Raffi Hovasapian

Raffi Hovasapian

AP Practice Exam: Multiple Choice, Part I

Slide Duration:

Table of Contents

Section 1: Review
Naming Compounds

41m 24s

Intro
0:00
Periodic Table of Elements
0:15
Naming Compounds
3:13
Definition and Examples of Ions
3:14
Ionic (Symbol to Name): NaCl
5:23
Ionic (Name to Symbol): Calcium Oxide
7:58
Ionic - Polyatoms Anions: Examples
12:45
Ionic - Polyatoms Anions (Symbol to Name): KClO
14:50
Ionic - Polyatoms Anions (Name to Symbol): Potassium Phosphate
15:49
Ionic Compounds Involving Transition Metals (Symbol to Name): Co₂(CO₃)₃
20:48
Ionic Compounds Involving Transition Metals (Name to Symbol): Palladium 2 Acetate
22:44
Naming Covalent Compounds (Symbol to Name): CO
26:21
Naming Covalent Compounds (Name to Symbol): Nitrogen Trifluoride
27:34
Naming Covalent Compounds (Name to Symbol): Dichlorine Monoxide
27:57
Naming Acids Introduction
28:11
Naming Acids (Name to Symbol): Chlorous Acid
35:08
% Composition by Mass Example
37:38
Stoichiometry

37m 19s

Intro
0:00
Stoichiometry
0:25
Introduction to Stoichiometry
0:26
Example 1
5:03
Example 2
10:17
Example 3
15:09
Example 4
24:02
Example 5: Questions
28:11
Example 5: Part A - Limiting Reactant
30:30
Example 5: Part B
32:27
Example 5: Part C
35:00
Section 2: Aqueous Reactions & Stoichiometry
Precipitation Reactions

31m 14s

Intro
0:00
Precipitation Reactions
0:53
Dissociation of ionic Compounds
0:54
Solubility Guidelines for ionic Compounds: Soluble Ionic Compounds
8:15
Solubility Guidelines for ionic Compounds: Insoluble ionic Compounds
12:56
Precipitation Reactions
14:08
Example 1: Mixing a Solution of BaCl₂ & K₂SO₄
21:21
Example 2: Mixing a Solution of Mg(NO₃)₂ & KI
26:10
Acid-Base Reactions

43m 21s

Intro
0:00
Acid-Base Reactions
1:00
Introduction to Acid: Monoprotic Acid and Polyprotic Acid
1:01
Introduction to Base
8:28
Neutralization
11:45
Example 1
16:17
Example 2
21:55
Molarity
24:50
Example 3
26:50
Example 4
30:01
Example 4: Limiting Reactant
37:51
Example 4: Reaction Part
40:01
Oxidation Reduction Reactions

47m 58s

Intro
0:00
Oxidation Reduction Reactions
0:26
Oxidation and Reduction Overview
0:27
How Can One Tell Whether Oxidation-Reduction has Taken Place?
7:13
Rules for Assigning Oxidation State: Number 1
11:22
Rules for Assigning Oxidation State: Number 2
12:46
Rules for Assigning Oxidation State: Number 3
13:25
Rules for Assigning Oxidation State: Number 4
14:50
Rules for Assigning Oxidation State: Number 5
15:41
Rules for Assigning Oxidation State: Number 6
17:00
Example 1: Determine the Oxidation State of Sulfur in the Following Compounds
18:20
Activity Series and Reduction Properties
25:32
Activity Series and Reduction Properties
25:33
Example 2: Write the Balance Molecular, Total Ionic, and Net Ionic Equations for Al + HCl
31:37
Example 3
34:25
Example 4
37:55
Stoichiometry Examples

31m 50s

Intro
0:00
Stoichiometry Example 1
0:36
Example 1: Question and Answer
0:37
Stoichiometry Example 2
6:57
Example 2: Questions
6:58
Example 2: Part A Solution
12:16
Example 2: Part B Solution
13:05
Example 2: Part C Solution
14:00
Example 2: Part D Solution
14:38
Stoichiometry Example 3
17:56
Example 3: Questions
17:57
Example 3: Part A Solution
19:51
Example 3: Part B Solution
21:43
Example 3: Part C Solution
26:46
Section 3: Gases
Pressure, Gas Laws, & The Ideal Gas Equation

49m 40s

Intro
0:00
Pressure
0:22
Pressure Overview
0:23
Torricelli: Barometer
4:35
Measuring Gas Pressure in a Container
7:49
Boyle's Law
12:40
Example 1
16:56
Gas Laws
21:18
Gas Laws
21:19
Avogadro's Law
26:16
Example 2
31:47
Ideal Gas Equation
38:20
Standard Temperature and Pressure (STP)
38:21
Example 3
40:43
Partial Pressure, Mol Fraction, & Vapor Pressure

32m

Intro
0:00
Gases
0:27
Gases
0:28
Mole Fractions
5:52
Vapor Pressure
8:22
Example 1
13:25
Example 2
22:45
Kinetic Molecular Theory and Real Gases

31m 58s

Intro
0:00
Kinetic Molecular Theory and Real Gases
0:45
Kinetic Molecular Theory 1
0:46
Kinetic Molecular Theory 2
4:23
Kinetic Molecular Theory 3
5:42
Kinetic Molecular Theory 4
6:27
Equations
7:52
Effusion
11:15
Diffusion
13:30
Example 1
19:54
Example 2
23:23
Example 3
26:45
AP Practice for Gases

25m 34s

Intro
0:00
Example 1
0:34
Example 1
0:35
Example 2
6:15
Example 2: Part A
6:16
Example 2: Part B
8:46
Example 2: Part C
10:30
Example 2: Part D
11:15
Example 2: Part E
12:20
Example 2: Part F
13:22
Example 3
14:45
Example 3
14:46
Example 4
18:16
Example 4
18:17
Example 5
21:04
Example 5
21:05
Section 4: Thermochemistry
Energy, Heat, and Work

37m 32s

Intro
0:00
Thermochemistry
0:25
Temperature and Heat
0:26
Work
3:07
System, Surroundings, Exothermic Process, and Endothermic Process
8:19
Work & Gas: Expansion and Compression
16:30
Example 1
24:41
Example 2
27:47
Example 3
31:58
Enthalpy & Hess's Law

32m 34s

Intro
0:00
Thermochemistry
1:43
Defining Enthalpy & Hess's Law
1:44
Example 1
6:48
State Function
13:11
Example 2
17:15
Example 3
24:09
Standard Enthalpies of Formation

23m 9s

Intro
0:00
Thermochemistry
1:04
Standard Enthalpy of Formation: Definition & Equation
1:05
∆H of Formation
10:00
Example 1
11:22
Example 2
19:00
Calorimetry

39m 28s

Intro
0:00
Thermochemistry
0:21
Heat Capacity
0:22
Molar Heat Capacity
4:44
Constant Pressure Calorimetry
5:50
Example 1
12:24
Constant Volume Calorimetry
21:54
Example 2
24:40
Example 3
31:03
Section 5: Kinetics
Reaction Rates and Rate Laws

36m 24s

Intro
0:00
Kinetics
2:18
Rate: 2 NO₂ (g) → 2NO (g) + O₂ (g)
2:19
Reaction Rates Graph
7:25
Time Interval & Average Rate
13:13
Instantaneous Rate
15:13
Rate of Reaction is Proportional to Some Power of the Reactant Concentrations
23:49
Example 1
27:19
Method of Initial Rates

30m 48s

Intro
0:00
Kinetics
0:33
Rate
0:34
Idea
2:24
Example 1: NH₄⁺ + NO₂⁻ → NO₂ (g) + 2 H₂O
5:36
Example 2: BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O
19:29
Integrated Rate Law & Reaction Half-Life

32m 17s

Intro
0:00
Kinetics
0:52
Integrated Rate Law
0:53
Example 1
6:26
Example 2
15:19
Half-life of a Reaction
20:40
Example 3: Part A
25:41
Example 3: Part B
28:01
Second Order & Zero-Order Rate Laws

26m 40s

Intro
0:00
Kinetics
0:22
Second Order
0:23
Example 1
6:08
Zero-Order
16:36
Summary for the Kinetics Associated with the Reaction
21:27
Activation Energy & Arrhenius Equation

40m 59s

Intro
0:00
Kinetics
0:53
Rate Constant
0:54
Collision Model
2:45
Activation Energy
5:11
Arrhenius Proposed
9:54
2 Requirements for a Successful Reaction
15:39
Rate Constant
17:53
Arrhenius Equation
19:51
Example 1
25:00
Activation Energy & the Values of K
32:12
Example 2
36:46
AP Practice for Kinetics

29m 8s

Intro
0:00
Kinetics
0:43
Example 1
0:44
Example 2
6:53
Example 3
8:58
Example 4
11:36
Example 5
16:36
Example 6: Part A
21:00
Example 6: Part B
25:09
Section 6: Equilibrium
Equilibrium, Part 1

