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Raffi Hovasapian

Raffi Hovasapian

AP Practice Exam: Multiple Choice, Part II

Slide Duration:

Table of Contents

I. 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
II. 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
III. 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
IV. 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
V. 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
VI. 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
VII. 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
VIII. 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
IX. 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
X. 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
XI. 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
XII. 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
XIII. 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
XIV. 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
XV. 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
XVI. 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 (5)

1 answer

Last reply by: Professor Hovasapian
Sun Feb 22, 2015 7:37 PM

Post by Shih-Kuan Chen on February 21, 2015

Professor, where can I learn more about flame color and colors of ions and compounds?

0 answers

Post by Mwongera Mwarania on October 27, 2013

Sorry, on re-listening the recording, you have explained that e0 is not like Enthalpy: you don't multiply with. You only reverse the sign if you reverse the reaction. Please ignore my earlier question.

0 answers

Post by Mwongera Mwarania on October 27, 2013

Q.57: Why did you not multiply the e0 by 3 for the second step in accordance with the Hess' law?

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 II

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 0:12
    • Multiple Choice 42
    • Multiple Choice 43
    • Multiple Choice 44
    • Multiple Choice 45
    • Multiple Choice 46
    • Multiple Choice 47
    • Multiple Choice 48
    • Multiple Choice 49
    • Multiple Choice 50
    • Multiple Choice 51
    • Multiple Choice 52
    • Multiple Choice 53
    • Multiple Choice 54
    • Multiple Choice 55
    • Multiple Choice 56
    • Multiple Choice 57
    • Multiple Choice 58
    • Multiple Choice 59
    • Multiple Choice 60
    • Multiple Choice 61

Transcription: AP Practice Exam: Multiple Choice, Part II

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

We are going to continue our multiple-choice practice, and we left off at problem #41, so we are going to start with problem #42: let's just jump right on in.0004

OK, when the equation above is balanced and all coefficients reduced to lowest whole-number terms, the coefficient of OH- is going to be...0013

All right, so 42: this is just a simple balancing issue, so when you balance this out, the coefficient of the OH- is going to be 3, so 42 is C.0021

OK, #43: A sample of 61.8 grams of H3BO3 (a weak acid) is dissolved in 1,000 grams of water to make a 1 molal solution.0034

Which of the following would be the best procedure to determine the molarity of the solution? Assume no additional information is available.0046

OK, of these choices, the best way to check the molarity of the situation...because molarity is moles per liter, you need the total volume, which is in the denominator of molarity; it actually is going to be D.0055

You need to take a measurement of the total volume of the solution.0068

That is it--nice and straightforward.0072

OK, #44: A rigid metal tank contains oxygen gas (rigid, so there is no volume change).0076

Which of the following applies to the gas in the tank when additional oxygen is added at a constant temperature?0084

That is the important part: additional oxygen is added at a constant temperature.0091

"The volume of the gas increases"--no, it's rigid.0097

"The pressure of the gas decreases"--no, you are actually injecting gas into the system, so the pressure actually increases.0100

"The average speed of the gas molecules remains the same"--this is where it is a little bit tricky.0107

Now, a constant temperature: that is the secret here--a temperature constant implies that the average kinetic energy is unchanged.0114

Remember, temperature is a measure of the energy of the actual molecules: if the temperature doesn't change, the average energy of the molecules doesn't change.0125

So, that is unchanged.0134

Well, that means one-half the mass times the average velocity squared is unchanged; this means that the velocity, the average velocity, is unchanged.0140

So, yes, C is the answer for #44.0151

OK, #27 What is the hydrogen ion concentration in a .05 Molar HCn?--the Ka for HCn is 5.0x10-10.0156

OK, so again, this is a weak acid problem; we do have to use an ICE chart, but again, the numbers are going to be really, really simple, so there doesn't have to be any complicated calculations.0168

Let's go ahead and run through what we normally run through.0181

The number of species in solution: well, we have HCn in solution, and we have H2O in solution.0186

