Download CHEMISTRY 133 LECTURE / STUDY GUIDE FOR R.H. LANGLEY

CHEMISTRY 133
LECTURE / STUDY GUIDE
FOR
R.H. LANGLEY
Note: Make sure you read the introduction.
© 2016 Sevagram Enterprises
without written permission
No part of this book may be reproduced, in any form or by any means,
INTRODUCTION
Note: If you did not get this until after the beginning of the term, do not just start using it
where you currently are in class. You should, at the very minimum, look over all
comments from the beginning of these notes. You do not want to miss any of the
comments, because one of them may be the most important one for you.
This guide is to supplement the lecture. Its usefulness will depend on how much you add.
This package outlines the class notes and gives additional comments. Not everything said in
class will be in these notes; you will need to add extra material. Numerous studies show that
the more you add the more helpful class notes are. There is additional information in these
notes to help you understand the material, many of these items go beyond what is in the
lecture, but you have it here to help you. You should intersperse class notes and problems in
the appropriate places. Simply reading these notes or reading what someone else has written,
is a total waste of your time. What you write in here and what you do here is the only way to
make these notes of any benefit to you.
This guide will not replace the Syllabus. Everything you need to know to get through this
class is in the Syllabus. Many students have had their course grade significantly lowered by
not reading the Syllabus. This booklet will help you implement many of the suggestions
given in the Syllabus. Anytime that you seem to be having trouble in class you should reread the Syllabus and/or the notes contained in this supplement.
You may use a calculator on exams. You must use your own calculator. A
fancy/complicated calculator is not required. If you buy one and are going to use it for
Chemistry 134 get one that does square roots, logs, and lns. You should use the same
calculator on the homework that you will use on the test.
You may use a programmable calculator. However, you should be aware of the fact that on a
recent exam, with approximately equal numbers of students using programmable and
nonprogrammable calculators, there was a significant difference in their average grades. The
average for those students using programmable calculators was 49.50, while that of students
using nonprogrammable calculators was 66.56. Thus, students using a programmable
calculator averaged over 17 points below the other students.
THERE IS NO SUBSTITUTE FOR CLASS ATTENDANCE – DO NOT MISS CLASS.
During a regular semester, expect each absence to lower your grade on the next test by 10-12
points. During a summer session, expect this value to double. The basis for these
predictions is class performance on the exams during the past several terms.
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Note: Enter notes on these pages or, less effectively, on notebook paper to insert into the
appropriate place, recopying may help.
Note: All chapter numbers, section numbers, problems etc. are keyed to N.J. Tro,
Chemistry, Pearson/Prentice Hall, 2008.
Note: If at any point you begin to feel overwhelmed, you should see the instructor.
CHAPTER 1
1.1 (This, and all other section numbers, refer to the appropriate sections in BLBMWS.)
Read
1.2-1.3 – Classification of Matter/Properties of Matter
Matter:
Mass:
Note: While mass and weight are technically different, they are interchangeable in this
course.
Matter normally occurs in one of three states:
1.
2.
3.
Law of Conservation of Mass
It is possible to describe Matter by changes/properties:
Physical Change:
Chemical Change (Chemical Reaction):
Energy
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Intensive Property:
Extensive Property:
Matter may occur as a pure substance or as a mixture.
Pure Substance:
Mixture:
Element:
Compound:
Mixture:
Heterogeneous Mixture:
Homogeneous Mixture (Solution):
1.4 Units of Measurement
Units: SI
NOTE: you are responsible for the base units in the textbook.
BASE UNITS:
Mass
Amount of a substance
Length
Electric Current
Time
Temperature
Luminous Intensity
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A combination of units may be necessary:
Speed =
Sometimes it is necessary to use the same unit more than once:
Area =
Temperature scales: °F, °C, K (no degree symbol on K)
K = °C + 273.15
(or K = °C + 273)
If the base unit is too large or too small for convenient use, it is possible to add a
multiplier.
A prefix indicates the presence of a multiplier.
i.e., 1 000 000 m =
NOTE: you are responsible for prefixes, except as discussed in class.
KNOW:
M = mega
=
k = kilo
=
d = deci
=
c = centi
=
m = milli
=
 = micro
=
n = nano
=
p = pico
=
In order to simplify expressing some numbers use scientific notation.
2 457 000 000 = 2.457 × 109 0.000 000 239 8 = 2.398 × 10–7
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NOTE:
1.00 Liter =
NOTE:
You should also know:
1.00 mL =
All SI-SI conversions (mostly from prefixes)
All common English-English length and mass relations
K = °C + 273.15
(or K = °C + 273)
One English-SI conversion for length, and one for mass
Note: A very common, and expensive (point wise), mistake is to try to learn more than these
do. Study for the test; do not waste your time on unnecessary conversions.
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
Note: If you have not completed problems for the previous section, do so as soon as
possible.
1.5 Uncertainty in Measurements
All calculated values consist of two parts: a number and a unit.
BOTH must be present to receive full credit on an exam.
Exact Numbers:
Measured Numbers:
Example: Multiple measurements:
6.75
6.74
6.77
6.76
6.73
6.76
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Significant Figures:
Rules:
Rules for calculations:
Exact numbers have no effect on the number of significant figures.
Multiplication & Division –
4.2 x 16.3
1456
Addition & Subtraction –
55.1|5
23.2|
27.1|3
Do not worry about rounding (as long as you are reasonable). DO NOT make the mistake of
rounding too soon.
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1.6 Dimensional Analysis
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
The key to the ease/difficulty of this course is the Factor-Label Method (Dimensional
Analysis). The sooner you confine your problem solving to this method the easier the
remainder of the course will be. Resist any attempt to solve problems by other methods.
You may find that you are more comfortable with other methods, but they will let you down
in the end. For example, you will not be able to do most of Chapter 6 without knowing the
Factor-Label Method, and if you wait until then to learn it you will be impossibly far behind.
Do not waste your time learning extra conversions. The time that these extras will save you
is time lost from studying important material for the exam. A common example is to learn
the prefixes (as required) followed by a number of superfluous conversions such as: 10 mm
= 1 cm, 100 cm = 1 m, 1000 mm = 1 m, 10 cm = 1 dm, and 10 dm = 1 m. While these
conversions are correct, once you have learned the prefixes none of these are necessary, and
a waste of your time. You should spend your time studying for the test not learning
superfluous conversions.
Example: How many cm are in 2 inches?
Example: How many meters are in 2.00 miles?
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Density =
Note: For this course Density = Mass/Volume is a definition NOT a mathematical
relationship. Using it as a mathematical relation may make a few Chapter 1 problems
easier – but it will put you significantly behind in preparing for other chapters.
Example: What is the density of a liquid for which 2.00 ft3 weighs 125 lbs?
Example: Express this density as g/cm3.
Example: How many grams do 500. mL of a liquid weigh if the density of the liquid
is 1.53 g/cm3?
Example: What is the volume of a 125 g sample of a liquid if the density of the liquid
is 0.8372 g/cm3?
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
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Study Guide Example – not covered in class:
Try to solve this problem without looking at the solution.
How many millimeters thick is a piece of aluminum foil if a piece measuring 18.0
inches by 20.0 inches weighs 9.875 grams. The density of aluminum is 2.70 g / cm3.
9.875g 
cm 3  1in 