46m

Intro
0:00
Equilibrium
1:32
Introduction to Equilibrium
1:33
Equilibrium Rules
14:00
Example 1: Part A
16:46
Example 1: Part B
18:48
Example 1: Part C
22:13
Example 1: Part D
24:55
Example 2: Part A
27:46
Example 2: Part B
31:22
Example 2: Part C
33:00
Reverse a Reaction
36:04
Example 3
37:24
Equilibrium, Part 2

40m 53s

Intro
0:00
Equilibrium
1:31
Equilibriums Involving Gases
1:32
General Equation
10:11
Example 1: Question
11:55
Example 1: Answer
13:43
Example 2: Question
19:08
Example 2: Answer
21:37
Example 3: Question
33:40
Example 3: Answer
35:24
Equilibrium: Reaction Quotient

45m 53s

Intro
0:00
Equilibrium
0:57
Reaction Quotient
0:58
If Q > K
5:37
If Q < K
6:52
If Q = K
7:45
Example 1: Part A
8:24
Example 1: Part B
13:11
Example 2: Question
20:04
Example 2: Answer
22:15
Example 3: Question
30:54
Example 3: Answer
32:52
Steps in Solving Equilibrium Problems
42:40
Equilibrium: Examples

31m 51s

Intro
0:00
Equilibrium
1:09
Example 1: Question
1:10
Example 1: Answer
4:15
Example 2: Question
13:04
Example 2: Answer
15:20
Example 3: Question
25:03
Example 3: Answer
26:32
Le Chatelier's principle & Equilibrium

40m 52s

Intro
0:00
Le Chatelier
1:05
Le Chatelier Principle
1:06
Concentration: Add 'x'
5:25
Concentration: Subtract 'x'
7:50
Example 1
9:44
Change in Pressure
12:53
Example 2
20:40
Temperature: Exothermic and Endothermic
24:33
Example 3
29:55
Example 4
35:30
Section 7: Acids & Bases
Acids and Bases

50m 11s

Intro
0:00
Acids and Bases
1:14
Bronsted-Lowry Acid-Base Model
1:28
Reaction of an Acid with Water
4:36
Acid Dissociation
10:51
Acid Strength
13:48
Example 1
21:22
Water as an Acid & a Base
25:25
Example 2: Part A
32:30
Example 2: Part B
34:47
Example 3: Part A
35:58
Example 3: Part B
39:33
pH Scale
41:12
Example 4
43:56
pH of Weak Acid Solutions

43m 52s

Intro
0:00
pH of Weak Acid Solutions
1:12
pH of Weak Acid Solutions
1:13
Example 1
6:26
Example 2
14:25
Example 3
24:23
Example 4
30:38
Percent Dissociation: Strong & Weak Bases

43m 4s

Intro
0:00
Bases
0:33
Percent Dissociation: Strong & Weak Bases
0:45
Example 1
6:23
Strong Base Dissociation
11:24
Example 2
13:02
Weak Acid and General Reaction
17:38
Example: NaOH → Na⁺ + OH⁻
20:30
Strong Base and Weak Base
23:49
Example 4
24:54
Example 5
33:51
Polyprotic Acids

35m 34s

Intro
0:00
Polyprotic Acids
1:04
Acids Dissociation
1:05
Example 1
4:51
Example 2
17:30
Example 3
31:11
Salts and Their Acid-Base Properties

41m 14s

Intro
0:00
Salts and Their Acid-Base Properties
0:11
Salts and Their Acid-Base Properties
0:15
Example 1
7:58
Example 2
14:00
Metal Ion and Acidic Solution
22:00
Example 3
28:35
NH₄F → NH₄⁺ + F⁻
34:05
Example 4
38:03
Common Ion Effect & Buffers

41m 58s

Intro
0:00
Common Ion Effect & Buffers
1:16
Covalent Oxides Produce Acidic Solutions in Water
1:36
Ionic Oxides Produce Basic Solutions in Water
4:15
Practice Example 1
6:10
Practice Example 2
9:00
Definition
12:27
Example 1: Part A
16:49
Example 1: Part B
19:54
Buffer Solution
25:10
Example of Some Buffers: HF and NaF
30:02
Example of Some Buffers: Acetic Acid & Potassium Acetate
31:34
Example of Some Buffers: CH₃NH₂ & CH₃NH₃Cl
33:54
Example 2: Buffer Solution
36:36
Buffer

32m 24s

Intro
0:00
Buffers
1:20
Buffer Solution
1:21
Adding Base
5:03
Adding Acid
7:14
Example 1: Question
9:48
Example 1: Recall
12:08
Example 1: Major Species Upon Addition of NaOH
16:10
Example 1: Equilibrium, ICE Chart, and Final Calculation
24:33
Example 1: Comparison
29:19
Buffers, Part II

40m 6s

Intro
0:00
Buffers
1:27
Example 1: Question
1:32
Example 1: ICE Chart
3:15
Example 1: Major Species Upon Addition of OH⁻, But Before Rxn
7:23
Example 1: Equilibrium, ICE Chart, and Final Calculation
12:51
Summary
17:21
Another Look at Buffering & the Henderson-Hasselbalch equation
19:00
Example 2
27:08
Example 3
32:01
Buffers, Part III

38m 43s

Intro
0:00
Buffers
0:25
Buffer Capacity Part 1
0:26
Example 1
4:10
Buffer Capacity Part 2
19:29
Example 2
25:12
Example 3
32:02
Titrations: Strong Acid and Strong Base

42m 42s

Intro
0:00
Titrations: Strong Acid and Strong Base
1:11
Definition of Titration
1:12
Sample Problem
3:33
Definition of Titration Curve or pH Curve
9:46
Scenario 1: Strong Acid- Strong Base Titration
11:00
Question
11:01
Part 1: No NaOH is Added
14:00
Part 2: 10.0 mL of NaOH is Added
15:50
Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added
22:19
Part 4: 50.0 mL of NaOH is Added
26:46
Part 5: 100.0 mL (Total) of NaOH is Added
27:26
Part 6: 150.0 mL (Total) of NaOH is Added
32:06
Part 7: 200.0 mL of NaOH is Added
35:07
Titrations Curve for Strong Acid and Strong Base
35:43
Titrations: Weak Acid and Strong Base

42m 3s

Intro
0:00
Titrations: Weak Acid and Strong Base
0:43
Question
0:44
Part 1: No NaOH is Added
1:54
Part 2: 10.0 mL of NaOH is Added
5:17
Part 3: 25.0 mL of NaOH is Added
14:01
Part 4: 40.0 mL of NaOH is Added
21:55
Part 5: 50.0 mL (Total) of NaOH is Added
22:25
Part 6: 60.0 mL (Total) of NaOH is Added
31:36
Part 7: 75.0 mL (Total) of NaOH is Added
35:44
Titration Curve
36:09
Titration Examples & Acid-Base Indicators

52m 3s

Intro
0:00
Examples and Indicators
0:25
Example 1: Question
0:26
Example 1: Solution
2:03
Example 2: Question
12:33
Example 2: Solution
14:52
Example 3: Question
23:45
Example 3: Solution
25:09
Acid/Base Indicator Overview
34:45
Acid/Base Indicator Example
37:40
Acid/Base Indicator General Result
47:11
Choosing Acid/Base Indicator
49:12
Section 8: Solubility
Solubility Equilibria

36m 25s

Intro
0:00
Solubility Equilibria
0:48
Solubility Equilibria Overview
0:49
Solubility Product Constant
4:24
Definition of Solubility
9:10
Definition of Solubility Product
11:28
Example 1
14:09
Example 2
20:19
Example 3
27:30
Relative Solubilities
31:04
Solubility Equilibria, Part II

42m 6s

Intro
0:00
Solubility Equilibria
0:46
Common Ion Effect
0:47
Example 1
3:14
pH & Solubility
13:00
Example of pH & Solubility
15:25
Example 2
23:06
Precipitation & Definition of the Ion Product
26:48
If Q > Ksp
29:31
If Q < Ksp
30:27
Example 3
32:58
Solubility Equilibria, Part III

43m 9s

Intro
0:00
Solubility Equilibria
0:55
Example 1: Question
0:56
Example 1: Step 1 - Check to See if Anything Precipitates
2:52
Example 1: Step 2 - Stoichiometry
10:47
Example 1: Step 3 - Equilibrium
16:34
Example 2: Selective Precipitation (Question)
21:02
Example 2: Solution
23:41
Classical Qualitative Analysis
29:44
Groups: 1-5
38:44
Section 9: Complex Ions
Complex Ion Equilibria

43m 38s

Intro
0:00
Complex Ion Equilibria
0:32
Complex Ion
0:34
Ligan Examples
1:51
Ligand Definition
3:12
Coordination
6:28
Example 1
8:08
Example 2
19:13
Complex Ions & Solubility

31m 30s

Intro
0:00
Complex Ions and Solubility
0:23
Recall: Classical Qualitative Analysis
0:24
Example 1
6:10
Example 2
16:16
Dissolving a Water-Insoluble Ionic Compound: Method 1
23:38
Dissolving a Water-Insoluble Ionic Compound: Method 2
28:13
Section 10: Chemical Thermodynamics
Spontaneity, Entropy, & Free Energy, Part I