Of those two things that can dissociate, the 5.0x10-10 Ka for HCn...this is the dominant equilibrium, so what we are going to have is the following.0192

We are going to have: HCn is in equilibrium with H+ + Cn-.0203

Well, they tell us that the...so we are going to have an Initial; we are going to have a Change; and we are going to have an Equilibrium.0210

They tell us we have a .05 Molar, so before anything happens, there is none of this, and there is none of that.0215

I'm sorry, I shouldn't say they don't matter; I'll use 0; we use straight lines for things that actually are irrelevant.0221

A certain amount is going to dissociate; a certain amount is going to show up; and a certain amount is going to show up.0228

We are looking for x, because that is going to be the hydrogen ion concentration.0232

0.05-x, x, and x; so they tell us that the Ka, which is 5.0x10-10, equals this times that, divided by that.0236

We get x2 over 0.05-x, which is just about equal to x2 over 0.05.0253

OK, and when I do that calculation, I end up with (so now we go over here; we end up with) x2=0.25x10-10; that means x is equal to the hydrogen ion concentration (right?--x is hydrogen ion concentration); it is equal to 0.5x10-5.0264

Square root: just basically take the square root of .25 and the square root of 10-10, and then, when you convert this to scientific notation, you end up with (oh, these stray lines are crazy! OK) 5.0x10-6.0292

That is D, so our answer for 45 is D.0311

Again, set up your ICE chart; the numbers are really easy to work with--nice and straightforward.0316

OK, so #46 says: Which of the following occurs when excess concentrated NH3 is mixed thoroughly with .1 Molar copper nitrate, aqueous?0322

OK, so we have a copper nitrate solution; a copper nitrate solution is going to be a colored solution.0336

It is a transition metal: most transition metals are colored--they absorb visible light.0343

And again, like I mentioned earlier, the copper nitrate dissolves; the water molecules surround the copper (2 of them, 4 of them, 6 of them...however many); there is a certain color that that complex ion absorbs.0349

When you put NH3 into this solution, well, the NH3...a couple of molecules of ammonia may actually replace a couple of molecules of water surrounding the copper ion.0363

What happens is: the color actually gets darker or changes or gets lighter; so, in this case, the best choice would be: The color of the solution turns from a light blue to a dark blue; so the answer would be C.0374

OK, #47: When hafnium metal is heated in an atmosphere of chlorine gas, the product of the reaction is found to contain 62.2% hafnium by mass and 37.4% chlorine by mass; what is the empirical formula of the compound?0390

OK, so this is a straight empirical formula problem where you convert percentages to grams; you can go to moles and divide by the least number, and you should be able to get some whole numbers.0410

Let's go ahead and do this really quickly: Hafnium--we have 62.2%, is 62.2 grams, times 1 mole of hafnium is roughly 179 grams; so I end up with about...one-third; so let's go ahead and put .34 (I'll round it slightly up, because this is a little bit more than that).0419

When I do chlorine, that is going to be...they say 37.4%, which is 37.4 grams times 1 mole/35.45 grams, so you are looking at about 1.0448

So now, we go ahead and divide by that to get 1; 0.34...you get 3; there is your answer, HFCl3; it looks like C is your final answer for #47.0464

OK, #48 says: if 87.5% of a sample of pure iodine-131 decays in 24 days, what is the half-life of iodine-131?0482

Well, we know what the half-life of something is: that is the amount of time it takes for half of a sample to vanish.0498

OK, this 87.5%...again, on these AP exams, you are going to be given very easy numbers to work with.0505

Here is what happens: I start with a sample; OK, half of it decays; now I am left with 50%.0512

Well, now half of that sample decays; well, half of 50% is 25%, so now, I am at 75% of the original sample (is gone).0522

Now, half of...so now I have 25% of the sample left; well, half of that 25% is 12 and a half percent; so after three cycles of half-life, half-life, half-life, 75 plus another 12.5 puts me at 87.5.0533