 2.70g  2.54cm 
2

1

2
 18.020.0in
 0.01  m 
–2


 = 1.57 × 10 mm
 c  0.001 
Note that no equations appear in the solution, and the only information needed that
was not in the problem were simple conversions. This is how a factor label
problem should be set-up. As with any factor label problem, you could have
started with any step, and it is still correct. Any step(s) that you added, or any
equations you used, or any conversions you did elsewhere, or any different
conversions that you used were a waste of your time.
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
Note: The final review for the test needs to be as close to test conditions as possible.
Thus, you should work the problems during this final review by yourself and
“closed book.”
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Reminder: Enter notes on these pages or, less effectively, on notebook paper to insert into
the appropriate place, recopying may help.
Reminder: If at any point you begin to feel overwhelmed, you should see the instructor.
CHAPTER 2
2.1 The Atomic Theory of Matter
(This, and all other section numbers, refer to the appropriate sections in BLBMWS.)
Dalton's Atomic Theory
1.
2.
3.
4.
5.
Law of definite proportions (Law of constant composition)
Law of multiple proportions
2.2 The Discovery of Atomic Structure
Particle
charge
(Coulombs)
Electron
Proton
Neutron
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mass
(Grams)
Rutherford (1911)
2.3 The Modern View of Atomic Structure
Atomic Number:
Atom:
A chemical symbol designates an element – an abbreviation of one or two letters.
1st (or only) must be capitalized.
2nd (if present) must not be capitalized.
NOTE: Be careful on exams.
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Examples:
4
3
Chlorine
Cl 12
Cl
Oxygen
O
Area 1:
Area 2:
Area 3:
Area 4
Mass Number
Isotope:
This information must be given to you; it is NOT normally given on
the Periodic Table. (Many students get into trouble by assuming that
the mass number is present.)
For Cl, the common values are 35 and 37
2.4 Atomic Weights
The masses will be very important. A mass scale has been set up to deal with atoms.
The basic unit is the Atomic Mass Unit (amu). (1 amu = 1 Dalton)
Atomic Mass Unit
Atomic Mass (Weight):
Example: Natural lead contains 1.4% lead-204 (203.973 amu), 24.1% lead-206
(205.9745 amu), 22.1% lead-207 (206.9759 amu), and 52.4% lead-208 (207.9766
amu). Calculate its atomic weight.
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Study Guide Example – not covered in class:
Fill in the gaps in the following table:
Symbol
Protons
Neutrons
Electrons
Net Charge
__ 122 Sb 3 __
__________
__________
__________
__________
__________
____74____
___110____
____71____
__________
__________
__________
___125____
____80____
____2+____
__________
____90____
___142____
__________
____4+____
__ 122 Sb 3 __
____51____
____71____
____54____
____–3____
__ 184 W 3 _
____74____
___110____
____71____
____+3____
_ 207 Pb 2 __
____82____
___125____
____80____
____2+____
__ 232 Th 4 _
____90____
___142____
____86____
____4+____
Answers:
Symbol
Protons
Neutrons
Electrons
Net Charge
2.5 Periodic Table
The symbols of the elements are gathered together on the Periodic Table (a summary
of the various elements grouped by their properties).
Groups (Families):
Period:
The elements on the left side (except H) are _______________
The elements on the right side are _______________
The elements between these are called _______________
Metal:
Nonmetal:
Metalloid (Semimetal):
NOTE: Hydrogen is often an exception.
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Certain columns (families) have names.
Alkali Metal:
Alkaline Earth Metal:
Halogen:
Noble Gas:
2.6-2.7 Chemical Substances
Molecule:
For elements, a few of these are very important.
KNOW:
Hydrogen
Chlorine
Nitrogen
Bromine
Oxygen
Iodine
Fluorine
You should not only know that these elements are diatomic, but you should
also know their symbols and their names (spelled correctly).
Note: You should precede to Nomenclature I on the website and begin studying from it.
See the notes in the text also.
Allotrope:
Compounds may contain ions instead of molecules. If they do contain ions, there
must be equal numbers of positive and negative charges.
Isolated ions always have a charge written: Cl–
OH–
Na+
Compounds are neutral so there is no charge written even though ions are
present: NaCl
Metals usually react with nonmetals to produce ions not molecules.
Ion:
Cation:
Anion:
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Monatomic Ion:
Polyatomic Ion:
Polyatomic ions are like molecules except for the fact that they
have charges.
2.8-2.9 Nomenclature
Note: The nomenclature covered in these two sections will be distributed over the first three
exams. For this reason, you only need to know a portion of this material by the next exam.
If you have not begun Nomenclature I on the website, you should do so immediately. This
guide will let you know when you need to move on to the other nomenclature sections on the
website.
Inorganic Nomenclature
Note: Plan to look at nomenclature daily. This takes time, it is not possible to learn all of the
nomenclature that you will need by "cramming," and it needs to be in long-term memory.
Note: You will only be responsible for the names put on the board in class, but they will be
comprehensive.
Note: It will be necessary for you to learn the names of certain elements, the names,
and charges of certain ions, along with the rules for predicting the charges of
some elements. You will also need to learn how to put everything together in
a compound. Simple memorization is not sufficient; you must practice naming
compounds.
Note: If you have not started Nomenclature I on the website, do so as soon as possible.
Compounds:
Molecular Formula (Chemical Formula):
Structural Formula:
Empirical Formula:
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Inorganic Nomenclature
See Nomenclature I on Website
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Reminder: Enter notes on these pages or, less effectively, on notebook paper to insert into
the appropriate place, recopying may help.
Reminder: If at any point you begin to feel overwhelmed, you should see the instructor.
CHAPTER 3
3.1 Chemical Equations
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Reactant:
Product:
Example:
H2(g) + Cl2(g)  2 HCl(g)
Equations must be balanced – they are not much help otherwise.
Balancing equations – General method
1. Set up reactants and products (If all are not given then they must be
determined) – use correct formulas- do not change the formulas.
C4H10 + O2 
___ C4H10 + ___O2
___ CO2 + ___ H2O
2. Pick an element and balance it (postpone balancing elements appearing
more than once on a side).
i.e., C (or H,) but not O
___C4H10 + ___O2 
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4 CO2 + ___ H2O
3. Adjust another element ("forced")
In this case H
___C4H10 + ___O2 
___ CO2 + ___ H2O
4. Repeat step 3 until all elements are done.
___C4H10 + ___O2 
___ CO2 + ___ H2O
(Fractions are allowed temporarily, but not in the final solution.)
5. Clear fractions (if present)
___ C4H10 + ___ O2 
___ CO2 + ___ H2O
NOTE: Also, make sure the coefficients are reduced to their lowest whole number values.
6. Check to make sure that it really is balanced.
(An often-neglected step)
NOTE: If it is not balanced, check for obvious mistakes.
NOTE: If not found quickly it is probably best to start over (sometimes with another
element).
Study Guide Example – not covered in class:
Write balanced chemical equations for each of the following reactions. a. Water
reacts with phosphorus trichloride to produce hydrogen chloride and H3PO3.
b. Water reacts with diboron trisulfide to produce H3BO3 (boric acid) and hydrogen
sulfide. c. Zinc metal (Zn) reacts with hydrochloric acid to produce hydrogen gas and
zinc chloride (ZnCl2). d. PH3 burns in oxygen gas to produce water and diphosphorus
pentoxide. e. silicon dioxide reacts with carbon at high temperatures to produce
elemental silicon and carbon monoxide gas.
Answers:
Note: this problem also tests your nomenclature.
a. 3 H2O + PCl3
b. 6 H2O + B2S3
c. Zn + 2 HCl
d. 4 PH3 + 8 O2
e. SiO2 + 2 C
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