56m 28s

Intro
0:00
Spontaneity, Entropy, Free Energy
2:25
Energy Overview
2:26
Equation: ∆E = q + w
4:30
State Function/ State Property
8:35
Equation: w = -P∆V
12:00
Enthalpy: H = E + PV
14:50
Enthalpy is a State Property
17:33
Exothermic and Endothermic Reactions
19:20
First Law of Thermodynamic
22:28
Entropy
25:48
Spontaneous Process
33:53
Second Law of Thermodynamic
36:51
More on Entropy
42:23
Example
43:55
Spontaneity, Entropy, & Free Energy, Part II

39m 55s

Intro
0:00
Spontaneity, Entropy, Free Energy
1:30
∆S of Universe = ∆S of System + ∆S of Surrounding
1:31
Convention
3:32
Examining a System
5:36
Thermodynamic Property: Sign of ∆S
16:52
Thermodynamic Property: Magnitude of ∆S
18:45
Deriving Equation: ∆S of Surrounding = -∆H / T
20:25
Example 1
25:51
Free Energy Equations
29:22
Spontaneity, Entropy, & Free Energy, Part III

30m 10s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:11
Example 1
2:38
Key Concept of Example 1
14:06
Example 2
15:56
Units for ∆H, ∆G, and S
20:56
∆S of Surrounding & ∆S of System
22:00
Reaction Example
24:17
Example 3
26:52
Spontaneity, Entropy, & Free Energy, Part IV

30m 7s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:29
Standard Free Energy of Formation
0:58
Example 1
4:34
Reaction Under Non-standard Conditions
13:23
Example 2
16:26
∆G = Negative
22:12
∆G = 0
24:38
Diagram Example of ∆G
26:43
Spontaneity, Entropy, & Free Energy, Part V

44m 56s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:56
Equations: ∆G of Reaction, ∆G°, and K
0:57
Example 1: Question
6:50
Example 1: Part A
9:49
Example 1: Part B
15:28
Example 2
17:33
Example 3
23:31
lnK = (- ∆H° ÷ R) ( 1 ÷ T) + ( ∆S° ÷ R)
31:36
Maximum Work
35:57
Section 11: Electrochemistry
Oxidation-Reduction & Balancing

39m 23s

Intro
0:00
Oxidation-Reduction and Balancing
2:06
Definition of Electrochemistry
2:07
Oxidation and Reduction Review
3:05
Example 1: Assigning Oxidation State
10:15
Example 2: Is the Following a Redox Reaction?
18:06
Example 3: Step 1 - Write the Oxidation & Reduction Half Reactions
22:46
Example 3: Step 2 - Balance the Reaction
26:44
Example 3: Step 3 - Multiply
30:11
Example 3: Step 4 - Add
32:07
Example 3: Step 5 - Check
33:29
Galvanic Cells

43m 9s

Intro
0:00
Galvanic Cells
0:39
Example 1: Balance the Following Under Basic Conditions
0:40
Example 1: Steps to Balance Reaction Under Basic Conditions
3:25
Example 1: Solution
5:23
Example 2: Balance the Following Reaction
13:56
Galvanic Cells
18:15
Example 3: Galvanic Cells
28:19
Example 4: Galvanic Cells
35:12
Cell Potential

48m 41s

Intro
0:00
Cell Potential
2:08
Definition of Cell Potential
2:17
Symbol and Unit
5:50
Standard Reduction Potential
10:16
Example Figure 1
13:08
Example Figure 2
19:00
All Reduction Potentials are Written as Reduction
23:10
Cell Potential: Important Fact 1
26:49
Cell Potential: Important Fact 2
27:32
Cell Potential: Important Fact 3
28:54
Cell Potential: Important Fact 4
30:05
Example Problem 1
32:29
Example Problem 2
38:38
Potential, Work, & Free Energy

41m 23s

Intro
0:00
Potential, Work, Free Energy
0:42
Descriptions of Galvanic Cell
0:43
Line Notation
5:33
Example 1
6:26
Example 2
11:15
Example 3
15:18
Equation: Volt
22:20
Equations: Cell Potential, Work, and Charge
28:30
Maximum Cell Potential is Related to the Free Energy of the Cell Reaction
35:09
Example 4
37:42
Cell Potential & Concentration

34m 19s

Intro
0:00
Cell Potential & Concentration
0:29
Example 1: Question
0:30
Example 1: Nernst Equation
4:43
Example 1: Solution
7:01
Cell Potential & Concentration
11:27
Example 2
16:38
Manipulating the Nernst Equation
25:15
Example 3
28:43
Electrolysis

33m 21s

Intro
0:00
Electrolysis
3:16
Electrolysis: Part 1
3:17
Electrolysis: Part 2
5:25
Galvanic Cell Example
7:13
Nickel Cadmium Battery
12:18
Ampere
16:00
Example 1
20:47
Example 2
25:47
Section 12: Light
Light

44m 45s

Intro
0:00
Light
2:14
Introduction to Light
2:15
Frequency, Speed, and Wavelength of Waves
3:58
Units and Equations
7:37
Electromagnetic Spectrum
12:13
Example 1: Calculate the Frequency
17:41
E = hν
21:30
Example 2: Increment of Energy
25:12
Photon Energy of Light
28:56
Wave and Particle
31:46
Example 3: Wavelength of an Electron
34:46
Section 13: Quantum Mechanics
Quantum Mechanics & Electron Orbitals

54m

Intro
0:00
Quantum Mechanics & Electron Orbitals
0:51
Quantum Mechanics & Electron Orbitals Overview
0:52
Electron Orbital and Energy Levels for the Hydrogen Atom
8:47
Example 1
13:41
Quantum Mechanics: Schrodinger Equation
19:19
Quantum Numbers Overview
31:10
Principal Quantum Numbers
33:28
Angular Momentum Numbers
34:55
Magnetic Quantum Numbers
36:35
Spin Quantum Numbers
37:46
Primary Level, Sublevels, and Sub-Sub-Levels
39:42
Example
42:17
Orbital & Quantum Numbers
49:32
Electron Configurations & Diagrams

34m 4s

Intro
0:00
Electron Configurations & Diagrams
1:08
Electronic Structure of Ground State Atom
1:09
Order of Electron Filling
3:50
Electron Configurations & Diagrams: H
8:41
Electron Configurations & Diagrams: He
9:12
Electron Configurations & Diagrams: Li
9:47
Electron Configurations & Diagrams: Be
11:17
Electron Configurations & Diagrams: B
12:05
Electron Configurations & Diagrams: C
13:03
Electron Configurations & Diagrams: N
14:55
Electron Configurations & Diagrams: O
15:24
Electron Configurations & Diagrams: F
16:25
Electron Configurations & Diagrams: Ne
17:00
Electron Configurations & Diagrams: S
18:08
Electron Configurations & Diagrams: Fe
20:08
Introduction to Valence Electrons
23:04
Valence Electrons of Oxygen
23:44
Valence Electrons of Iron
24:02
Valence Electrons of Arsenic
24:30
Valence Electrons: Exceptions
25:36
The Periodic Table
27:52
Section 14: Intermolecular Forces
Vapor Pressure & Changes of State

52m 43s

Intro
0:00
Vapor Pressure and Changes of State
2:26
Intermolecular Forces Overview
2:27
Hydrogen Bonding
5:23
Heat of Vaporization
9:58
Vapor Pressure: Definition and Example
11:04
Vapor Pressures is Mostly a Function of Intermolecular Forces
17:41
Vapor Pressure Increases with Temperature
20:52
Vapor Pressure vs. Temperature: Graph and Equation
22:55
Clausius-Clapeyron Equation
31:55
Example 1
32:13
Heating Curve
35:40
Heat of Fusion
41:31
Example 2
43:45
Phase Diagrams & Solutions

31m 17s

Intro
0:00
Phase Diagrams and Solutions
0:22
Definition of a Phase Diagram
0:50
Phase Diagram Part 1: H₂O
1:54
Phase Diagram Part 2: CO₂
9:59
Solutions: Solute & Solvent
16:12
Ways of Discussing Solution Composition: Mass Percent or Weight Percent
18:46
Ways of Discussing Solution Composition: Molarity
20:07
Ways of Discussing Solution Composition: Mole Fraction
20:48
Ways of Discussing Solution Composition: Molality
21:41
Example 1: Question
22:06
Example 1: Mass Percent
24:32
Example 1: Molarity
25:53
Example 1: Mole Fraction
28:09
Example 1: Molality
29:36
Vapor Pressure of Solutions

37m 23s

Intro
0:00
Vapor Pressure of Solutions
2:07
Vapor Pressure & Raoult's Law
2:08
Example 1
5:21
When Ionic Compounds Dissolve
10:51
Example 2
12:38
Non-Ideal Solutions
17:42
Negative Deviation
24:23
Positive Deviation
29:19
Example 3
31:40
Colligatives Properties