So, after 3 cycles...the original sample--half goes away; what is left over--another half goes away; what is left over--another half goes away; at that point, I have 87.5% that has decayed.0552

Well, I have three cycles; 24 days to go through three cycles; that means each cycle is 8 days long, so the half-life of iodine-131 decay is 8 days.0568

Take a look at the percentage, and chances are it is either going to be 1, 2, 3, or 4; and then, just take the number of days or hours or minutes or seconds and divide by the number of cycles.0582

We went through one cycle, two cycles, three cycles; 24 days; 3 into 24 is 8 days; so 48 is B.0592

I hope that makes sense.0604

OK, #49: Which of the following techniques is most appropriate for the recovery of solid potassium nitrate from an aqueous solution of potassium nitrate?0606

If you have an aqueous solution, and you need to recover the solid, it's very simple: just boil off the water--so "evaporation to dryness"--49 is E.0615

OK, #50--let's see: #50 says: In the periodic table, as the atomic number increases from 11 to 17, what happens to the atomic radius?0628

As you move from left to right on the periodic table, atomic radius (not ionic radius, atomic radius) decreases.0641

As the positive charge gets bigger, even though you are adding one electron, you are actually going to be sucking in the electrons a little bit more.0649

So, as you move from left to right (from your perspective, left to right on the periodic table), the atomic radius gets smaller.0656

The answer is D: it decreases only.0662

OK, #51--let's see what we have: Which of the following is a correct interpretation of the results of Rutherford's experiment, in which gold atoms were bombarded with alpha particles?0667

Well, you remember the Rutherford scattering experiment: he shot alpha particles (which are helium nuclei) at a thin piece of gold--through a thin piece of gold--and very few of the alpha particles actually ended up bouncing backward.0680

Most of the alpha particles just passed right on through.0697

What that means is that the positive charge of the atom is concentrated in a very small region.0701

An atom like this--a positive charge is actually concentrated in a very tiny part of it; most of the alpha particles just pass through the atom.0706

So, 51 is E.0715

OK, #52: Under which of the following sets of conditions could most O2 be dissolved in H2O?0722

OK, if you are trying to dissolve a gas in a liquid, what you need is high pressure, low temperature.0730

So, here: high pressure--let me see: we look at 5 and 5 atmospheres (that is A and B); low temperature--between 80 and 20, we are going to take the 20 degrees, so we will take B.0737

52 is B; so, again, when you need to dissolve a gas in a liquid, you need conditions of high pressure, but you need conditions of low temperature.0751

If the liquid actually is a high temperature, it is going to be bubbling, bubbling, bubbling; it is going to be all excited; high temperature is actually going to force the gas out.0761

Low temperature means the gas can actually occupy a space in that liquid.0769

All right, let's see--#53: OK, gases W and X react in a closed, rigid vessel to form gases Y and Z according to the equation above; the initial pressure of W is 1.2 atm, and that of X is 1.6 atm.0775

No Y or Z is initially present; the experiment is carried out at constant temperature; what is the partial pressure of Z when the partial pressure of W has decreased to 1 atmosphere?0796

OK, so let's work this one out; it is actually very, very simple.0809

W gas, plus X gas, goes to Y gas plus Z gas.0814

They tell me this is initially at 1.2 atm, and this is at 1.6 atm.0825

It is 0 and 0 here; so at the end, when this is dropped down to 1 atmosphere, they want to know what is the concentration of...yes, what is the partial pressure of Z.0832

OK, now, notice what they said...equation...container...carried out...OK, here is what is important: the experiment is carried out at constant temperature.0851

Because it is carried out at constant temperature, that means nothing actually changes in terms of pressure.0860

So, the beginning total pressure, which is 1.2 plus 1.6 (which is a total of 2.8)--that is going to actually end up being the final pressure: 2.8.0865

Now, the final pressure consists of the pressure of the W, the X, the Y, and the Z.0877

Now, pressure is the same as concentration (right?--pressure and concentration are just different sides of the same coin): now, this pressure went from 1.2 to 1.0.0884

