Raffi Hovasapian

Raffi Hovasapian

AP Practice Exam: Multiple Choice, Part III

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 (3)

2 answers

Last reply by: Professor Hovasapian
Fri May 8, 2015 1:15 AM

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 III

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 0:16
    • Multiple Choice 62
    • Multiple Choice 63
    • Multiple Choice 64
    • Multiple Choice 65
    • Multiple Choice 66
    • Multiple Choice 67
    • Multiple Choice 68
    • Multiple Choice 69
    • Multiple Choice 70
    • Multiple Choice 71
    • Multiple Choice 72
    • Multiple Choice 73
    • Multiple Choice 74
    • Multiple Choice 75

Transcription: AP Practice Exam: Multiple Choice, Part III

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

We are going to continue our discussion of the AP practice exam: this is the third part of the multiple choice.0004

We left off on problem 61, so let's continue on with problem #62.0011

OK, so #62 says: The reaction represented above has an equilibrium constant equal to 3.7x104; which of the following can be concluded from this information?0017

OK, now, let's see...so we take a look at this reaction: we have acetic acid reacting with the cyanide ion, and it forms the hydrocyanic acid, the HCn, plus the acetate ion.0030

So basically, the H has jumped from the acetic acid to the cyanide ion.0047

Or, another way of looking at it: the cyanide ion has taken the H away from the acetic acid.0052

The best information that is given here is the equilibrium constant, 3.7x104; that is a huge equilibrium constant.0057

Now, a high equilibrium constant means that, at equilibrium, the reaction is all the way to the right, meaning that there is virtually no reactant left; it is all product.0069

So, as such, basically what this says...these choices, A, B, C, D, and E: A is the best choice--it says that the cyanide ion is a stronger base than the acetate, because the cyanide has taken the H.0079

A stronger base: it takes the H.0095

And because it's such a high equilibrium constant, that means it is virtually all HCn and all acetate, and no acetic acid and Cn-.0099

So, for #62, A is our best choice: 62, A.0109

OK, #63: Let's see, the graph above shows the results of a study of the reaction of X, with a large excess of Y, to yield Z.0116

The concentration of X and Y were measured over a period of time; according to the results, which of the following can be concluded about the rate law for the reaction under the conditions studied?0128

OK, so Y actually doesn't matter altogether this much, because they said it is mostly excess; and you notice the Y is just sort of a steady concentration.0139

That doesn't really tell us anything; and because they tell us that it is in excess, it is not really going to show up; so we are going to be concerned mostly with what is happening with X.0150

OK, so this is a plot of concentration versus time.0159

Now, remember, with rate laws, when we have time and concentration data, what we do is: we create three plots.0164

We create a plot of time versus concentration, time versus logarithm of concentration, and one over concentration against time; so we have three plots.0172

Depending on which one is a straight line, it will tell us what the order of the reaction is.0184

Here, unfortunately, all we have is just the concentration and the time; well, it's not a straight line--as you can see, it drops down that way; it's curved; so we definitely know it's not a 0 order reaction.0189

Well, here what you want to look at is concentration and time; and what you notice--we start with a .2 Molar concentration at time 0.0207

Well, at the first time increment (which is 1), it has dropped down to .1 molarity; so it has dropped by half.0216

At the next equal time increment (2), now the concentration is .5; at the next equal time increment (which is 3), it has dropped down to .5; so, for equal time increments, the concentration always drops by half.0225

This says that the half-life is constant; in other words, the half-life doesn't depend on concentration.0243

The only time that the half-life doesn't depend on concentration is for a first-order reaction: the half-life is 0.693, divided by K, which is the rate constant.0253

That means that every equal time interval, half the reaction is gone; half the concentration vanishes; half the concentration vanishes.0269

And, based on these numbers (.2 to .1 to .05, from 0 to 1 to 2 to 3), equal time intervals--half of the concentration at the beginning of that time interval vanishes.0280

So, because that is constant, I know that this is a first-order reaction; so in this particular case, B is our answer.0298

So again, when you see a graph of time versus concentration (or Y versus X, I guess, so concentration versus time; I never know which one goes first), take a look at the values themselves, and recognize: again, these are multiple choice questions, so they are not going to be altogether that strange.0308

If you were to be given a problem like this on the free response, then you might be given a couple of different graphs.0326

But here, since they only gave you one graph of time and concentration, chances are they are going to be asking you something about the half-life.0332