34m 11s

Intro
0:00
Colligative Properties
1:07
Boiling Point Elevation
1:08
Example 1: Question
5:19
Example 1: Solution
6:52
Freezing Point Depression
12:01
Example 2: Question
14:46
Example 2: Solution
16:34
Osmotic Pressure
20:20
Example 3: Question
28:00
Example 3: Solution
30:16
Section 15: Bonding
Bonding & Lewis Structure

48m 39s

Intro
0:00
Bonding & Lewis Structure
2:23
Covalent Bond
2:24
Single Bond, Double Bond, and Triple Bond
4:11
Bond Length & Intermolecular Distance
5:51
Definition of Electronegativity
8:42
Bond Polarity
11:48
Bond Energy
20:04
Example 1
24:31
Definition of Lewis Structure
31:54
Steps in Forming a Lewis Structure
33:26
Lewis Structure Example: H₂
36:53
Lewis Structure Example: CH₄
37:33
Lewis Structure Example: NO⁺
38:43
Lewis Structure Example: PCl₅
41:12
Lewis Structure Example: ICl₄⁻
43:05
Lewis Structure Example: BeCl₂
45:07
Resonance & Formal Charge

36m 59s

Intro
0:00
Resonance and Formal Charge
0:09
Resonance Structures of NO₃⁻
0:25
Resonance Structures of NO₂⁻
12:28
Resonance Structures of HCO₂⁻
16:28
Formal Charge
19:40
Formal Charge Example: SO₄²⁻
21:32
Formal Charge Example: CO₂
31:33
Formal Charge Example: HCN
32:44
Formal Charge Example: CN⁻
33:34
Formal Charge Example: 0₃
34:43
Shapes of Molecules

41m 21s

Intro
0:00
Shapes of Molecules
0:35
VSEPR
0:36
Steps in Determining Shapes of Molecules
6:18
Linear
11:38
Trigonal Planar
11:55
Tetrahedral
12:45
Trigonal Bipyramidal
13:23
Octahedral
14:29
Table: Shapes of Molecules
15:40
Example: CO₂
21:11
Example: NO₃⁻
24:01
Example: H₂O
27:00
Example: NH₃
29:48
Example: PCl₃⁻
32:18
Example: IF₄⁺
34:38
Example: KrF₄
37:57
Hybrid Orbitals

40m 17s

Intro
0:00
Hybrid Orbitals
0:13
Introduction to Hybrid Orbitals
0:14
Electron Orbitals for CH₄
5:02
sp³ Hybridization
10:52
Example: sp³ Hybridization
12:06
sp² Hybridization
14:21
Example: sp² Hybridization
16:11
σ Bond
19:10
π Bond
20:07
sp Hybridization & Example
22:00
dsp³ Hybridization & Example
27:36
d²sp³ Hybridization & Example
30:36
Example: Predict the Hybridization and Describe the Molecular Geometry of CO
32:31
Example: Predict the Hybridization and Describe the Molecular Geometry of BF₄⁻
35:17
Example: Predict the Hybridization and Describe the Molecular Geometry of XeF₂
37:09
Section 16: AP Practice Exam
AP Practice Exam: Multiple Choice, Part I

52m 34s

Intro
0:00
Multiple Choice
1:21
Multiple Choice 1
1:22
Multiple Choice 2
2:23
Multiple Choice 3
3:38
Multiple Choice 4
4:34
Multiple Choice 5
5:16
Multiple Choice 6
5:41
Multiple Choice 7
6:20
Multiple Choice 8
7:03
Multiple Choice 9
7:31
Multiple Choice 10
9:03
Multiple Choice 11
11:52
Multiple Choice 12
13:16
Multiple Choice 13
13:56
Multiple Choice 14
14:52
Multiple Choice 15
15:43
Multiple Choice 16
16:20
Multiple Choice 17
16:55
Multiple Choice 18
17:22
Multiple Choice 19
18:59
Multiple Choice 20
20:24
Multiple Choice 21
22:20
Multiple Choice 22
23:29
Multiple Choice 23
24:30
Multiple Choice 24
25:24
Multiple Choice 25
26:21
Multiple Choice 26
29:06
Multiple Choice 27
30:42
Multiple Choice 28
33:28
Multiple Choice 29
34:38
Multiple Choice 30
35:37
Multiple Choice 31
37:31
Multiple Choice 32
38:28
Multiple Choice 33
39:50
Multiple Choice 34
42:57
Multiple Choice 35
44:18
Multiple Choice 36
45:52
Multiple Choice 37
48:02
Multiple Choice 38
49:25
Multiple Choice 39
49:43
Multiple Choice 40
50:16
Multiple Choice 41
50:49
AP Practice Exam: Multiple Choice, Part II

32m 15s

Intro
0:00
Multiple Choice
0:12
Multiple Choice 42
0:13
Multiple Choice 43
0:33
Multiple Choice 44
1:16
Multiple Choice 45
2:36
Multiple Choice 46
5:22
Multiple Choice 47
6:35
Multiple Choice 48
8:02
Multiple Choice 49
10:05
Multiple Choice 50
10:26
Multiple Choice 51
11:07
Multiple Choice 52
12:01
Multiple Choice 53
12:55
Multiple Choice 54
16:12
Multiple Choice 55
18:11
Multiple Choice 56
19:45
Multiple Choice 57
20:15
Multiple Choice 58
23:28
Multiple Choice 59
24:27
Multiple Choice 60
26:45
Multiple Choice 61
29:15
AP Practice Exam: Multiple Choice, Part III

32m 50s

Intro
0:00
Multiple Choice
0:16
Multiple Choice 62
0:17
Multiple Choice 63
1:57
Multiple Choice 64
6:16
Multiple Choice 65
8:05
Multiple Choice 66
9:18
Multiple Choice 67
10:38
Multiple Choice 68
12:51
Multiple Choice 69
14:32
Multiple Choice 70
17:35
Multiple Choice 71
22:44
Multiple Choice 72
24:27
Multiple Choice 73
27:46
Multiple Choice 74
29:39
Multiple Choice 75
30:23
AP Practice Exam: Free response Part I

47m 22s

Intro
0:00
Free Response
0:15
Free Response 1: Part A
0:16
Free Response 1: Part B
4:15
Free Response 1: Part C
5:47
Free Response 1: Part D
9:20
Free Response 1: Part E. i
10:58
Free Response 1: Part E. ii
16:45
Free Response 1: Part E. iii
26:03
Free Response 2: Part A. i
31:01
Free Response 2: Part A. ii
33:38
Free Response 2: Part A. iii
35:20
Free Response 2: Part B. i
37:38
Free Response 2: Part B. ii
39:30
Free Response 2: Part B. iii
44:44
AP Practice Exam: Free Response Part II

43m 5s

Intro
0:00
Free Response
0:12
Free Response 3: Part A
0:13
Free Response 3: Part B
6:25
Free Response 3: Part C. i
11:33
Free Response 3: Part C. ii
12:02
Free Response 3: Part D
14:30
Free Response 4: Part A
21:03
Free Response 4: Part B
22:59
Free Response 4: Part C
24:33
Free Response 4: Part D
27:22
Free Response 4: Part E
28:43
Free Response 4: Part F
29:35
Free Response 4: Part G
30:15
Free Response 4: Part H
30:48
Free Response 5: Diagram
32:00
Free Response 5: Part A
34:14
Free Response 5: Part B
36:07
Free Response 5: Part C
37:45
Free Response 5: Part D
39:00
Free Response 5: Part E
40:26
AP Practice Exam: Free Response Part III

28m 36s

Intro
0:00
Free Response
0:43
Free Response 6: Part A. i
0:44
Free Response 6: Part A. ii
3:08
Free Response 6: Part A. iii
5:02
Free Response 6: Part B. i
7:11
Free Response 6: Part B. ii
9:40
Free Response 7: Part A
11:14
Free Response 7: Part B
13:45
Free Response 7: Part C
15:43
Free Response 7: Part D
16:54
Free Response 8: Part A. i
19:15
Free Response 8: Part A. ii
21:16
Free Response 8: Part B. i
23:51
Free Response 8: Part B. ii
25:07
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Lecture Comments (14)

1 answer

Last reply by: Professor Hovasapian
Sun May 19, 2019 11:23 PM

Post by Radhika Narala on May 12, 2019

hello,
do you happen to have recent ap multiple choice chem tests after  2014

1 answer

Last reply by: Professor Hovasapian
Fri Apr 7, 2017 6:56 PM

Post by Iwin Yeung on March 10, 2017

Can you do videos for the other practice exams from collegeboard?

1 answer

Last reply by: Professor Hovasapian
Thu Apr 7, 2016 2:15 AM

Post by Nihal Celik on April 5, 2016

Where can I get a visual copy of this practice exam?

1 answer

Last reply by: Professor Hovasapian
Sat Aug 15, 2015 12:58 AM

Post by Jim Tang on August 12, 2015

Where is reaction mechanism in this series?