Well, this is a 1:1 mole ratio; so, if .2 moles of this was used up, I can think of pressure as just moles.0895

.2 moles of this was used up: that means this pressure dropped down to 1.4.0903

Well, this is 1:1; if .2 moles was used up, that means .2 moles of this was formed, and .2 moles of this was formed.0908

So now, we want the total pressure to be 2.8.0919

Well, 1, 2.4, 2.6, 2.8; that is how you work out this problem; the idea is constant temperature--that means pressure doesn't change: the initial pressure equals the final pressure.0924

You use the mole ratios, and you use the fact that pressure is like concentration: if this pressure drops from 1.2 to 1.0, since this is 1:1, this one drops, also, from 1.6 to 1.40.0937

Well, if this loses .2 atmospheres, there is gain of .2 atmospheres here, gain here; as long as the sum of those equals the sum of this, everything is fine.0951

So, 1.2 is the answer: so we get: 53 is equal to A.0963

OK, #54: Let's see, which of the following changes alone would cause a decrease in the value of Keq for the reaction represented above?0972

OK, so they say that we have: 2 NO gas, plus O2 gas, is in equilibrium with 2 NO2; and they tell me that this ΔH is less than 0, which means that it is exothermic.0989

Exothermic means that it releases heat, so I'm going to write +heat; heat is one of the products.1006

Now, which of the following changes alone would cause a decrease in the value of Keq for the reaction represented above?1012

Let's see: if you decrease the temperature...no, if you decrease the temperature, it is going to want to increase the temperature, so the reaction is going to go to the right.1021

Decreasing the value of Keq means pushing the equilibrium to the left, backward towards the reactants: the only thing here that does that--if you increase the temperature, which means increasing the heat, that means you are going to increase one of the products, which is heat.1031

Well, the system is going to offset; it is going to offset by wanting to move to the left.1049

A reaction that moves to the left: that means the equilibrium constant is actually--all things being equal, the equilibrium constant will decrease.1055

That is what you have recognize: a decrease in the value of Keq means that more of the reactant is going to be favored.1067

In this particular case, because it's exothermic, by increasing the heat--increasing the temperature--you are forcing the reaction to move to the left.1074

I hope that makes sense; so here, 54--the best answer is B.1086

OK, #55: According to the balanced equation above, how many moles of HI would be necessary to produce 2.5 moles of I2, starting with 4 moles of KMnO4 and 3 moles of H2SO4?1092

OK, this is basically just a limiting reactant problem; so we need to find out...we need to just look at the equation to see...which is which.1110

Let me actually write this equation out...you know what, that is fine; we can just work right from the page; we don't really need to write anything out--there is nothing too complicated going on here.1119

We want to produce 2.5 moles of I2, starting with 4 moles of KMnO4.1130

OK, so we have 4 moles of KMnO4; how many moles of HI?...and 3 moles of H2SO4.1136

OK, so we have 3 moles of H2SO4; for that, 2 moles of KMnO4 are necessary; well, we have 4 moles of KMnO4, so that is not a problem.1145

3 moles of H2SO4 requires 10 moles of HI (right?--because there is a 10 in front of the HI).1155

10 moles of HI gives 5 moles of I2; in order to produce 2.5 moles of I2, which is half of 5, I need half of 10.1162

Therefore, I need 5 moles of HI; so the answer is D.1171

That is it: just look at the mole ratio; see which one is the limiting reactant (in this case, the H2SO4); the rest is just mole ratios.1175

And again, the numbers are very, very simple.1183

OK, #56: A yellow precipitate forms when .5 Molar NaI is added to a .5 Molar solution of which of the following ions?1186

OK, so you have sodium iodide: sodium is not going to do anything--it is going to be soluble with anything; iodide ion is not going to mix with chromate, sulfate, or hydroxide, and it is going to be soluble with zinc; so A is your final answer.1196

Lead iodide: that is the one that is the yellow precipitate.1210

OK, #57: Let's see, according to the information above, what is the standard reduction potential for the half-reaction M3+ + 3 electrons goes to M, solid?1215