3 HCl + H3PO3
2 H3BO3 + 3 H2S
H2 + ZnCl2
6 H2O + 2 P2O5
Si + 2 CO
18
3.2 Types of Reactions
Be able to classify the various types.
Some reactions may fit into more than one category; it will not be important
which one you choose (as long it is one of the choices).
Combination Reactions –
Decomposition Reactions –
Displacement Reactions –
Metathesis Reactions –
Neutralization Reactions –
Combustion Reactions –
3.3 Formula Weights
Molecular (Formula) Weight:
Example: N2
Example: HCN
amu H
amu C
amu N
amu HCN
Example: [Cu(NH3)4]SO4·.2H2O
1 Cu
4N
16 H
1S
6O
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amu
amu
amu
amu
amu
19
________
amu
An alternative to formulas is percent composition. It may be derived from a formula,
or more importantly, a formula may be derived from it.
Example: C6H12O6
FW = 180.158 amu
%C=
%H=
%O=
Note: the same answers would result if grams and moles appeared instead of amu
and atoms.
Reminder: If you have had trouble with any of the preceding problems, you should get help
immediately.
3.4 Avogadro’s Number and the Mole
The SI unit for the quantity of a substance is a mole.
Mole:
Avogadro's Number:
NOTE: MOST PROBLEMS FROM THIS POINT ON WILL INVOLVE MOLES IN
AT LEAST ONE STEP.
NOTE: AVOGADRO'S NUMBER IS USED ONLY RARELY IN CALCULATIONS.
IT IS NECESSARY ONLY WHEN THE NUMBER OF ATOMS OR
MOLECULES IS ASKED FOR OR GIVEN.
Example: How many moles are present in 1.20 × 1025 silver atoms?
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Example: If one N2 molecule weighs 28.01 amu, calculate the number of grams in
one mole of N2 molecules.
Molar Mass:
Example: How many grams of CO2 are in 4.00 moles of CO2?
Example: How many oxygen atoms are in 22.0 grams of CO2?
3.5 Empirical Formulas from Analysis
Determining the formula from the composition
Example: Analysis of a sample of a gas found 2.34g of nitrogen and 5.34g of
oxygen. What was the formula of this gas?
Use N and O atoms not N2 and
O2 molecules. Molecules do
not contain other molecules.
NOTE: DO NOT CONFUSE THIS RATIO WITH LATER.
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Example: Analysis of a sample of Vitamin C found it to contain: 40.9% carbon,
4.58% hydrogen and the remainder was oxygen. Its formula weight is about
180g/mole. What are its empirical and molecular formulas?
BE CAREFUL TO MAINTAIN SUFFICIENT SIG. FIGS.
Note: this method will always give the empirical formula; only the additional information
(MW) allows the molecular formula to be determined.
Example: A 1.778 g sample of an unknown solid was burned in oxygen, and 2.842 g
of CO2, 0.2609 g of H2O, and 0.4056 g of N2 were produced. The FW of the
unknown was about 740 g/mole. The compound contained C, H, N, and O.
Determine its molecular formula.
Note: Most students believe the preceding type of question to be the most difficult one
possible on Exam 1. Do not expect it to be perfectly clear immediately in class.
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Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
3.6 Quantitative Information from Balanced Equations
Calculations involving chemical reactions
Stoichiometry:
NOTE:
1. You must have a balanced chemical equation.
2. The reactants/products are related by a mole ratio.
Example: If 40.0 g of Cl2 and excess H2 are combined, HCl will be produced. How
many grams of HCl will form?
3.7 Limiting Reactants
Limiting Reagent (Limiting Reactant):
Example: The following reaction will produce bismuth(III) fluoride:
2 Bi(s) + 3 F2(g)  2 BiF3(s)
During one experiment, the reaction mixture contained 183.9 grams of bismuth and
105.6 grams of fluorine. What was the maximum number of grams of BiF3 formed?
HINT: Anytime the quantities of more than one reactant are given, it is probably a L. R.
problem.
Do not confuse with earlier
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Study Guide Example – not covered in class:
You are a bicycle manufacturer. You have an inventory of 3165 handlebars, 6000
wheels, and 4175 frames. a. How many bicycles can you make before you have to
order more parts? b. Which part will you need to order first? c. How much of each
of the parts will you still have on hand when you order new parts?
a. You could make 3165 bicycles from the handlebars, or 3000 bicycles from the
wheels, or 4175 bicycles from the frames. The smallest value is 3000 bicycles – thus
the wheels behave like a limiting reagent, and manufacture must stop not matter how
many handlebars and frames are remaining.
b. You will need to order wheels first, because they are the limiting part.
c. You will have 3165 – 3000 = 165 handlebars, 6000 – 2(3000) = 0 wheels, and 4175
– 3000 = 1175 frames.
Various things may occur in a reaction to prevent the reactants from complete
conversion to products.
Actual Yield:
Theoretical Yield:
Percent Yield =
Example: The heating of a 25.0 g sample of calcium oxide with excess hydrogen
chloride produces water and 37.5 g of calcium chloride. What is the percent yield?
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Reminder: You are not expected to understand the lecture material while you are in the
lecture room, you are expected to have some understanding (or specific questions)
when the next lecture begins.
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
Reminder: The final review for the test needs to be as close to test conditions as possible.
Thus, you should work the problems during this final review by yourself and