And, in this case, the numbers confirm that the half-life is constant: for every equal time interval, there is a 50% depletion of the concentration of reactant, since the beginning of that time interval.0339

The only reaction that has a constant half-life is the first-order reaction: the 0 order and the second-order are not like that; so 63 is B.0352

OK, let's see: #64 says: An equal number of moles of He, Ar, and Ne are placed in a glass vessel at room temperature; if the vessel has a pinhole-sized leak, which of the following will be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the mixture has effused?0370

They put a bunch of gases in this vessel, and after a certain amount of time, they measure how much gas is in the vessel (that means some has escaped); and here is what they find.0397

Well, helium, argon, neon: so argon is a bigger atom than neon; neon is bigger than helium, which means that...the heavier something is, the slower it moves (the slower it effuses).0407

Therefore, helium is going to escape faster, because it is the smallest; neon will escape the next fastest, because it is next smallest; and argon will escape the slowest.0426

Well, at the end, because helium has escaped the fastest, there is less helium than there is neon; there is less neon than there is argon.0436

Therefore, the partial pressure of argon is going to be greater than the partial pressure of neon, which is going to be greater than the partial pressure of helium.0447

As far as the choices that they gave us: helium--less than neon, less than argon; so it is A.0461

That is how you work that out: a heavier molecule escapes slower; a lighter molecule escapes faster.0468

A heavier molecule moves slower--a heavier atom/a heavier molecule moves slower; a lighter molecule or atom moves faster.0475

OK, #65: Which of the following compounds is not appreciable soluble in water, but is soluble in dilute hydrochloric acid?0484

The answer is A, magnesium hydroxide, and here is why: magnesium hydroxide is a solid; magnesium 2+, plus 2 OH-; it is not very soluble.0494

This equilibrium is mostly over here--it is mostly solid.0510

What happens, though, is: when you put it in acid, the acid reacts with the OH- that has dissolved a little bit, and it forms water.0513

Well, when you add acid, it uses up this, so it depletes that; when this is depleted, this reaction needs to create more OH-, needs to create more OH-, needs to create more OH-, as it is being produced.0527

As it is being produced, it is being eaten up by the hydrogen, so eventually, all of this dissolves.0541

So, in acidic solution, magnesium hydroxide is soluble, because the hydroxide ion that does dissolve a little bit is used up in this neutralization reaction.0547

OK, #66: When solid ammonium chloride is added to water at 25 degrees Celsius, it dissolves, and the temperature of the solution decreases.0559

Which of the following is true for the values of ΔH and ΔS for the dissolving process?0573

OK, so let's deal with ΔH first: they say that the temperature decreases--in other words, when I touch the solution, it gets colder; that means it is sucking heat from the environment.0578

That means that ΔH is greater than 0; it is endothermic.0588

Or...endothermic...that means it is absorbing heat; I'll write it this way: Endothermic--that means that the ΔH is greater than 0 (it is positive).0595

Now, we are going from a solid to aqueous: a solid is a highly ordered thing; aqueous, less ordered.0608

Something that is less ordered has higher entropy, so ΔS is greater than 0--it is positive.0618

A is ΔH positive, ΔS positive, so our answer is A.0626

That is it--nice and straightforward; just reason it out; this means this, this means this, this means this.0632

OK, #67: Wait, that is number...yes, #67--it says: What is the molar solubility in water of Ag2CrO4?0640

OK, #67: so, Ag2CrO4 dissociates as 2 Ag+ + CrO42-, and they tell me that the Ksp is equal to 8x10-12.0657

Well, I know the Ksp for this: it is equal to the silver ion, raised to the 2 power because of that 2, times the CrO42- ion raised to the 1 power.0677

Well, for every chromate produced, twice as much silver is produced.0689

So, if I call the chromate x, well, that means 2x of silver is produced (right?--when this dissociates, it produces 1 mole of chromate and 2 moles of silver ion, so if I produce x amount of chromate, I produce 2x amount of silver ion).0703

Now, I put these into here; so I write the Ksp, which they gave me as 8x10-12, equals 2x squared times x, which is equal to 4x squared times x, which is equal to 4x cubed.0725

That gives me x3; when I divide both sides by 4, I get 2x10-12; and x is equal to the cube root of 2x10-12.0746