1 answer

Last reply by: Professor Hovasapian
Sun Aug 24, 2014 11:41 PM

Post by John K on August 24, 2014

Professor is this AP course + the practice test advisable for taking SAT subject test?

2 answers

Last reply by: Quazi Niloy
Fri Jul 4, 2014 1:05 PM

Post by Quazi Niloy on June 23, 2014

Professor does this practice test meet the requirements and standards of the current AP Chemistry exam?

0 answers

Post by Professor Hovasapian on July 18, 2012

Link to the AP Practice Exam:

http://apcentral.collegeboard.com/apc/public/repository/chemistry-released-exam-1999.pdf

Take good Care

Raffi

AP Practice Exam: Multiple Choice, Part I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Multiple Choice 1:21
    • Multiple Choice 1
    • Multiple Choice 2
    • Multiple Choice 3
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    • Multiple Choice 33
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    • Multiple Choice 35
    • Multiple Choice 36
    • Multiple Choice 37
    • Multiple Choice 38
    • Multiple Choice 39
    • Multiple Choice 40
    • Multiple Choice 41

Transcription: AP Practice Exam: Multiple Choice, Part I

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we are going to begin actually going through an AP exam completely.0004

We are going to start with the multiple-choice section.0010

Before I launch into answering the questions on the multiple-choice section, I want to talk to you, just briefly, about it.0012

Now, you are going to have about 90 minutes to do 75 multiple-choice questions; you are going to get 1 point for every one that you get right, and you are going to be subtracted a quarter point for the ones that you miss.0020

Now, you should know that you could actually get a raw score (which is the number you get correct, minus the number you miss) of about 50; so you could literally skip one-third of the multiple choice questions, and still end up with a 5.0033

You can answer about 35 or 40 only of the multiple choice questions and still end up with a 4; so don't think like you have to answer absolutely every single one.0048

If you are not exactly sure about one, move on: chances are that your score will still be sufficiently high to keep you in the 4-5 range.0058

Having said that, let's just go ahead and get started.0067

We are going to be going through the 1999 AP practice exam; so hopefully, you have it in front of you; I have it in front of me here.0071

We are just going to run through it; OK.0079

The first question gives you a choice of activation energy, free energy, ionization energy, kinetic energy, and lattice energy: so those are your answer choices, and we are going to...the questions that follow actually ask you to identify which energy is being talked about.0082

#1: The energy required to convert a ground-state atom in the gas phase to a gaseous positive ion: so, this one is the ionization energy, so this one is C.0102

If you remember the ionization energy, it is if you have some ground-state gaseous atom; you hit it with a certain amount of energy, and you convert it to a gaseous phase ion.0118

You knock off one electron, and it is written like this (this is the first ionization energy--the second ionization energy, and so on, will just be this thing when you knock off another electron).0129

That is it; #1 is C.0141

OK, #2: The energy change that occurs in the conversion of an ionic solid to a widely-separated gaseous ion; OK, now this is one of the things that, in our particular course, we did not discuss formally: this is called the lattice energy.0143

The lattice energy is basically the energy that is keeping an ionic compound together.0160

When you have, let's say, sodium and chlorine, they come together; the chlorines separate; they each steal an electron from the sodium, and then you have a chloride ion, which is negatively charged, and sodium, which is positively charged.0167

They combine, because positive and negative attract; when they combine, they actually are more energetically stable--remember, we said all systems seek the lowest energy.0180

Ionic compounds, because of the positive and negative charge, release energy; so that is the lattice energy--it's the amount of energy that is necessary to keep that together.0192

In order to break that apart--to separate them as far away as possible--to separate sodium ion from chloride ion--that is called the lattice energy, because these ionic compounds are composed of a crystal lattice (positive/negative, positive/negative, arranged in a particular arrangement).0201

OK, so #3: The energy in a chemical or a physical change that is available to do useful work: OK, that is the important word here, "useful work": that is free energy.0219

OK, that is B; and you remember free energy: the reason we called it free energy--it is energy available to actually do work, not random energy, which would be heat.0231

OK, so these are reasonably straightforward; and again, this is how it's going to be on the multiple choice exam.0243

I would say a good 75% of the multiple choice questions are things that you either know or don't know, and can answer pretty quickly.0248

There are some calculations; now, of course, you can't use a calculator on this part of the exam--all you are given is a periodic table--but that is all you need.0257

Any calculations that you do need to make are actually really, really simple numerical calculations--things like .5, .25, 2, 4--numbers that you can multiply pretty quickly.0264

OK, #4: The energy required to form the transition state in a chemical reaction: so you remember, when we go from reactants to products, we have to pass over a transition state: that is called the activation energy.0275

Activation energy is A, so 4 is A.0287

OK, let's see--let's move on.0292

#5 gives you a series of electron diagrams (you know, the ones with the arrows up and down); it gives you a choice of 5 of them, and now we are going to ask you some questions about which one is which.0299

We have to identify.0314

#5 says it represents an atom that is chemically unreactive; well, if you take a look at the choices..."unreactive"--generally, you are looking at a noble gas configuration; you are looking for the s and the p to be completely filled, so s2, p6.0317

The only one in there that is s2, p6, is D: 1s2, 2s2, 2p6.0332

The answer to #5 is D.0338

OK, "represents an atom in an excited state": so, in an excited state means that an atom...the normal electron configurations, when we do them--they all go into the lowest energy; we go from the lowest energy, and we work our way up.0342

An excited state means an atom has been kicked up from its ground state energy level (whatever that may be) to a higher level; so it is going to have orbitals lower than it that may not be filled.0355

Well, the only one that satisfies that one is A: you have the 1s, which is empty; you have the 2s with the one electron in it; that is an excited state atom.0366

#6 is A.0375

#7: "represents an atom that has 4 valence electrons": OK, 4 valence electrons--remember, the valence electrons are all of the electrons in the highest primary shell.0381

In here, we have...let's see: 2s, no; B is 2s2, 2p4...no, I'm sorry, 2s2, 2p2; there you go.0393

C is the answer here, and the reason is because the valence shell is the primary electronic level 2: 2s, 2p: there are 2 in the s orbitals; there are 2 in the p; that gives you a total of 4 valence electrons; so #7 is C.0405

OK, #8: "represents an atom of a transition metal": well, a transition metal is the one where the d orbitals are actually filled--in the periodic table, that middle section with the 10 columns.0423

Those are the transition metals; so here, you are looking for something that has filled d orbitals, and that is going to be E.0436

#8 is E; OK.0445

Let's see: so now, questions 9 through 12 refer to an aqueous solution containing a 1:1 mole ratio of the following pairs of substances; assume all concentrations are 1 Molar.0452

So, we have ammonia-ammonium chloride; phosphoric acid and sodium dihydrogen phosphate; hydrochloric acid and NaCl; NaOH and NH3; and NH3 and acetic acid.0465

Those are our pairs of...a solution with a pair of those species; those are our choices.0479

And again, they said they are in a 1:1 mole ratio.0485

So, #9: we are looking for the solution with the lowest pH; OK.0488

Now, lowest pH means most acidic.0494

OK, now we look over here: as far as the most acidic is concerned, the only thing that I see that is the strongest acid is HCl.0507

And then, we look at NaCl; well, the sodium chloride, when they actually break up in solution into sodium and chloride--they are going to be spectator ions; they don't actually do anything.0518

They are not going to act as bases; they are not going to affect the equilibrium concentration of the hydrogen ion; so, in this case, the answer is C.0529

Let me put a little box around that: so lowest pH=most acidic.0539

OK, #10: the most nearly neutral solution: OK, in this particular case, the most nearly neutral solution is going to be E, the NH3 and the acetic acid.0544

All right, you might think that it...well, let's just sort of run through the possibilities.0561

It is not going to be the NaOH and the NH3, because they are both bases; it is not going to be C, the HCl and the NaCl, because that is acidic; the H3PO4 is a weak acid, and the sodium dihydrogen phosphate--that is going to dissociate.0568

H2PO4- is also a weak acid, so you are probably looking at an acidic solution there.0586

NH3 and NH4Cl: well, this one--you might think that it is A, but as it turns out, let's take a look at why it's not A.0593

Now, if I take NH3 in solution, I'm going to get something like this: the equilibrium for this involves NH4+ plus OH, the standard dissociation (or association) equilibrium of a weak base--in this case, NH3.0609

Well, we know (or we should know) that the Kb for this is a lot less than 1.0628

Well, the Kb for this is equal to the NH4+ concentration, times the OH- concentration, divided by the NH3 concentration.0637

But they said that the NH3 and the NH4Cl are in a 1:1 ratio, so even though there is going to be some dissociation, the Kb is really, really small, so it's not going to be that much.0654

So, for all practical purposes, the ammonium concentration and the ammonia concentration are going to be the same, so what you are left with is: the concentration of hydroxide is actually going to be equal to the Kb (which is going to be somewhere on the order of 10-5 and 10-6).0664