“closed book.” In addition, work though a practice exam under “exam conditions”
about 24 hours before the scheduled exam time. This will give you sufficient time
to correct your studying.
END OF MATERIAL FOR EXAM 1
Note: As soon as you finish with this test, begin Nomenclature II on the website.
Note: If you feel that, you did not do as well as you would have liked on the exam, you
should begin immediately preparing for the retest. Do not wait until you get the test
back before you start looking at the material and do not use the test you just
completed as a study guide. Also, make sure you find out if there is a curve, before
you make any decisions about this class. Persons with B’s or better in the class have
dropped the class because they did not bother to find out.
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The following is the key to an Exam 1 that was used in the past. The format has been slightly
altered to allow the addition of comments. The answers are given in italics and the
comments will be given in italics and in parentheses. Comments are based on the class that
took this exam.
EXAM 1
CHEMISTRY 133
SIL 1, 2, 3
R.H. Langley
(There are not a lot of directions – make sure you follow them.)
Put your answers in the appropriate blanks.
Name (Print) ___________________
Make sure your exam has 10 different questions.
Section _____
1.(10) Define each of the following, as they apply to this course, in a sentence or two.
(a) Chemical Property Any change based upon a change in composition.
(b) Group
A vertical column on the periodic table.
(c) Metathesis Reaction
A double displacement reaction.
(d) Empirical FormulaThe simplest formula giving the lowest ratio of the atoms
present.
(e) Density
Mass/Volume
(Approximately 50% of the students did not pass this question. Less than 5% got all 10
points. A question is worth a letter grade, time must be spent on the definitions. In addition,
you cannot understand the material if you do not understand the terms.)
2.(10) Name or give the formula (symbol) for each of the following.
(a) Arsenic
____As____
(f) Bi ______Bismuth_________________
(b) Oxygen
____O_____ (g) IF7 ___Iodine heptafluoride__________
(c) Antimony trichloride____SbCl3_ (h) SiBr4
__Silicon tetrabromide_____
(d) Disulfur dibromide ___S2Br2___ (i) Cl2O5
_Dichlorine pentaoxide_____
(e) Hydrochloric acid ___HCl____ (j) NH3 ________Ammonia______________
(Only 5% of the students got all 10 of these correct. This was the best performance on any of
the nomenclature questions all semester. A low performance on this question (below an 8)
indicates that significantly more time is required. One student managed to lose a total of 131
points to nomenclature over the semester (22% of the possible points for the term), the best
student only lost 18 points (3%) during the term. The average student lost 77 points (13%).
Something this important requires significant amounts of study time, and the students in this
particular class did not study enough nomenclature. Do not forget – spelling counts.)
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Exam 1
3.(10) The Morgan silver dollar, which contains 90. % silver, has a mass of 26.73 g. When
the coin was first minted in 1878, silver was worth $1.18 per troy ounce (31.103 g).
What was the value of the silver in the silver dollar?
(From Problem ....)
(On a regular exam, this
(This answer is only worth 2 points.)
would be a homework
(Make sure the answer is in this blank.)
problem)
___$0.91___
(This is the only
26.73g  90%  TroyOz.  $1.18  = $0.91
 100%  31.103g  TroyOz
totally correct
answer – it must.)
look exactly like
(As with any factor label problem the sequence is not
this or $.91.
important. For full credit ALL steps, including the,
Additional
“trivial” ones must be included. Thus, for example,
conversions
converting 90 % to 0.90 in your head will not be given
could be
credit. You will receive credit for the work shown, not what
added,
you did in your head. While this may seem extreme, this
but are
will get you in the habit for later on in the term.)
unnecessary.)
4.(10) A cesium (Cs) atom has a diameter of about 4.7 Å. How many cesium atoms would
be needed to equal 1.0 cm?
(From Problem ....)
(Again, on a regular exam this would
(Make sure the answer in the blank.)
be a homework problem.)
(Wrong number of significant figures = –1)
_2.1 × 107 atoms__
1.0cm  0.01  1Å10  1atom  = 2.1 × 107 atoms
 c  10 m  4.7Å 
(This is the proper
number of significant
figures.)
(The Å to m conversion was given to the students
with the exam, they did not need to memorize it.)
(Even though this was directly out of the homework 15% of the students got 0 points for it,
and less than one fourth of the class got full credit. It is bad enough to get 0 points on a
problem (a letter grade loss on the exam), but a 0 is worse when it is on a homework
problem. A score of 0 on a homework problem indicates no studying effort on the part of the
student (whether this is true or not). Also, if a student understood the problem so poorly,
then they should have been in the instructor’s office (if they cared at all about their grade).)
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Exam 1
5.(10) How many vitamin C (C6H8O6)molecules are in a 500-mg tablet of vitamin C.
(From Problem ...)
(Three chapters on this test =
(No units = –1)
three homework problems.)
21
_1.71 × 10 molecules_
23



500mg  0.001  1moleC 6 H 8 O 6  6.022x10 molecules  = 1.709 × 1021 molecules
 m  176.12gC 6 H 8 O 6  1moleC 6 H 8 O 6

21
(An answer of 1.70 × 10
(Answer to be rounded
molecules would also be acceptable.)
to the proper number of
significant figures.)
C
6(12.01)
H
8(1.008)
O
6(16.00)
176.12 g/mole
(It is not necessary to show this, but it is safer.)
(On the average, people did worse on this homework problem than on the preceding one.
Over 20 % of the students received 0 points on this problem. The comments on question 4
apply here.)
(At this point, the students taking the exam have had two “memorization” problems
(definitions and nomenclature) and three homework problems. This amounts to a total of 50
points based only on things that the student should have seen before, and discussed with the
instructor if there was any problem in understanding the material.)
6.(10) The density of germanium, Ge, is 5.35 g/cm3. How many pounds does 2.00 ft3 of
germanium weigh?
3
3
 1lb 
3  12in   2.54cm   5.35g 
 = 667.9827 lbs
2.00ft 



3
 1ft   1in   cm  453.59g 
(To be rounded)
__668 lbs__
(A common mistake in the calculation is in the two cubed
conversions. Many people enter the first conversion into their
calculator as 12 /1. This is incorrect; it should be entered as 123/1
(technically as 123/13).
(This is a factor label problem, thus the order is irrelevant. In addition, if you learned
different conversions in place of the ones used above they will be acceptable. A common
problem occurs when a student attempts to learn more than the required number of
conversions. A student learning too many conversions often gets them mixed. Nearly 30% of
the students taking this test could not do a single conversion for this problem (i.e., they did
not even write down that 12 inches = 1 foot).)
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Exam 1
7.(10) Fill in the gaps in the following table:
18
Symbol
9
F
_6Li1+_
_224Ra
3
32
16
S2
88
Protons
_9___
__3__
_88__
_16__
Neutrons
_9___
__3__
_136_
_16__
Electrons
_9___
__2__
_88__
__18_
Mass Number _18__
__6__
_224_
_32__
(One person managed to get a 0 on this problem.)
8.(10) Balance the following equations. Enter ALL coefficients, EVEN ONES.
(a) _1_ H2 + _1_ Cl2  _2_ HCl
(b) _1_ XeF6 + _3_ H2O  _1_ XeO3 + _6_ HF
(c) _3_ Ra(OH)2 + _2_ H3VO4  _1_ Ra3(VO4)2 + _6_ H2O
(d) _2_ C2H6 + _7_ O2  _4_ CO2 + _6_ H2O
(e) Methane reacts with fluorine to form carbon tetrafluoride and hydrogen fluoride.
1 CH4 + 4 F2  1 CF4 + 4 HF
(Only 15% of the students balanced all five equations correctly. The most common mistakes
were not following directions (leaving out the ones), not checking, and not knowing the
nomenclature for the last equation.)
(For the class as a whole this was their best page.)
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Exam 1
9.(10) How many grams of ClO2 could be formed if 9.010 g of H2C2O4, 9.810 g of H2SO4,
and 12.200 g of KClO3 are mixed and allowed to react as follows:
H2C2O4 + H2SO4 + 2 KClO3  K2SO4 + 2 H2O + 2 CO2 + 2 ClO2


9.010gH 2 C 2 O 4  1molH 2 C 2 O 4  = 0.10006889 = 0.100068859
1
 90.038gH 2 C 2 O 4 
6.7150 g ClO2_
(Correct sig. figs.)


9.810gH 2SO 4  1molH 2SO 4  = 0.100020391 = 0.100020391
1
 98.08gH 2 SO 4 
 1molKClO 3  0.099550391
 =
= 0.049775195
2
 122.55gKClO 3 
(Limiting Reagent)



0.099550391molKClO 3  2molClO 2  67.453gClO 2  = 6.71497 g ClO2
 2molKClO 3  1molClO 2 
(Extra sig. figs.)
(Most people did not do well on this problem because they did not recognize it as a limiting
reagent problem. Even without this recognition, everyone should have been able to get some
credit by simply converting everything to moles. Many people simply left the problem blank.
Leaving a problem blank is the only way to guarantee no partial credit.)
12.200gKClO 
3
(All molecular weights should have at least as many significant figures as the grams given.)
10.(10) Under certain conditions the compound S4N4 may be explosive. One way to destroy
this compound safely is by the following reaction:
S4N4 + 6 NaOH + 3 H2O  Na2S2O3 + 2 Na2SO3 + 4 NH3
In one test reaction, 1.000 gram of S4N4 generated 0.298 grams of NH3. What was
the percent yield in the test reaction?




