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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 01_03.JPG 01_02.JPG 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 02_03.JPG 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+ 08_T01.JPG 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 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture.