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Slide
1
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Chapter 15
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Solutions
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Slide
Chapter 15
2
Table of Contents
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
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Solubility
Solution Composition: An Introduction
Solution Composition: Mass Percent
Solution Composition: Molarity
Dilution
Stoichiometry of Solution Reactions
Neutralization Reactions
Solution Composition: Normality
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Slide
2
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Chapter 15
3
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What is a Solution?
• Solution – homogeneous mixture
 Solvent – substance present in largest
amount
 Solutes – other substances in the solution
 Aqueous solution – solution with water as the
solvent
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Section 15.1
4
Solubility
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Various Types of Solutions
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Section 15.1
5
Solubility
4
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Solubility of Ionic Substances
•
Ionic substances breakup into individual
cations and anions.
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Section 15.1
6
Solubility
5
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Solubility of Ionic Substances
•
Polar water molecules interact with the positive
and negative ions of a salt.
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Section 15.1
7
Solubility
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Solubility of Polar Substances
•
Ethanol is soluble in water because of the
polar OH bond.
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Section 15.1
8
Solubility
7
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Solubility of Polar Substances
•
Why is solid sugar soluble in water?
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Section 15.1
9
Solubility
8
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Substances Insoluble in Water
•
•
Nonpolar oil does not interact with polar water.
Water-water hydrogen bonds keep the water
from mixing with the nonpolar molecules.
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Section 15.1
10
Solubility
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How Substances Dissolve
•
•
•
A “hole” must be made in the water structure
for each solute particle.
The lost water-water interactions must be
replaced by water-solute interactions.
“like dissolves like”
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Section 15.1
11
Solubility
10
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Concept Check
___________________________________
Which of the following solutes will generally not
dissolve in the specified solvent? Choose the best
answer. (Assume all of the compounds are in the
liquid state.)
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a)
b)
c)
d)
CCl4 mixed with water (H2O)
NH3 mixed with water (H2O)
CH3OH mixed with water (H2O)
N2 mixed with methane (CH4)
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Section 15.2
12
Solution Composition: An Introduction
11
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•
The solubility of a solute is limited.
 Saturated solution – contains as much
solute as will dissolve at that
temperature.
 Unsaturated solution – has not
reached the limit of solute that will
dissolve.
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Section 15.2
13
Solution Composition: An Introduction
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•
Supersaturated solution – occurs when a
solution is saturated at an elevated
temperature and then allowed to cool but
all of the solid remains dissolved.
 Contains more dissolved solid than a
saturated solution at that temperature.
 Unstable – adding a crystal causes
precipitation.
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Section 15.2
14
Solution Composition: An Introduction
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•
•
Solutions are mixtures.
Amounts of substances can vary in different
solutions.
 Specify the amounts of solvent and
solutes.
 Qualitative measures of concentration
 concentrated – relatively large amount
of solute
 dilute – relatively small amount of solute
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Section 15.3
15
Solution Composition: Mass Percent
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Mass percent =
Mass percent =
___________________________________
mass of solute
 100%
mass of solution
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grams of solute
 100%
grams of solute + grams of solvent
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Slide
Section 15.3
16
Solution Composition: Mass Percent
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Exercise
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What is the percent-by-mass concentration of
glucose in a solution made my dissolving 5.5 g
of glucose in 78.2 g of water?
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6.6%
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[5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6% glucose
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Slide
Section 15.4
17
Solution Composition: Molarity
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•
Molarity (M) = moles of solute per
volume of solution in liters:
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moles of solute
M = Molarity =
liters of solution
3 M HCl =
___________________________________
6 moles of HCl
2 liters of solution
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Slide
Section 15.4
18
Solution Composition: Molarity
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Exercise
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You have 1.00 mol of sugar in 125.0 mL of
solution. Calculate the concentration in units
of molarity.
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8.00 M
___________________________________
1.00 mol / (125.0 / 1000) = 8.00 M
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Slide
Section 15.4
19
Solution Composition: Molarity
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Exercise
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A 500.0-g sample of potassium phosphate
is dissolved in enough water to make 1.50 L
of solution. What is the molarity of the
solution?
1.57 M
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500.0 g is equivalent to 2.355 mol K3PO4 (500.0 g /
212.27 g/mol). The molarity is therefore 1.57 M
(2.355 mol/1.50 L).
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Slide
Section 15.4
20
Solution Composition: Molarity
___________________________________
Exercise
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You have a 10.0 M sugar solution. What
volume of this solution do you need to have
2.00 mol of sugar?
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0.200 L
___________________________________
2.00 mol / 10.0 M = 0.200 L
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Slide
Section 15.4
21
Solution Composition: Molarity
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Exercise
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Consider separate solutions of NaOH and KCl
made by dissolving 100.0 g of each solute in
250.0 mL of solution. Calculate the
concentration of each solution in units of
molarity.
10.0 M NaOH
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[100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 M NaOH
5.37 M KCl
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[100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 M KCl
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Slide
Section 15.4
22
Solution Composition: Molarity
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Concept Check
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You have two HCl solutions, labeled Solution A and
Solution B. Solution A has a greater concentration than
Solution B. Which of the following statements are true?
a)
b)
c)
d)
If you have equal volumes of both solutions,
Solution B must contain more moles of HCl.
If you have equal moles of HCl in both solutions,
Solution B must have a greater volume.
To obtain equal concentrations of both solutions,
you must add a certain amount of water to
Solution B.
Adding more moles of HCl to both solutions will make
them less concentrated.
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22
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Slide
Section 15.4
23
Solution Composition: Molarity
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Concentration of Ions
•
For a 0.25 M CaCl2 solution:
CaCl2 → Ca2+ + 2Cl–
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Ca2+:
Ca2+