1.000gS 4 N 4  1molS 4 N 4  4molNH 3  17.031gNH 3  = 0.36967 g
 184.28gS 4 N 4  1molS 4 N 4  1molNH 3 
___80.6 %___
(Extra sig. fig.)
 0.298g 

x100% = 80.600% (Leaving out the × 100% [doing it in
 0.36967g 
your head] will result in a point
deduction)
(This was question with the lowest average grade on this test. Over a third of the class
received 0 points. If they had simply written out the definition of percent yield they would
have gotten 2 points – never leave a question blank).
(As with all tests, the average grade on this test was a C. Point wise a C on this test was a
56.)
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Note: It is exceedingly important that, while it is fresh in your mind, you make sure that you
have the correct solution to every problem on the first exam (and re-test). Not doing
so will have a strong negative influence on your grade on the final exam.
Note: If your grade on the test is not what you want, you should immediately refer to the
syllabus and/or see the instructor.
Reminder: If you feel that, you did not do as well as you would have liked on the exam,
you should begin immediately preparing for the retest. Do not wait until you get the
test back before you start looking at the material and do not use the test you just
completed as a study guide. Also, make sure you find out if there is a curve, before
you make any decisions about this class. Persons with B’s or better in the class have
dropped the class because they did not bother to find out.
Note: Many people get the same grade on the retest because they continue to make the
same studying mistakes. If you do not fix whatever it was that caused you to get a
low grade, you will continue getting the same grades. See the syllabus for
suggestions for improving your study habits.
Note: A few people do not do as well on the retest as the original test. These people
have consciously, or subconsciously, used the first test as a study guide. You must
prepare for the retest like a fresh test.
Reminder: Enter notes on these pages and/or on notebook paper inserted later in the
appropriate place, recopying may help.
Reminder: Begin on Nomenclature II on the website.
Reminder: If at any point you begin to feel overwhelmed, you should see the instructor.
CHAPTER 10
Note: The textbook gives a large number of additional gas law equations. You only need
to know four. Learning additional gas laws will not help you on the exam, and they
take away from your study time. In addition, if you attempt to learn unnecessary
material you are more likely to make mistakes.
10.1 Gases
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
States of Matter
Condensed Phases:
Intermolecular Forces:
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10.2 Pressure
Pressure =
Units of Pressure =
Pascal (Pa) =
Standard Atmosphere:
Note: You need to know the atm to torr (mmHg) conversion (Note: 1 mmHg = 1 torr)
10.3-10.4 Gas Laws
Boyle's Law – P and V
Charles's Law – V and T
Avogadro's Law – V and n
Other combinations are possible – however they lead to the same overall
relationships.
Boyle's Law
Charles's Law
Avogadro's Law
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Temperature Problem:
Absolute zero = 0 K = –273.15 °C
NOTE:
For all gas laws the temperatures must be expressed on an absolute
temperature scale. (Normally Kelvins are used (Not °K).)
TK = T°C + 273(.15)
The Molar Volume is the volume of 1 mole of gas at STP.
=
22.41L
mole
at 0°C and 1 atm.
Boyle's, Charles's and Avogadro's Laws may be combined.
KNOW:
R=
PV = nRT
0.08206L • atm
mol • K
(Ideal Gas Equation = Ideal Gas Law)
For others rearrange and cancel what does not change.
R=
PV
nT
P1 V1
PV
=R= 2 2
n 2 T2
n 1 T1
Combined Gas Law
NOTE: A common mistake is to use 22.4 L/mol under conditions that are not STP.
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Example: A sample of a gas occupies 25 liters under a pressure of 1.0 atm. What
will be the volume of the gas after increasing the pressure to 2.0 atm?
Example: A 500.0 mL sample of a gas is confined at a pressure of 1 atm at 25°C.
What temperature is necessary for the sample to occupy 600.0 mL?
Example: A sample of krypton gas occupies 50.0 L at 27°C and 1000. torr.
Determine its volume at 0°C and 1.10 atm.
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Example: How many moles of helium gas are present in a 20.0 L container if the
pressure is 14.0 atm at a temperature of 27°C?
Study Guide Example – not covered in class:
In 1937, the German Zeppelin Hindenburg exploded while landing in New Jersey.
The Hindenburg contained 7.1 × 106 ft3 of hydrogen gas. If, at the time of the
explosion, the hydrogen was at a temperature of 20.°C at a pressure of 745 torr, how
many grams of hydrogen were present?
PV
=
RT




