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Wednesday, October 24th
•Review
•Quiz
•Finish Chapter 6,
•Begin Chapter 7
Frequency: CD
Remaining Scedule
Month
October
November
December
Holidays:
Day
Topic
22
24
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Finish Chapter 6
Quiz #7, Review Chapter 6, Lecture on Chapter 7.1-7.4
Lecture on Chapter 7.4-7.5, In Class Group Work
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Lecture on Chapter 7..6-7.8
Quiz #8, Lecture on Chapter 7.9
In Class Group Work
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7
9
Test #2
Lecture on Chapter 8.1-8.3
Lecture on Chapter 8.4, 8.6 In Class Group Work
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14
16
No Class
Quiz #9, Lecture on Chapter 10.1
In Class Group Work
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No Class
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28
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Lecture on Chapter 10.1-10.3
Quiz #10 Lecture on Chapter 10.4-10.5
In class Group Work
3
5
7
Lecture on Chapter 10.6 In Class Group Work
Test #3
Final Exam Review
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12
Final Exam- 3pm class, 11:30-1:00
Final Exam- 8am class, 1:15- 2:45
Monday, November 12- Veteran’s Day
November 21-25- Thanksgiving Break
2
Review
3
Practice
Balance the equations for the combustion
of propane (C3H8).
4
Avogadro’s Number as an Equality
Avogadro’s number (6.02 x 1023) can be
written as an
equality and two conversion factors.
Equality:
Conversion Factors:
Subscripts State Atoms and Moles
Molar Mass
Molar mass is
 the mass of one mole of a
substance.
 the number of grams that
equals the atomic mass
of that element.
Molar mass is rounded to
the tenths (0.1 g) place for
use in this text.
Quiz
• 7 questions, 2 pts each
1. What are the molar coefficients
needed to balance the following
equation?
Al (s) + HCl (aq)  AlCl3 (aq) + H2 (g)
a)
b)
c)
d)
© 2013 Pearson Education, Inc.
1:4:1:2
2:3:2:3
2:6:2:3
4:6:4:3
Copyright © 2011 Pearson Education Inc.
2. How many moles of oxygen can be
produced from one mole of KClO3 in
the reaction KClO3  KCl + O2?
a)
b)
c)
d)
© 2013 Pearson Education, Inc.
1 mol
1.5 mol
2 mol
3 mol
Copyright © 2011 Pearson Education Inc.
3. The number of C atoms in 0.500
mole C is_______.
a)
b)
c)
d)
2.50 × 1023 C atoms
3.01 × 1023 C atoms
5.01 × 1023 C atoms
6.00 × 1023 C atoms
© 2013 Pearson Education, Inc.
Copyright © 2011 Pearson Education Inc.
4. How many moles of O are there in 2
moles of glucose (C6H12O6)?
a) 2 moles
b) 24 moles
c) 6 moles
d) 12 moles
© 2013 Pearson Education, Inc.
Copyright © 2011 Pearson Education Inc.
5. What is the molar mass of S3 if the
molar mass of sulfur is 32.0?
a) 64 g
b) 32 g
c) 94 g
d) 10.7 g
© 2013 Pearson Education, Inc.
Copyright © 2011 Pearson Education Inc.
6. What is the percent yield if 50 g of
CO2 were collected and 200g were
expected?
a)
b)
c)
d)
© 2013 Pearson Education, Inc.
50%
25%
75%
100%
Copyright © 2011 Pearson Education Inc.
7. Which of the following reactions
can be classified as decomposition?
a)
b)
c)
d)
CuCO3 (s)  CuO (s) + CO2 (g)
4Fe (s) + 3O2 (g)  2Fe2O3 (s)
Mg (s) + 2AgNO3 (aq)  Mg(NO3)2 (aq) + 2Ag (s)
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
© 2013 Pearson Education, Inc.
Copyright © 2011 Pearson Education Inc.
Continuing on in
Chapter 6
16
Learning Check
If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2
can be produced?
Solution
If 48.2 grams of Ca are mixed with 31.0 grams
of N2, how many grams of Ca3N2 can be
produced?
Step 1 State the given and needed quantities.
Analyze the Problem.
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of
N2, how many grams of Ca3N2 can be produced?
Step 2 Use coefficients to write mole–mole
factors; write molar mass factors, if needed.
1 mole of Ca = 40.1 g of Ca
1 mole of N2 = 28.0 g of N2
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of
N2, how many grams of Ca3N2 can be produced?
Step 2 Use coefficients to write mole–mole
factors; write molar mass factors, if needed.
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
Step 3 Calculate the number of moles of product
from each reactant and determine the limiting
reactant.
The limiting reactant is Ca, which produces 0.402 mole Ca3N2.
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of
N2, how many grams of Ca3N2 can be produced?
Use the molar mass to convert the smaller
number of moles of product to grams.
Reaction Conditions
There are three conditions required for a
chemical reaction to occur.
1. Collision: The reactants must collide.
2. Orientation: The reactants must align
properly to .
break and form bonds.
3. Energy: The collision must provide the
energy of .
activation.
Activation Energy
 Activation energy is the amount of energy
required to break the bonds between
atoms of the reactants.
 If the energy of a collision is less than the
activation energy, the molecules bounce
apart without reacting.
 Many collisions occur, but only a few
actually lead to the formation of product.
Activation Energy
The activation energy is the energy
needed to convert reacting molecules
into products.
Heat of Reaction
The heat of reaction
 is the amount of heat
absorbed or released
during a reaction.
 is the difference between
the energy of breaking
bonds in the reactants
and forming bonds in the
products.