1 × 0.25 M = 0.25 M
–
 Cl : 2 × 0.25 M = 0.50 M Cl–.
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Slide
Section 15.4
24
Solution Composition: Molarity
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Concept Check
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Which of the following solutions contains
the greatest number of ions?
a)
b)
c)
d)
400.0
300.0
200.0
800.0
mL of 0.10
mL of 0.10
mL of 0.10
mL of 0.10
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M NaCl.
M CaCl2.
M FeCl3.
M sucrose.
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Slide
Section 15.4
25
Solution Composition: Molarity
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___________________________________
Let’s Think About It
•
Where are we going?

•
___________________________________
To find the solution that contains the greatest
number of moles of ions.
___________________________________
How do we get there?


Draw molecular level pictures showing each
solution. Think about relative numbers of ions.
How many moles of each ion are in each
solution?
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Section 15.4
26
Solution Composition: Molarity
25
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Notice
•
The solution with the greatest number of
ions is not necessarily the one in which:
 the volume of the solution is the
largest.
 the formula unit has the greatest
number of ions.
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Section 15.4
27
Solution Composition: Molarity
26
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Standard Solution
•
A solution whose concentration is
accurately known.
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Slide
Section 15.4
28
Solution Composition: Molarity
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To Make a Standard Solution
•
•
•
Weigh out a sample of solute.
Transfer to a volumetric flask.
Add enough solvent to mark on flask.
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Section 15.5
29
Dilution
28
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•
•
•
The process of adding water to a
concentrated or stock solution to achieve
the molarity desired for a particular
solution.
Dilution with water does not alter the
numbers of moles of solute present.
Moles of solute before dilution = moles of
solute after dilution
M1V1 = M2V2
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Section 15.5
30
Dilution
29
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Diluting a Solution
•
•
Transfer a measured amount of original solution
to a flask containing some water.
Add water to the flask to the mark (with swirling)
and mix by inverting the flask.
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Section 15.5
31
Dilution
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Concept Check
A 0.50 M solution of sodium chloride in an open
beaker sits on a lab bench. Which of the
following would decrease the concentration of
the salt solution?
Add water to the solution.
Pour some of the solution down the sink drain.
Add more sodium chloride to the solution.
Let the solution sit out in the open air for a
couple of days.
e) At least two of the above would decrease the
concentration of the salt solution.
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___________________________________
a)
b)
c)
d)
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Section 15.5
32
Dilution
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31
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Exercise
___________________________________
What is the minimum volume of a 2.00 M
NaOH solution needed to make 150.0 mL of
a 0.800 M NaOH solution?
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60.0 mL
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M1V1 = M2V2
(2.00 M)(V1) = (0.800 M)(150.0 mL)
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Section 15.6
33
Stoichiometry of Solution Reactions
32
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Steps for Solving Stoichiometric Problems Involving Solutions
1. Write the balanced equation for the reaction. For
reactions involving ions, it is best to write the net
ionic equation.
2. Calculate the moles of reactants.
3. Determine which reactant is limiting.
4. Calculate the moles of other reactants or
products, as required.
5. Convert to grams or other units, if required.
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Section 15.6
34
Stoichiometry of Solution Reactions
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Concept Check (Part I)
___________________________________
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What precipitate will form?
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lead(II) phosphate, Pb3(PO4)2

___________________________________
What mass of precipitate will form?
1.1 g Pb3(PO4)2
___________________________________
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Slide
Section 15.6
35
Stoichiometry of Solution Reactions
___________________________________
Let’s Think About It
•
Where are we going?