OK, we need to look at what they gave us; we need to rearrange it in such a way that, when we add it, we get what we want, which is: M3+ + 3 electrons goes to M(s).1229

All right, now notice what they want: the M3+ is on the left; what they give us--the M3+ is on the right; so I'm going to flip that first equation.1240

I'm going to write that first equation as (oops, let me write--this is #57)...I'm going to write: 3 Ag, solid, plus M3+, goes to M, solid, plus 3 Ag+.1249

Now, the potential for that is +2.46, but since I flipped the equation, now the electric potential is going to be -2.46 volts.1273

Now, I have an Ag+ on the right, and the other equation they give me has an Ag+ on the left; so I'm going to leave that as written.1283

Ag+ + an electron goes to Ag, solid.1291

Let's see: the electrostatic potential for that is positive 0.80, because I haven't done anything to it.1308

However, I am going to go ahead and put a 3 in front of the Ag, 3 in front of the electron, and 3 in front of this Ag, because the final equation that I am looking for--I need the Ag's to cancel.1317

This cancels with that; that cancels with that; and I am left with: M3+ + 3 electrons goes to M, solid--which is exactly what I wanted.1327

Now, I just add these two, and I end up with -1.66 volts, which is A.1339

That is it; I just need to arrange this in such a way...now, the only thing you have to remember here is: notice, I change this second equation by actually multiplying everything by a 3.1350

However, this is not like enthalpy: reduction potential is an intrinsic property--it doesn't depend on how much of something you have; it is just a property.1361

So, even though I put 3 here, I didn't multiply the electrostatic potential by 3; the only thing that I change when I change electrostatic potentials is the sign, if I flip the equation.1373

Flip the equation--just go positive/negative, negative/positive; but by adding coefficients in order to make things cancel, I do not change the electrostatic potential.1385

It is not like enthalpy or entropy or free energy; OK, so very, very important: electrostatic potential is an intrinsic property--does not depend on the amount that you have.1394

If this were enthalpy and we were doing Hess's Law, yes, you would have to multiply this by 3; OK.1403

Let's see, where are we?--we are at #58.1411

OK, #58: On a mountaintop, it is observed that water boils at 90 degrees Celsius, not at 100 degrees Celsius, as at sea level.1415

This phenomenon occurs, because on the mountaintop...of our choices, C is the best choice.1425

Equilibrium water vapor pressure equals the atmospheric pressure at a lower temperature.1433

Remember what boiling point is: the boiling point is the temperature at which the vapor pressure is equal to the atmospheric pressure.1439

As you go higher in elevation, the atmospheric pressure drops; there is less atmosphere, so there is less pressure.1448

Well, because there is less pressure, you reach vapor pressure at a lower temperature.1454

That is why boiling point decreases as you go higher in elevation; so C is the best answer here.1460

OK, #59: Let's see, a 40-milliliter sample of a .25 Molar KOH is added to 60 milliliters of a .15 Molar BaOH2.1468

What is the molar concentration of OH- in the resulting solution?--assume that the volumes are additive.1479

OK, so we need to calculate the total moles of OH- and divide by the total volume.1485

All right, so we know that potassium hydroxide dissociates as 1:1: one mole of potassium hydroxide produces one mole of hydroxide ion.1491

Let's go ahead and do that one: that is going to be 40 milliliters, times 0.25 millimoles per milliliter (moles per liter--I can do millimoles per milliliter, or centimoles per centiliter, as long as I do it consistently); so I end up with 10 millimoles of OH-.1501

Now, barium hydroxide (BaOH2): when it dissociates, it produces 1 mole of barium 2+, but it produces 2 mol of hydroxide.1524

Here, I have 60 milliliters times 0.15 millimole per milliliter, and then I multiply that by 2 moles of OH- per one mole of barium hydroxide; I end up with 18 millimoles.1534

