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

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

Plato’s Solids

Earth WaterFire Air

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.

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

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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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

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.

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

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

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 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

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

Solid-Liquid Transformations

endothermic

exothermic

Liquid-Gas Transformations

endothermic

exothermic

Phase Transition Summary

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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)

++

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.

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 EnergyEnergy required to Remove

First Electron

Increasing Metallic Character

Incr

easi

ng

Met

alli

c C

har

acte

r

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

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

Positron Emission Tomography

Nuclear Fission

n0

1U

92

235U

92

236+ +Kr

36

91Ba

56

1423 n

0

1+

Nuclear Fusion

Monoatomic Ions

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

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

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!!

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

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

H2O + CO2 HCO3– + H+

HbH+ + O2 HbO2 + H+tissues lungs

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)

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

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

Equilibriumrate of forward reaction = rate of reverse reaction

Equilibriumconcentration of reactants andproducts no longer changing

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

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:

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

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

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)

*Expired air is a mixture of alveolar air and inspired air.

*

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.

H2O + CO2 HCO3– + H+

HbH+ + O2 HbO2 + H+tissues lungs

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Dissolving NaCl in H2O

Like Dissolves Like

H2O (polar)

CH2Cl2 (nonpolar)

Add I2 (nonpolar)

Add Ni(NO3)2 (ionic)

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

Solubility as a Function of Temperature

Colloids

Osmosis

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are needed to see this picture.

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O NOH

C

O

O

O

C O

C

O

O

CH3

OH

CH3

OH

H

O

C

O

CH3

O

C O

CH3

CH3

CH3

CH3

Docetaxel Structure

1

2

3

4

5

67

8 9

10

11 CH3

CH3