•
___________________________________
To find the mass of solid Pb3(PO4)2 formed.
How do we get there?






___________________________________
What are the ions present in the combined solution?
What is the balanced net ionic equation for the
reaction?
What are the moles of reactants present in the
solution?
Which reactant is limiting?
What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?
___________________________________
___________________________________
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___________________________________
Slide
Section 15.6
36
Stoichiometry of Solution Reactions
___________________________________
Concept Check (Part II)
___________________________________
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
___________________________________
___________________________________
 What is the concentration of nitrate ions
left in solution after the reaction is
complete?
___________________________________
___________________________________
0.27 M
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Slide
Section 15.6
37
Stoichiometry of Solution Reactions
___________________________________
___________________________________
Let’s Think About It
•
Where are we going?

•
___________________________________
To find the concentration of nitrate ions left in
solution after the reaction is complete.
___________________________________
How do we get there?


What are the moles of nitrate ions present in the
combined solution?
What is the total volume of the combined
solution?
___________________________________
___________________________________
___________________________________
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___________________________________
Slide
Section 15.6
38
Stoichiometry of Solution Reactions
___________________________________
Concept Check (Part III)
___________________________________
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
___________________________________
___________________________________
 What is the concentration of phosphate
ions left in solution after the reaction is
complete?
___________________________________
___________________________________
0.011 M
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Section 15.6
39
Stoichiometry of Solution Reactions
38
___________________________________
___________________________________
Let’s Think About It
•
Where are we going?

•
___________________________________
To find the concentration of phosphate ions left in
solution after the reaction is complete.
___________________________________
How do we get there?




What are the moles of phosphate ions present in
the solution at the start of the reaction?
How many moles of phosphate ions were used
up in the reaction to make the solid Pb3(PO4)2?
How many moles of phosphate ions are left over
after the reaction is complete?
What is the total volume of the combined
solution?
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___________________________________
___________________________________
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39
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Section 15.7
40
Neutralization Reactions
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•
•
•
An acid-base reaction is called a
neutralization reaction.
Steps to solve these problems are the same
as before.
For a strong acid and base reaction:
H+(aq) + OH–(aq)  H2O(l)
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___________________________________
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Section 15.7
41
Neutralization Reactions
___________________________________
Concept Check
___________________________________
For the titration of sulfuric acid (H2SO4) with
sodium hydroxide (NaOH), how many moles
of sodium hydroxide would be required to
react with 1.00 L of 0.500 M sulfuric acid?
___________________________________
___________________________________
1.00 mol NaOH
___________________________________
___________________________________
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Section 15.7
42
Neutralization Reactions
41
___________________________________
___________________________________
Let’s Think About It
•
Where are we going?

•
___________________________________
To find the moles of NaOH required for the
reaction.
___________________________________
How do we get there?




What are the ions present in the combined
solution? What is the reaction?
What is the balanced net ionic equation for the
reaction?
What are the moles of H+ present in the solution?
How much OH– is required to react with all of the
H+ present?
___________________________________
___________________________________
___________________________________
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Section 15.8
43
Solution Composition: Normality
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___________________________________
Unit of Concentration
•
•
•
One equivalent of acid – amount of acid
that furnishes 1 mol of H+ ions.
One equivalent of base – amount of base
that furnishes 1 mol of OH ions
Equivalent weight – mass in grams of 1
equivalent of acid or base.
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
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Section 15.8
44
Solution Composition: Normality
43
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
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Section 15.8
45
Solution Composition: Normality
44
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
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___________________________________
Slide
Section 15.8
46
Solution Composition: Normality
___________________________________
Normality = N =
___________________________________
number of equivalents
equivalents
equiv
=
=
1 liter of solution
liter
L
___________________________________
•
To find number of equivalents:
N  V=
___________________________________
equiv
 L = equiv
L
___________________________________
___________________________________
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Section 15.8
47
Solution Composition: Normality
46
___________________________________
___________________________________
Concept Check
___________________________________
If Ba(OH)2 is used as a base, how many
equivalents of Ba(OH)2 are there in 4 mol
Ba(OH)2?
a)
b)
c)
d)
___________________________________
___________________________________
2
4
8
16
___________________________________
___________________________________
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48
47
___________________________________
Section 15.8
Solution
Chapter
15Composition:
Homework Normality
___________________________________
Homework
___________________________________
• Reading assignment
– Pages 475 through 502
___________________________________
• Homework Problems
– Questions and problems 3, 5, 7, 11, 17, 19,
21, 23, 25, 27, 31, 35, 37, 39, 41, 43, 45, 47,
49, 51, 57, 59, 61, 71, 73, 85, 87.
___________________________________
___________________________________
• Due on
___________________________________
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48