Or, I could have just done this first calculation and gotten 9 and just multiplied by 2, because 1 mole of this produces 2 moles.1556

So now, I have a total of 28 millimoles of OH-, and I divide by my total volume, which is 40 milliliters plus 60 milliliters, which is 100 milliliters; so I get a molarity of 0.28 M.1563

Now, you notice: if you remember, my m has a line on top of it; the molarity symbol that is used nowadays is a capital M; so you will see it as .28 M.1583

There you go; that is your answer, and .28 is C.1594

#59 is C; all right.1599

Just a couple more to go here for this lesson: let's see, #60 says: A .03 mole sample of NH4NO3 is placed in a 1-liter evacuated flask, which is then sealed and heated.1603

The NH4NO3 decomposes completely according to the balanced equation above.1621

The total pressure in the flask is measured at 400...I'm sorry, the total pressure of the flask is measured at 400 Kelvin; it is closest to which of the following?--and they give you the value of R; OK.1627

Here, we need to find the total number of moles of gas that is produced; we have a solid, and it is evacuated, so the pressure is 0 to begin with, but you are producing N2 gas and H2O gas.1644

OK, well, let's see: we have .03 moles...so let's go ahead and write our equation here.1657

We have: NH4NO3--that is a solid--and it is going to form 1 mole of N2 gas and 2 moles of H2O gas--2 moles of that.1665

Well, if I have 0.03 moles of this to start off with, 1:1 ratio, it is going to end up forming .03 moles of this; 1:2--it is going to end up forming .06 moles of that; the total number of gas that is going to be floating around in that container is going to be 0.09.1683

So now, we use our PV=nRT; the total pressure is equal to nRT over V; number of moles is 0.09; 0.082 is what they give us for R; the temperature is going to be 400; and then, it is in a one-liter flask.1705

When you multiply all of this out (which you should be able to do just on sight): .09, .08, times 400...you are looking at about 2.95; the answer is going to be 3 atmospheres.1730

So the answer is A; so 60 is A--that is it.1745

You want to find the total moles of gas, and then just use the ideal gas law to find the pressure.1750

OK, #61: For the reaction of ethylene represented above, ΔH is -1323 kilojoules.1756

What is the value of ΔH if the combustion produced liquid water, rather than water vapor?1766

ΔH for the phase change of H2O gas to H2O liquid is -44 kilojoules per mole.1773

OK, so we have this reaction: C2H4 + 3 O2 gas forms 2 moles of CO2 and 2 moles of H2O.1781

They tell me that the ΔH for this is equal to -1323 kilojoules.1794

Notice, it is not kilojoules per mole; it is kilojoules.1800

OK, now, this H2O is as a gas; well, they are saying, "What would happen to the ΔH if, instead of forming water vapor, it actually formed liquid water?"1803

Well, they are telling me that the ΔH for the phase change of H2O gas to H2O liquid...they are telling me that the ΔH for this process is equal to -44 kilojoules per mole.1815

Here we have to watch a couple of things: we have to watch the direction, and we have to watch the units.1838

This one is in kilojoules; this is in kilojoules per mole.1843

Well, we have 2 moles of water that are going to go from gas to liquid; OK.1846

Here is what this looks like: the final ΔH is going to be the sum of the ΔH's, so we are going to start with -1323 kilojoules, and we are going to add...1852

Now, in going from a gas to a liquid, you are actually giving off heat, which is what this negative 44 says; so, we are going to add the heat given off in going from gas to liquid.1865

But notice: we have 2 moles of water (let's get rid of these crazy lines that are showing up all over again), so we have -44 kilojoules per mole, times 2 moles, because we have 2 moles of water that is going from gas to liquid.1879

That does that; when we actually do this, 2 times -44 is -88; -1323 plus a -88...our final answer is -1411 kilojoules, and that is E; so 61 is E.1904

OK, thank you for joining us for the second part of the multiple choice; we have one more section to go.1926

So, thank you for joining us here at Educator.com.1931

We'll see you for the next lesson; goodbye.1933

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