That is my answer: 67--that would be E; there you go.0762

We spent a fair amount of time on solubility.0770

OK, #68: Now, in this particular course in AP Chemistry, I did not discuss the bonding of solids in networks (for example, metallic bonding, ionic solids, things like that); that was one of the things that I left to you, because it was reasonably straightforward.0774

It was mostly informational--just stuff that you sort of have to know; there is nothing to explain, so much--nothing to sort of describe what is going on--no real calculations.0797

It was just one of those things that I felt that, if I had covered it, it would have just been sort of going through a laundry list of information; so that is the reason I didn't cover it, but it will show up in a couple of questions here, like this one.0808

In which of the following processes are covalent bonds broken?0822

OK, iodine solid to iodine gas--those are not covalent bonds that hold the iodine molecules together; those are London dispersion forces.0826

CO2 solid to CO2 gas--no; NaCl solid to NaCl liquid--no; diamond to carbon gas--yes.0838

Diamond is a covalent network solid; diamond is a bunch of carbons that are covalently bonded; gas--gaseous carbon--that means all of the bonds have been broken, and there are free carbon atoms just bouncing around all over the place--very high temperatures, but the answer is D.0848

The rest are essentially intermolecular forces; OK.0869

69: What is the final concentration of barium ions in a solution, when 100 milliliters of a .1 Molar barium chloride is mixed with 100 milliliters of a .05 Molar H2SO4?0873

OK, well, here is where we have to be careful: so let's do this--barium chloride dissolves into barium 2+, plus 2 Cl-; OK.0888

We have 100 milliliters, times 0.1 millimoles per milliliter; that gives us 10 millimoles of barium +...barium 2+; I'm sorry.0903

OK, so now, they say they mix it with 100 milliliters of a .05 Molar H2SO4.0920

What I have to recognize here is that this 10 millimoles of barium that is floating around in solution is now going to be bound up with sulfate, because barium sulfate is insoluble.0928

In other words, this barium ion that is floating around is going to react with the sulfate to form this solid barium sulfate (and remember, an arrow going down means that it is a...).0938

This is a 1:1 relationship, so we need to find out how many moles of sulfate there is.0951

Well, 100 milliliters, times 0.05 millimole per milliliter: that is going to give me, if I'm not mistaken, 5 millimoles of sulfate.0956

Well, if I have 5 millimoles of sulfate and 10 millimoles of barium, that means that 5 millimoles of the sulfate is going to react with 5 millimoles of the barium.0975

So, if 5 millimoles of the barium ends up reacting, I am going to be left with 5 millimoles left over, of barium floating around.0987

What is it that they want?--they want to know the final concentration; so millimoles over the total volume: I had 100 milliliters and 100 milliliters, so now I have a total volume of 200 milliliters; and that, when I do that, is going to equal...so 5/200 is 0.025 Molar.0997

That is it: so here, the crux of the problem is recognizing that barium sulfate is insoluble.1021

When you have some barium ions, there is a certain number of moles of that; you have some sulfate ion, a certain number of moles of that; you know that it is a 1:1 ratio.1028

Again, this is chemistry; chemistry is always going to involve some kind of a reaction, some kind of an equation; so when you write this equation, it's 1:1; 5 moles reacts with 5 moles; that leaves 5 moles left over.1036

Now, you have a total volume of 200 milliliters, so the molarity is .025 Molar.1049

OK, #70--we are almost done: we have 5 more questions to go.1054

OK, when 100 milliliters of a 1 Molar sodium phosphate is mixed with 100 milliliters of a 1 Molar AgNO3, a yellow precipitate forms, and silver ion becomes negligibly small.1060

Which of the following is a correct listing of the ions remaining in solution, in order of increasing concentration (lowest to highest)?1074

OK, so let's see: so let's do Na3PO4; 3 Na+ + PO43-.1083

OK, so 100 milliliters times 1 mole per liter...1 millimole per milliliter...100 millimoles...OK; here is where you are basically going to have to look at mole ratios.1101

1 mole of this produces 3 moles of that and 1 mole of that (right? 1 mole of sodium phosphate produces 3 moles of sodium ion and 1 mole of phosphate ion).1117

Well, now let's do AgNO3: AgNO3--one mole of this produces 1 mole of silver and 1 mole of nitrate.1131

This is just a mole ratio problem; well, the silver ion and the phosphate ion are going to mix: silver--three silvers--plus one phosphate is going to go to Ag3PO4 solid.1146