When you take the negative log of that, you are going to get the pOH.0681

Well, the pOH is not going to be anywhere near neutral; but E, where you have ammonia (which is a weak base) and acetic acid (which is a weak acid), they are in a 1:1 molar ratio, so that one actually will end up neutralizing completely and giving you a pH close to about 7.0685

So, E is the best answer for #10.0707

OK, let's see: #11: A buffer at a pH of greater than 8...so we are looking for a buffer that is basic.0711

Let's write down here (remember what a buffer is?): A buffer is a weak acid plus its conjugate base, or it is a weak base plus its conjugate acid.0730

Well, when you have a weak acid and its conjugate base, you are going to get a solution that is acidic.0760

That is going to be a pH less than 7; we don't know where--I mean, we can adjust the pH any way we want; that is the whole idea of a buffer--you choose the pH beforehand, and then you create the solution.0765

If this is going to be...but they want the buffer with a pH that is going to be greater than 8, so you are looking for a weak base and its conjugate acid.0775

As it turns out, A satisfies that; so 11 is A: the weak base is ammonia; its conjugate acid is ammonium; the chloride is irrelevant.0784

OK, #12: A buffer at a pH less than 6: so now, this is the one--this is the buffer--that is going to be acidic.0796

So here, we are looking for a weak acid, and we are looking for its conjugate base.0807

As it turns out, the only one that satisfies that is B.0814

Phosphoric acid is a weak acid; its conjugate base...remember how we get a conjugate base--we just pull off one of the hydrogen ions; it is going to be the H2PO4-, and that is your conjugate base.0819

B is the combination.0831

OK, let's see: let's move on to questions 13 through 16.0833

OK, so questions 13 through 16 refer to the following descriptions of bonding in different types of solids.0844

The key word here is "solids."0850

OK, well, let's just run through it; so #13: cesium chloride--cesium chloride is an ionic compound; it is made of cesium cations and chloride anions; and that is going to be A.0854

It is a lattice of positive and negative ions held together by electrostatic forces.0875

So, it will always be that for ionic compounds: bonding in an ionic solid is actually pretty easy--it's just positive and negative charges stuck together--so no worries there.0880

#14: Gold--OK, so gold is a metal, and it consists of closely-packed lattice with delocalized electrons throughout.0892

#14 is going to be B; so what we mean when we say this: one of the things that makes metals such good conductors of electricity is precisely because the electrons are delocalized.0903

So, if I have a bunch of gold atoms--a piece of gold--the electrons aren't necessarily attached to any one gold atom; they are delocalized.0917

They are actually free to move: that is why metals conduct electricity, because the electrons can actually move through the sample (or over the surface of the sample, anyway).0928

When you see delocalized electrons, that represents metallic bonding.0938

OK, #09 carbon dioxide--all right, carbon dioxide--well, if you remember carbon dioxide...the Lewis structure for carbon dioxide is this.0944

Our choices would be: Strong multiple covalent bonds, including pi bonds (yes, we have pi bonds here; a double bond represents one pi bond), with weak intermolecular forces.0958

Well, CO2 is a nonpolar molecule (right?--the dipole moments--they actually cancel out: it is nonpolar), so this one would be D; that is the best choice.0968

16: We have (let me see) strong single covalent bonds with weak intermolecular forces.0983

The same thing: when you look at methane, if you remember the Lewis structure for methane, you have something that looks like a little tripod here, and the dipole moments of each individual CH bond do cancel out, and when they cancel out, you get a nonpolar molecule.0989

These are single bonds, so the best answer would be C.1007

OK, so questions 17 and 18 refer to the following elements: we have lithium, nickel, bromine, uranium, and fluorine.1014

#17 says: Is a gas in its standard state at 298 Kelvin (298 Kelvin is room temperature, just about 25 degrees Celsius).1025

That one is going to be fluorine.1033

17 is E, fluorine.1038

Now, "reacts with water to form a strong base": lithium, nickel, bromine, uranium, fluorine--it reacts with water.1043

The alkali metals (you just know this), when you put them in water, react to form a strong base; they react violently, which is why lithium, sodium, things like that actually can't be stored in water, because they will react.1053

Let me go ahead and put 18 over here; 18 is going to be lithium, which is A, and just so you know what the reaction actually looks like...it's actually the reason why they call them alkali metals, because they form hydroxide in solution.1068

If you take lithium metal, and if you add it to water (which I am going to write as HOH), what ends up happening is...well, literally, what ends up happening is (I'm going to draw 2 of these and 2 lithiums): a hydrogen atom--one of these hydrogen atoms--steals an electron from lithium; another hydrogen atom steals another electron from lithium.1085

They form hydrogen gas (they come together to form hydrogen gas); it leaves 2 hydroxides free, and therefore, what you end up with--the lithium + and the hydroxide minus--they come together; you end up with lithium hydroxide in solution, which is basic; that is what happens.1114

OK, so let's see: Starting with #19, each of the questions or incomplete statements below is followed by 5 suggested answers or completions; select the one that is best in each case, and then fill in the corresponding oval on the answer sheet.1137

OK, so which of the following best describes the role of the spark from the spark plug in an automobile engine?1159

OK, of these choices, C is the best: the spark plug supplies some of the energy of activation for the combustion reaction.1167

Combustion reactions are highly exothermic--they want to move forward; they are thermodynamically favorable.1176

However, just because you put some gas, like some gasoline, in a container with oxygen doesn't mean that the reaction is going to happen.1185

They have a very high activation energy; the spark gets it the activation energy.1196

If I put oxygen and hydrogen in a container and don't do anything, well, they want to form water really badly, but they can't get over the activation energy.1203

When I ignite it--when I just do a little spark--that is what makes the reaction go forward.1212

The answer here is C; 19 is C.1218

OK, what mass of Au is produced when .05 moles of Au2S3 is reduced completely with excess H2?1224

OK, so let's write out the reaction for this: so we have: Au2S3 is reduced completely with excess H2.1235

The mass of Au...so they say that gold is actually formed; so our reaction is going to be...gold is formed, and then it's going to be H2S, and then we are going to have to balance this equation out, so we have ourselves 3, 2, 3; I think it's going to be 3, 2, and 3 (3 H2, 3 S, 3 S, 2 Au); yes, that is our balanced equation.1247

OK, so they say we have 0.0500: 0.0500 moles of the Au2S is reduced completely, so we want to produce...let me see, so for every 1 mole of this, what mass of Au?1273

For every one mole of this, we produce 2 moles (right? for every 1 mole, we produce 2 moles) of gold.1295

.05 moles of this produces 0.100 moles of gold; well, now we do 0.100 mol (and again, these numbers are very, very easy), times 197.0 grams per mole (which, again--you will have a periodic table available for you, so you will have molar masses, and this is a really simple calculation): .1 times 197 is just 19.7 grams.1301

19.7 grams is the answer; that makes it B.1333

Good; OK, now, when a solution of sodium chloride is vaporized in a flame, the color of the flame is yellow.1339

This is just one of those things that you should know: when we do flame testing (we actually didn't discuss this; it is part of the qualitative analysis that we discussed briefly, when we talked about separating ions)...when you have a solution of ions, or if a solution that you have separated out--separated the separate ions--you need to sort of quickly test to make sure what ion is in there.1350

As it turns out, sodium ion burns yellow when you dip a piece of metal (a paper clip or something) in the solution and put that in a flame.1374

The flame will actually turn yellow, because the heat is exciting the electrons up to higher energy levels; they fall back down; and they emit light, yellow light.1386

Sodium burns yellow, lithium burns red, and potassium will burn blue; so just sort of know that.1396

And again, it is just one of those things that just sort of comes up.1404

OK, #22: Of the following reactions, which involve the largest decrease in entropy?1410

OK, so let's see: the largest decrease in entropy...so we are looking for something that goes to a smaller number of gas particles, or maybe from a gas to a liquid, a gas to a solid, liquid to solid...things like that--something that becomes more ordered.1420

A decrease in entropy means something that becomes more ordered.1441

Of these, it looks like E is the best choice, because you have three moles of oxygen gas that are all of a sudden contained in this lanthanum oxide, so you have a gas going to a solid; so yes, our best choice is E, for #22.1445

OK, let's see: #23: A hot air balloon, shown above, rises; which of the following is the best explanation for this observation?1467

OK, the best explanation for this observation: air in a balloon is surrounded on the outside by normal air; things rise when their density is less than the density of the surrounding medium.1484

So, for example, ice floats because ice is less dense than water.1504

Metal sinks in water because metal is more dense than the water.1507

So, in this case, E is the best choice: because the air density inside the balloon is less than that of the surrounding air.1511

That is it; so 23--we have E.1518

OK, #24: The safest and most effective emergency procedure to treat an acid splash on skin is to do which of the following immediately?1524

24 is going to be: Flush the affected area with water and then with dilute sodium bicarbonate solution.1533

That is going to be D; the reason that is, is: you flush it with water to get rid of as much of it as possible, but if there is any excess, you want to use bicarbonate, which is a very, very weak base, to compensate for the acid.1540

You don't want to use any strong base (like a sodium hydroxide solution or anything like that), because you run the risk of going over the top.1554

A strong base will also do damage the same way that an acid will; they actually will do the same damage, so you want to use a weak base like sodium bicarbonate (baking soda).1562

That is the best: wash it off; put a little baking soda on there--it will neutralize everything; then wash it off again.1573

So, D for #24.1578

OK, #15 The cooling curve for a pure substance as it changes from a liquid to a solid is shown above; the solid and the liquid coexist at...this one--the answer is C.1581

The answer is C because all points between Q and S...OK, this is a heating curve; so basically, what it is telling you is: anytime you see a straight line on a heating curve, that means it is making the transition.1601

By "a straight line," I mean a horizontal line, where the temperature is not changing; so you notice, the temperature is vertical; the time is the horizontal axis.1617

The temperature is dropping, dropping, dropping, dropping, dropping; all of a sudden it hits Q, jumps up to R, and then it stays horizontal from R to S.1625

That means there is a phase change; that means--and since we are doing a drop in temperature, that means the liquid is going to a solid.1633

And, of course, we know as much, because that is what the question asks.1642

Well, in the transition from a liquid to a solid, both the liquid and the solid coexist; until all of the liquid is converted to a solid, they coexist; so, certainly they coexist between R and S.1645

Well, this from-Q-to-R part--as it turns out, you can drop the temperature, drop the temperature, drop the temperature of a liquid; and if you drop it kind of quickly, the molecules don't have time to arrange themselves completely in the lattice necessary for the solid.1657

You can actually drop the temperature below its normal freezing point; it is called supercooling.1675

Well, as you drop to a supercooling temperature, at a certain point they start to rearrange; so between Q and R, you still have...that means the ice is starting to form.1680

As the ice starts to form, the temperature of that ice will actually start to rise back up until it reaches the temperature where it is the normal freezing point.1692

This little supercooling part--you will often see that in heating curves--at least experimentally, you will see it.1704

Under ideal conditions, you wouldn't see that supercooling, because we would go very slowly, very slowly; you would just see a drop--or from your perspective, a drop, a horizontal, and then another drop.1711

That horizontal, again, is where it is making the transition from liquid to solid, or from liquid to vapor, or from vapor to liquid.1721

OK, temperature doesn't change at that point--only when it is actually completely solid, completely liquid, or completely gas--that is when the temperature changes, depending on its heat capacity.1729

We did discuss heating curves.1739

So in this case, yes, C would be the best answer.1742

OK, let's see: When the equation above is balanced, and all the coefficients are reduced to their lowest whole-number terms, the coefficient of O2 is...1745

All right, so let's do this one: 26.1754

We have: C10H12O4S, plus O2; OK, we are going to form CO2; we are going to form H2O; and, if you have sulfur, sulfur is also going to oxidize.1759

It is going react with the oxygen, so you are going to end up with SO2.1777

OK, so let's go ahead and do the carbon first; so we'll stick a 10 in here for the carbon.1782

We'll do the hydrogen next, which is 12 hydrogen here; so we'll put a 6 there.1789

And then...now, let's see what we have: we have one S; we have one S; so now, let's count the oxygens: oxygens--we have 20 oxygens, plus 6 oxygens, plus 2 oxygens, for a total of 28.1794

We have 28 oxygens on the right; we have 2 here; we have 4 here.1812

We want 28; if we get 28-4, that is 24; how do we get 24 out of this?--we put a 12 here...and what do they actually ask for?1819

What is the coefficient of O2?--the coefficient of O2 is 12, so 26 is C.1830

There you go: just a straight balancing problem--nothing too strange.1838

OK, #27: let's see, appropriate uses of a visible spectra photometer include which of the following?1842

Determining the concentration of a solution of copper nitrate (copper (2) nitrate); measuring the conductivity of a solution of potassium permanganate; determining which ions are present in a solution that may contain sodium, magnesium, and aluminum ion...1855

A visible light spectra photometer is used for the determination of a concentration of a particular species, so it would be 1 only, so 27--it would be A.1871

OK, by the way, I should let you know so that...we didn't actually really talk about this, because I didn't discuss coordination chemistry in this particular AP Chemistry course.1886

Coordination chemistry usually comes at the end; some teachers talk about it; some don't.1899

If you see any questions about coordination chemistry, whether it is naming or something like this, it may be anywhere from maybe 1 to 3 questions...well, 1 question--really, you are not going to see more than 1 or 2.1904

So, the reason I didn't want to discuss it is because, again, it is one of those things that, running through a practice exam, you can actually get the little bit of information that you need regarding coordination chemistry.1915

I didn't want to spend an entire lesson or two lessons on it; it wasn't worth it, which is the same reason why I decided not to discuss organic chemistry.1927

Anything that comes up in organic chemistry is going to be 2 questions, at most; and again, if you miss them--if you don't know them--it is not going to hurt you in absolutely any way at all.1934

I didn't want to actually spend a whole bunch of lesson time on that; I wanted to concentrate on the things that were absolutely important and essential.1946

You should know: transition metals like copper--basically, what happens is: when you--if you--were to dissolve this copper nitrate in water, the copper nitrate dissolves, but what ends up happening is that the copper...the water molecules actually arrange themselves around the copper ion: maybe 4 of them, maybe 6 of them, maybe 2 of them.1955

Transition metals are notorious for absorbing visible light, which is why transition metal solutions actually are of different colors (green, blue, orange, pink, things like that--really, really beautiful colors; that is the reason why).1977

We can use a visible light spectra photometer to measure how much of the light they actually absorb, based on their color; and that tells us the concentration.1991

That is a little sideline explanation of what it is that is going on here.2000

OK, 27...let's see, 28: OK, the melting point of magnesium oxide is higher than that of sodium fluoride; explanations for this observation include which of the following?2005

Magnesium is more positively charged than sodium; O2- is more negatively charged than F-; the O2 ion is smaller than F- ion.2020

Well, when we are talking about melting point, we are talking about the charges between; so, if I have magnesium oxide and I want to melt it, that means I am trying to break the magnesium and oxygen bond, which is ionic.2033

Well, the stronger the bond, the more energy I have to put into it to break that bond.2048

So, as it turns out, here it is going to be B: 1 and 2: magnesium is more positively charged; oxygen is more negatively charged.2052

Magnesium is more positively charged than sodium; oxygen is more negatively charged than fluoride; more positive charge, more electrostatic force, stronger bond, higher melting point; so this is going to be B.2061

OK, 29: let's see, the organic compound represented above is an example of...that is a keytone.2079

Again, this is...so this is going to be E, and just to let you know, a keytone consists of a carbonyl group (which is this thing) connected to two...this carbon is connected to 2 carbons.2088

There can be more carbons here, but it is connected to at least two carbons.2107

This is called a carbonyl group, and this whole thing...when this carbonyl group is attached to two carbons, at the very least, it is called a keytone.2111

Again, organic--it is not going to hurt you; besides, those of you that are going to continue on--you are going to discuss organic chemistry, for those of you that are biology majors and chemistry majors.2122

Not a problem--no worries about that.2135

OK, #18 Which of the following is true regarding the reaction represented above?2139

These things...it looks like they are asking about oxidation numbers, mostly; so let's go ahead and calculate some oxidation numbers here.2144

We have H2Se + 4 O2F2 goes to SeF6 + 2 HF + 4 O2 (I hope I copied that correctly).2151

Let's just do some oxidation numbers here: hydrogen is going to be +1; selenium is going to be...there are 2 of them, so it's +2, so selenium here is -22170

O2F2...this is a very unusual compound; the only atom that is actually more electronegative than oxygen is fluorine, so in this case, oxygen will not have a -2 oxidation state.2181

The fluorine will have a -1 oxidation state; and because there are two of them, the total charge is going to be -2.2192

Well, it has to be balanced by a +2, because it's a neutral...that means oxygen has a charge of +1.2202

You don't see that too often.2211

Fluorine--again (oops, this is not sulfur; this is selenium; this is selenium fluoride), fluoride is -1; there are 6 of them, so a total charge of -6; that means selenium is a +6.2213

This is going to be -1, +1, and 0 on that; so in this particular case, it looks like it's going to be D: the oxidation number of selenium changes from -2 to +6, so our answer is D.2229

OK, let's see--31: If the temperature of an aqueous solution of sodium chloride is increased from 20 degrees to 90 degrees Celsius, which of the following statements is true?2247

The density of the solution remains unchanged--no; the molarity of the solution remains unchanged--no.2264

Density--as you heat something up, it's going to expand, so the density is going to decrease; the molarity of the solution remains...it's going to expand, which means the total volume is going to change, so molarity doesn't stay.2274

The molality of the solution remains unchanged--yes: the molality is moles of solute per kilograms of solvent; the mass of water doesn't change--its volume changes.2285

The mass of the solution (of the water) doesn't change; its volume changes, so C is the answer here.2295

