic reaction: N2O4(g) + 4 H2(g)
2 NO2(g)
2 N2(g)+ 4 H2O(g) + 977 kJ
Activation Energy Diagram
Collisions and Rate of Chemical Reactions
Effect of a Catalyst
Catalyst provides an alternate
pathway that has a lower
activation energy.
Kinetics and Equilibrium
Equilibrium
rate of forward reaction
= rate of reverse reaction
Equilibrium
concentration of reactants and
products no longer changing
Equilibrium Constants
At equilibrium, a mathematical relationship exists between the
concentration(s) of the products and the concentration(s) of the
products - called the equilibrium constant expression.
Example: aA + bB
Kc =
cC + dD
[C]c [D]d
a
b
[A] [B]
[ ] = concentration in mol/liter
Other examples:
2SO2(g) + O2(g)
CH4(g) + 2H2S(g)
2SO3(g)
Kc =
CS2(g) + 4H2(g)
[SO3]2
[SO2]2 [O2]
Kc =
[CS2] [H2]4
[CH4] [H2S]2
Equilibrium Constant and Extent of Reaction
If Kc is large (>> 1) then equilibrium mixture is mostly products.
Examples:
2H2(g) + S2(g)
N2(g) + 3H2(g)
2H2S(g)
2NH3(g)
Kc = 1.1 x 107
Kc = 1.6 x 102
If Kc is small (<< 1) then equilibrium mixture is mostly reactants.
Examples:
PCl5(g)
N2(g) + O2(g)
PCl3(g) + Cl2(g)
2NO(g)
Kc = 1.2 x 10–2
Kc = 2 x 10–9
Changing Equilibrium Conditions
Le Châtelier's Principle
When a system at equilibrium is disturbed, the system will
shift in the direction that will reduce that stress.
Three types of stress:
(1) Concentration change
(2) Temperature change
(3) Volume change
Concentration change effects:
2SO2(g) + O2(g)
2SO3(g) + heat
Increase concentration: Shift in direction that uses what was added.
Add some SO2 (reactant)- shift toward products
Add some SO3 (product) - shift toward reactants
Decrease concentration: Shift in direction that replaces what was removed.
Remove some SO2 (reactant)- shift toward reactants
Remove some SO3 (product)- shift toward products
Changing Equilibrium Conditions
Le Châtelier's Principle
Temperature change effects:
2SO2(g) + O2(g)
2SO3(g) + heat
Addition of heat (increase temperature)
Shift toward reactants 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)
Addition of heat (increase temperature)
Shift toward products to use added heat
Removal of heat (decrease temperature)
Shift toward reactants to produce heat
Changing Equilibrium Conditions
Le Châtelier's Principle
Volume change effects: Requires that there is a difference in the
volume of the products and the volume of the reactants.
2SO2(g) + O2(g)
2SO3(g)
Decrease volume
Shift toward products (less moles of gas, less volume)
Increase volume
Shift toward reactants (more moles of gas, more volume)
2 NO2(g)
N2(g) + 2 O2(g)
Decrease volume
Shift toward reactants (less moles of gas, less volume)
Increase volume
Shift toward products (more moles of gas, more volume)
*
*Expired air is a mixture of alveolar air and inspired air.
Le Châtelier's Principle
Oxygen and Carbon Dioxide Transport in the Blood
Oxygen Transport
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.
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.
tissues
lungs
HbH+ + O2
HbO2 + H+
H2O + CO2
HCO3– + H+
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Dissolving NaCl in H2O
Like Dissolves Like
H2O (polar)
CH2Cl2 (nonpolar)
Add I2 (nonpolar)
Add Ni(NO3)2 (ionic)
Electrolyte Concentrations in Body Fluids
Equivalent (concentration unit)
= one mole of positive or
negative charge:
1 mole Na+ = 1 Eq,
1 mol Mg2+ = 2 Eq
Solubility as a Function of Temperature
Colloids
Osmosis
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