So, 3 moles of this, plus 1 mole of this, is going to form 1 mole of this.1167

So now, let's see what we have: here, I have 1 mole of the silver...100 milliliters...I actually don't have to use the 100 milliliters and the molarity--I can just use these numbers.1174

OK, I don't have to actually find out the number of moles; I can just work with mole ratios.1191

All of the silver is going to react with all of the phosphate.1195

Well, three moles of silver per one mole of phosphate...well, if I have 1 mole of silver, I need 3 moles to react with 1 mole of phosphate.1200

Therefore, all of the silver will be used up: so the silver ion...there is going to be nothing left.1213

The phosphate ion--well, 1 mole is produced; 1 mole is used; but if I have 1 mole of silver, I need 3 moles of silver for 1 mole of phosphate; therefore, since I only have 1 mole of silver, I'm only going to end up using 1/3 of the amount of phosphate.1222

Phosphate--I have 2/3 of that left over; and now, sodium--I have 3 moles left over, because it produced 3 moles of sodium; and the nitrate is 1 mole NO3-.1245

So, in terms of increasing concentration, I'm going to end up with PO4, and then the next highest concentration is going to be nitrate, and then the next highest concentration is going to be sodium.1271

That means A is our best answer.1292

Now, notice: I didn't have to use the 100 milliliters times the 1 Molar, 100 milliliters times the 1 Molar; I mean, I certainly recognize that this is 100 times 1 Molar of the sodium phosphate and 100 milliliters times 1 Molar of the silver nitrate.1297

I looked at those numbers; but because they were the same--the 100, 1; 100, 1; I realized that I could just go ahead and use 1 mole, straight forward.1313

100 milliliters times 1 millimole per milliliter gives me 100 millimoles; so I could use 100; I could use the numbers directly; but because it was the same, I could just pick a standard, and a standard is just 1--I could just use the mole ratios as is.1324

If those numbers were different--if it were something like 100 milliliters of a 1 Molar sodium phosphate and, let's say, 50 milliliters of a .5; then, I would actually have to use the specific number of millimoles, and I would have to do more conversions.1340

But here, because they were the same, I could just use 1, 3; 1, 1; 1, 1; 1, 1; 3 to 1.1353

I could just use them straight; I hope that makes sense.1360

OK, #71: Let's see, in a qualitative analysis for the presence of lead, iron, and copper ions in aqueous solution, which of the following will allow the separation of lead from the other ions at room temperature?1364

OK, let's see: we'll call this x; OK, so qualitative analysis: we discussed qualitative analysis briefly when we were talking about complex ions and ions in solution.1388

We talked about how to separate a collection of ions in a solution; so let's say you have lead, mercury, zinc, iron, copper, all floating around in a solution: you could add different things to the solution to precipitate out different ions.1414

In the case of lead, what we added was the first step in that qualitative analysis; if you look back in one of the lessons, I actually showed very, very carefully what the analysis was.1430

The first step was adding dilute hydrochloric acid to any collection of ions: that is sort of the first step in classical qualitative analysis.1441

You add dilute hydrochloric acid, and that automatically pulls out the lead ion and the mercury; in this case, because we have lead 2+, iron, and copper, you would add dilute hydrochloric acid.1451

So, in this case, B would be the best answer.1462

OK, #72: After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of 38%.1467

The correct value for the percentage of water in a hydrate is 51%: which of the following is the most likely explanation for this difference?1479

OK, the best way to think about this is: pick a sample size--I chose 10 grams--and then deal with that; so let's say I have a 10-gram sample.1487

If I heat up the sample to drive off the water, I'm going to end up with a certain mass--let's call it x grams.1500

OK, now, the percentage of water is going to be...so I have this 10-gram sample, which is a hydrate (it contains the sample and water attached to it); I drive off the water by heating it; I weigh it again; now it is going to weigh x grams.1508

Well, so the sample, in and of itself, would be the x grams: that is how much the sample weighs without any water attached to it.1523

Well, the water weight would be 10-x, because I had 10 grams to begin with, and I weighed out the dried sample; so 10-x is the water weight; over 10, times 100--that gives me the percentage of water.1532

OK, so now, they are telling me that this...well, you know what, I don't even need to multiply it by 100; I can just go ahead and just use decimals.1550

When they say 38%, that means this thing is equal to 0.38.1558

Now, they are saying what it should be is (oops, let's get rid of these stray lines)...they are telling me that what it should be is 51%.1564