ΔH =
 has the symbol ΔH.
ΔHproducts − ΔHreactants
Exothermic Reactions
In an exothermic reaction,
 the energy of the reactants
is greater than that of the
products and heat is
released along with the
products.
 the heat of reaction (ΔH)
value is written with a
negative sign (–) indicating
heat is released.
Endothermic Reactions
In an endothermic reaction,
 the energy of the reactants
is lower than that of the
energy of the products.
 heat is absorbed and used
to convert the reactants to
products and written with
a (+) sign.
Heat of Reaction Summary
Reaction
Type
Endothermic
Exothermic
Energy
Heat
Change
in Reaction ΔH
Heat absorbed Reactant side
Heat released
Product side
Sign of
+
–
Common Gases
Of the elements on the periodic table, some exist as a
gas at room temperature, these include
 the Noble Gases, Group 8A (18),
 H2, N2, O2, F2, Cl2, and
 many oxides of nonmetals such as CO, CO2, NO, NO2, SO2,
and SO3.
30
Gases and Environmental Concerns
Some gases are responsible for environmental and
health concerns including
 methane, CH4,
 chlorofluorocarbons (CFCs),
 nitrogen oxides found in smog, and
 volatile organic compounds (VOCs), such as
compounds found in paint thinners.
31
Kinetic Theory of Gases
A gas consists of small particles that
• move randomly with high velocities.
• have essentially no attractive (or
repulsive) forces toward each other.
• have a very small volume compared
to the volume of the containers they
occupy.
• are in constant motion.
• have kinetic energies that increase
with an increase in temperature.
32
Properties of Gases
Gases are described in terms of four
properties:
1. pressure (P),
2. volume (V),
3. temperature (T)
4. amount (n).
33
Gas Pressure
Gas pressure
• is the force acting on a specific area.
• has units of atm, mmHg, torr, lb/in.2, and
kilopascals(kPa).
1 atm
=
760 mmHg (exact)
1 atm
=
760 torr (exact)
1 atm
=
14.7 lb/in.2
1 atm
=
101.325 kPa
34
Practice
1.
What is 475 mmHg expressed in
atmospheres (atm)?
35
2. The pressure in a tire is 2.00 atm. What is this
pressure in mmHg?
Solution
1. What is 475 mmHg expressed in
atmospheres (atm)?
B. 0.625 atm
2. The pressure of a tire is measured as 2.00
atm. What is this pressure in mmHg?
B. 1520 mmHg
37
Atmospheric Pressure
Atmospheric pressure is the
pressure exerted by a column of air
from the top of the atmosphere to
the surface of the Earth.
• is about 1 atmosphere at sea
level.
• depends on the altitude and the
weather.
• is lower at high altitudes where
the density of air is less.
• is higher on a rainy day than on
a sunny day.
38
Boyle’s Law
Boyle’s law states that
• the pressure of a gas
is inversely related to
its volume when
temperature (T) and
amount of gas (n) are
constant.
• if the pressure (P)
increases, then the
volume (V)
decreases.
39
Solving for a Gas Law Factor
A sample of nitrogen gas (N2) has a
volume of 4.2 L at 1.0 atm. If the pressure
is decreased to 0.75 atm with no change in
the temperature or amount of gas, what
will be the new volume?
40
Solving for a Gas Law Factor
A sample of nitrogen gas (N2) has a volume of 4.2 L
at 1.0 atm. If the pressure is decreased to 0.75 atm
with no change in the temperature or amount of
gas, what will be the new volume?
Step 2 Rearrange the gas law equation to solve for the
unknown quantity.
P1V1 = P2V2
Boyle’s Law
To solve for V2 , divide both sides by P2.
41
Charles’s Law
Charles’s law states that,
• the Kelvin temperature of a
gas is directly related to the
volume of the gas,
• P and n are constant, and
• as the temperature of a gas
increases, the molecules
move faster and its volume
increases to maintain constant
P.
42
Practice
A balloon has a volume of 785 mL at 21 ˚C. If
the
temperature drops to 0 ˚C, what is the new
volume of the balloon (P constant)?
43
Solution
A balloon has a volume of 785 mL at 21 ˚C. If the
temperature drops to 0 ˚C, what is the new volume of
the balloon (P constant)?
-Rearrange the gas law equation to solve for
the unknown quantity.
-Substitute values into the gas law equation and the
table.
44
Gay-Lussac’s Law: P and T
Gay-Lussac’s law states
that
• the pressure exerted by
a gas is directly related
to the Kelvin
temperature.
• V and n are constant.
• an increase in
temperature increases
the pressure of a gas.
45
Practice
Solve for the final pressure of a gas with an
initial pressure of 1.20 atm at 75 ˚C when
cooled to −22 ˚C.
(V and n constant)
46
Solution
Solve for the final pressure of a gas with an
initial pressure of 1.20 atm at 75 ˚C when
cooled to −22 ˚C.
Step 3 Substitute values into the gas law
equation and calculate.
47
Combined Gas Law
The combined gas law uses Boyle’s Law, Charles’s
Law, and Gay-Lussac’s Law (n is constant).
48
Learning Check
A gas has a volume of 675 mL at 35 °C and
0.850 atm pressure. What is the volume (mL)
of the gas at −95 °C and a pressure of 802
mmHg (n constant)?
49
Solution
A gas has a volume of 675 mL at 35 °C and
0.850 atm pressure. What is the volume (mL) of
the gas at −95 °C and a pressure of 802 mmHg
(n constant)?
50
Next Week
Quiz - Written Chapter 7
Finish Chapter 7
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