Density changes, molarity changes...OK, that is 31.2305

Let's move on to #32: let's see, types of hybridization exhibited by the C atoms in propene...OK.2308

Let me go ahead and draw a propene atom: now again, this is one of those questions...so we didn't discuss organic chemistry; is it possible that a molecule like propene actually shows up in a Lewis structure problem that you have studied, maybe, in class, or in the problem sets?--maybe.2319

I mean, you have three carbons in there; each carbon is going to be attached to each other; yes, it is possible, but let me just go ahead and do it.2343

Propene is going to look like this.2351

Basically, what you have is a double bond, which is going to be sp2 hybridized (any time you see a double bond, the atoms are sp2 hybridized), and you have single bonds all the way around here (so sp3 hybridized).2361

In this case, it is going to be sp2 and sp3 hybridized; so it looks like the correct answer is D.2375

That is it: single bonds--sp3 hybridization; double bonds--sp2 hybridization; triple bonds--sp hybridization.2382

OK, #33: A 1-liter sample of an aqueous solution contains .1 moles of sodium chloride and .1 moles of calcium chloride.2391

What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the chloride as AgCl?2403

Assume that AgCl is insoluble.2411

OK, so basically, we have this: NaCl will dissociate into Na+ + Cl-, so that is going to be one source of the chloride.2414

Our other source is CaCl2; it is going to dissociate into calcium 2+, plus 2 Cl-.2425

So, for every mole of this, it is going to be 2 moles of chloride; so we need to calculate some moles here.2434

1 liter...so let's do this one...times 0.1 millimole per milliliter...actually, I don't need to do...this is 1 liter, so I can just leave it as mole per liter...so we end up with 0.1 moles of sodium chloride solution.2439

.1 moles of this produces .1 moles of chloride ion; so I have 0.1 mol of chloride ion that comes from the sodium chloride solution.2465

OK, 1-liter sample...let's see: so this one--again, they tell me I have...oh, the same thing; OK, 1 liter times 0.1 moles per liter equals 0.1 moles of the calcium chloride.2476

Well, one mole of this produces 2 moles of chloride; so what I end up with is 0.2 moles of chloride ion.2496

I end up with a total of 0.3 moles of chloride ion.2504

Now, AgCl: Ag + Cl- goes to AgCl, solid; well, the question is: What is the minimum number of moles of AgNO3?2510

Well, I need .3 moles of Ag, and AgNO3...one mole of that produces 1 mole of Ag+ plus one mole of NO3-; so I need 0.3 moles of AgNO3.2525

That is it; you could obviously have done this a little bit faster, maybe, but I want you to see where this comes from.2544

A certain amount of chloride from sodium chloride; a certain amount of chloride from magnesium chloride: you have to find the total number of moles of the chloride ion if you are trying to take all of that out of solution.2550

The relationship is 1:1 with silver and chloride, and the silver comes from silver nitrate, which is 1:1.2563

That is it; so the answer is C.2572

That is #33; OK.2576

#34: Questions 34 and 35 refer to an electrolytic cell that involves the following half-reaction.2579

Let's see: which of the following...so we have the AlF63-, plus 3 electrons, goes to aluminum metal, plus 6 fluoride ions.2588

So, in this case, it looks like the aluminum has been reduced from aluminum 3+ to aluminum 0.2601

Which of the following occurs in the reaction: AlF63- is reduced at the cathode--yes, oxidation occurs at the anode; reduction occurs at the cathode.2609

The equation as written, which is AlF63- + 3 electrons goes to aluminum + 6 F-: this is a reduction.2621

The oxidation state of fluorine doesn't change; aluminum is a +3; this is a 0; it is a reduction, and we know it is a reduction because the electrons are on the left-hand side.2639

This is a reduction; reduction happens at the cathode, so our answer for 34 is A.2651

OK, #21 A steady current of 10 amps is passed through an aluminum production cell for 15 minutes; which of the following is the correct expression for calculating the number of grams of aluminum produced?2659

1 faraday equals 96,500 coulombs: OK, this is an electrolytic cell; this is where we are going to pass the particular current through a cell for a certain amount of time to see how many grams we want.2673

OK, so let's start with the time: this is the conversion.2687

We have 15 minutes; we are going to multiply that by 60 seconds in 1 minute to cancel minutes; we are going to have 10 amps, so this is going to be 10 coulombs per second, times 1 mole of electrons is 96,500 coulombs.2692

We have 3 moles of electrons per 1 mole of aluminum, times 27.0 grams of aluminum per 1 mole.2722

Therefore, our answer is C, because C is 15 times 60 times 10, times 27 divided by 96,500, and 3 on the denominator; so 35 is C.2737

OK, now 36: the initial rate data in the table above were obtained for the reaction represented below; what is the experimental rate law for the reaction 2 NO + O2 gas goes to NO2?2753

OK, so we know that our rate law...rate is going to equal K times concentration of NO...2773

Well, actually, you know what, let me sort of run through it.2781

I take a look at...I start with the nitrogen oxide, and I notice experiment 1 is .1 Molar initial, and experiment 2 is .2 Molar; so this is the initial rate, so we are using the method of initial rates here.2786

It says that, for .1 Molar, the initial rate is 2.5x10-4; and for .2 Molar, it's 5.0x10-4; so this is...when I double the concentration, the rate doubles.2804

Well, that means that it is first order in NO.2819

Now, I go to the other reactant; I go to oxygen, which is in the second column; so now, I use experiment 2 and 3, because I'm going to hold the NO concentration fixed.2824

I notice, when I do .1, for a .1 concentration of O2, the rate is 5x10-4; and if I multiply that by 4 and go to .4 for an initial concentration, my rate goes to 8.0x10-3.2836

So, when I multiply the concentration by 4, I actually end up multiplying the rate by 16, which means that it is 4 squared.2857

Therefore, it is second order in O2; so that means our answer is B: the rate is K times NO times O2 squared.2868

OK, #37: The ionization energies for element X are listed in the table above; on the basis of the data, element X is most likely to be _____.2883

OK, so take a look at these energies of ionization: I have 580, 1815, 2740; the jump from 580 to 1815 is actually pretty high; however, notice the fourth and fifth: the jump from 2740 to 11,600 is huge.2892

So, chances are, the best that this represents is probably aluminum; and the reason it is aluminum is because of the following.2916

The 1s2, 2s2, 2p1...the first ionization energy that takes this electron is reasonably low.2928

Well, to go here, it is going to be a significant jump in order to start taking away these two electrons; and the 1815 and the 2740 represent a pretty significant jump.2940

And then, of course, the fourth and the fifth--the 1s2 electrons are just way too high--it's way too much energy, so just based on these numbers, aluminum is the best choice.2952

#38: Let's see, a molecule or an ion is classified as a Lewis acid if it accepts a pair of electrons to form a bond; that one is just something that you should know.2966

A Lewis acid is something that accepts a pair of electrons to form a bond.2978

OK. #39: A phase diagram for a pure substance is shown above; which point on the diagram corresponds to the equilibrium between solid and liquid phases at normal melting point?2983

Normal melting point--the atmospheric pressure is 760 millimeters, 1 atmosphere; so we go up to point C: that is where the phase difference is between the solid and the liquid--solid on the left, liquid in the middle, gas on the right.2998

So, C is the answer.3013

#24: We have: Of the following molecules, which has the largest dipole moment?3017

D--that is the one that is ionic; CO does have a dipole moment...well, HF...some people consider it ionic; some people, not; the largest difference in electronegativity between H and F is a bigger difference in electronegativity between C and O.3022

CO2 is nonpolar; O2 is nonpolar; F2 is nonpolar; so it is between A and D; D is actually bigger--fluorine is the most electronegative.3039

OK, and #41: After the equilibrium represented above is established, some pure O2 gas is injected into the equilibrium vessel at constant temperature.3049

After equilibrium is established, which of the following has a lower value, compared to its value at the original equilibrium?3061

OK, I'm at equilibrium, constant pressure; I put some O2 into this reaction vessel.3070

When I put some O2 into this reaction vessel, Le Chatelier's principle--it is going to shift to the left.3079

So, what it is actually going to do: is the Keq for the reaction going to change?--no, at a given temperature, the Keq stays the same.3086

The equilibrium position might change; in other words, the SO3, SO2, and O2...different concentrations; but the Keq does not change.3095

The total pressure in the reaction vessel...let me see, what is it saying?--which of the following has a lower value?3104

No, the total pressure does not go down; it actually goes up.3110

The amount of SO3 gas doesn't go down; the amount of O2 in the reaction doesn't go down, because you are injecting O2; the amount of SO2 in the reaction vessel--yes.3114

Since the reaction is going to move to the left, the SO2 is going to be used up--the excess that you use; the SO2 concentration that was there before anything happened--that is going to deplete, because the reaction is going to move to the left.3125

So, 41 is E.3139

OK, thank you for joining us here at Educator.com for this particular discussion of the multiple choice questions.3144

The next lesson, we will continue on with question 42.3150

Take care; goodbye.3153

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