What I should have is: 10-x, over 10, equals 0.51.1573

When I solve this for x, x equals 6.2; what I should have is x = 4.9.1583

In other words, when I measured the mass (or when this person doing the experiment measured the mass), they measured a mass that was higher than what should have been.1592

She got 38%; it should be 51%; this is the equation for how she got the 38%; this is the equation for the 51%.1602

Well, that means she measured a mass, after driving off the water, of 6.2 grams; the mass should have been 4.9 grams--it is heavier than it should have been.1613

Heavier than it should have been--that means it has more water; that means she didn't drive off all of the water.1623

Or, the other possibility: if she did drive off all of the water, after drying it, she didn't weigh it fast enough, and it ended up actually recollecting moisture.1630

There is actually still water attached to this, so B is the best answer for this.1641

The dehydrated sample absorbed moisture after heating: so she may have driven off the water, but before she got to weighing it, it actually reabsorbed, and therefore the mass that she measured was higher than it should have been.1647

It should have been 4.9; it ended up being 6.2, which gave her a lower percentage (38 instead of 51).1659

OK, 73 (we are almost done--it's kind of exciting!): OK, let's see: The volume of distilled water that should be added to 10 milliliters of a 6 Molar HCl in order to prepare a .5 Molar HCl solution is approximately what?1667

OK, this is a dilution problem; I'm going to use...OK, molarity 1 times volume 1 (or molarity initial...actually, let me use initial and final; I think it's better)...initial molarity times initial volume equals final molarity times final volume.1687

Well, the initial molarity is 6; the initial volume is 10 milliliters; and I can just leave it: 6, 10 milliliters, 6 millimoles per milliliter; so, I can just deal in milliliters; it's not a problem--I don't have to change to liters or anything like that.1717

The final molarity is 0.5; now, the final volume is 10+x: x is what we want--they want to know how much distilled water I have to add to this 10 milliliters, so it's 10+x.1739

There you go: you end up with 60=5+0.5x; you end up with 0.5x=55; x=55/.5, which is 110: your answer is 110 milliliters of distilled water, which is D; that is it.1755

OK, #74: Which of the following gases deviates most from ideal behavior?1779

The gas that will deviate most from ideal behavior is the gas that is the most polar.1785

SO2 happens to be the most polar; neon, no; CH4, no; N2, no; H2, no; so SO2, by process of elimination.1791

By process of elimination, you can get to SO2; however, just so you know, the Lewis structure is this.1801

There is a dipole moment here--the net dipole is this way: it is a polar molecule; it will deviate from ideal behavior.1811

So, the answer is A for 74.1819

OK, and #75, the last question of the multiple choice: All right, which of the following pairs of liquid forms the solution that is most ideal (most closely follows Raoult's Law)?1823

An ideal solution (in other words, a mixture of two different liquids) is when the liquids that you mix are the most alike.1835

In this list, A: C8H18: it is a hydrocarbon, nonpolar; H2O: hydrogen bonding and polar; no way.1847

You have the propanol, and you have water; they are kind of cool, but you have this long carbon chain on here; not really.1858

The propanol and the C8H18...not really.1866

Of the choices, C6H14, C8H18...they are practically identical; they are only off by two carbons; they are pure hydrocarbons; the answer is D.1870

75--the answer is D: an ideal solution--the one that obeys Raoult's Law, one that doesn't deviate positively or negatively (remember?--that is one of the last lessons that we discussed--Raoult's Law and deviation)--you want the ideal solution to be where the two solutions are the most alike.1881

So, for example, if I had something like water and methanol, water and methanol are very much alike: there is hydrogen bonding...even though there is that one carbon, they are actually quite alike.1900

Water and acetone: a great mixture--you have hydrogen bonding...I'm sorry, no: water and acetone--that will actually end up giving you deviation.1911

In this case, you want an ideal solution, so that they are alike, so that the interactions between particles of A among themselves and particles of B among themselves are the same as the interactions between the particles of A and B.1921

In this particular case, the C6H14 and the C8H18...the interactions among these and the interactions among these are virtually identical with the interactions among those.1938

That will give you the most ideal situation.1954

OK, that takes care of the multiple choice for this practice exam.1959

Thank you for joining us here at Educator.com, and for AP Chemistry.1963

Next lesson, we will start our discussion of the free response questions; take care; goodbye.1966

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