3
3
3
3
6
3
 745torr7.1x10 ft   1atm  12in   2.54cm   0.01   d   1L  2.0gH 2

 760torr  1ft   1in   c   0.1   1dm 3  1molH 2
L • atm 

293K  
  0.08216
mol
•
K




Answer:
n=
= 1.6 × 107 grams H2
(Both your factor label and nomenclature are
being tested.)
10.5-10.6 More Gas Laws
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
When dealing with gas mixtures:
ntotal =
Ptotal =
Dalton's Law of Partial Pressures:
Partial Pressures:
NOTE: This is the only significant gas law in this section.
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


Also of use is the Mole Fraction:
Xa =
X may also be related to pressure.
Xa =
Example: The mole fraction of nitrogen in air is 0.79, and the mole fraction of
oxygen is 0.20. What is the partial pressure of each of these gases, if the total
pressure is 2.00 atm.?
Example: (a) The vaporization of a sample of an unknown liquid produced 0.8001
grams of vapor. At a temperature of 99°C and a pressure of 750. mmHg, the vapor
occupied 0.250 L. What was the molecular weight of the unknown liquid?
(b) Analysis of the unknown liquid found the following: 24.2% C, 4.04% H,
and 71.7%Cl. Using these data and the above, determine the molecular
formula of this compound.
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Gas Stoichiometry
Need to be able to use mole relationships
Example: A 10.0 L container is filled with 600. g of ethane (C2H6) and 2080 g of
oxygen. When ignited the mixture reacts as follows:
2 C2H6(g) + 7 O2(g) 
4 CO2(g) + 6 H2O(g)
After the reaction, the mixture has a temperature of 150.°C. Determine the
partial pressure of each gas after the reaction, and the total pressure.
CO2
H2O
NOTE:
Since O2 is the limiting reagent then, by definition, it is entirely gone, so it can
yield no pressure.
O2
NOTE:
The C2H6 is a reactant, not a product, so its treatment must be different.
C2H6
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Example: Determine the mole fraction of each of the components after the
reaction described in the last example.
Two methods:
Pressure Method
atm
C2H6
atm
O2
atm
CO2
atm
H2O
atm

total 
Mole Method
moles
moles
moles
moles
moles
Total
10.7 Kinetic Molecular Theory
Kinetic Molecular Theory
Present explanation of the behavior of gases.
Basic Postulates:
1.
2.
3.
4.
Kinetic Energy (molecular speed)
The pressure is due to the number of molecules hitting the walls, and how
hard the molecules hit the walls.
Decrease V –
Increase n –
Increase T –
Two gases –
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Ideal Gas:
10.8 Applications of Kinetic Molecular Theory
Diffusion:
Effusion:
Graham's Law of Diffusion
NOTE: Rates will ALWAYS have a time unit in the denominator, i.e., mL/s or g/hr.
10.9 Real Gases
van der Waal's Equation:
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
Note: The textbook gives a large number of additional gas law equations. You only need to
know four. Learning additional gas laws will not help you on the exams, and they take away
from your study time for the exam. In addition, if you attempt to learn unnecessary material
you are more likely to make mistakes.
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
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Reminder: Enter notes on these pages and/or on notebook paper inserted later in the
appropriate place, (recopying may help).
Note: If you ask, most students who have already completed this class “which is the hardest
chapter?” most will say chapter 4. Plan to adjust your studying accordingly. Also,
consider, if this is the hardest chapter, the remaining chapters must be easier.
Reminder: If at any point you begin to feel overwhelmed, you should see the instructor.
CHAPTER 4
4.1 General Properties of Aqueous Solutions
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Aqueous Solution:
NOTE:
Water is assumed the solvent unless otherwise specified.
Solute:
Solvent:
Saturated:
Unsaturated:
Supersaturated:
Nonelectrolyte:
Electrolyte:
Dissociation:
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Types of electrolytes: acids, bases, and salts
Acid:
Base:
Strong Electrolyte:
Weak Electrolyte:
Two important types: weak acids and weak bases
There is only a partial separation into ions: HNO2(aq)  H+(aq) + NO2–(aq)
NOTE: You should see the Net Ionic Equations section on the website.
Reminder: You are not expected to understand the lecture material while you are in the
lecture room, you are expected to have some understanding (or specific questions)
when the next lecture begins.
4.2 Precipitation Reactions
Solubility rules:
Ionic compounds in water.
The rules depend on the strength of the ionic bonds, with a few exceptions.
1.
2.
3.
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Note: See the Solubility Rules on the website.
The insoluble materials will not dissolve (to a great degree); also, any reaction that
forms them will give a precipitate.
Various driving forces:
Form a gas
CO2, NH3, SO2, H2S
Form a weak/nonelectrolyte
H2O, a weak acid, or a weak base
Form a precipitate
See solubility rules
4.3 Acids, Bases, and Neutralization Reactions
Strong Acid:
KNOW:
Strong Base:
KNOW:
What are ionic compounds?
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Reactions may be written in more than one way.
Molecular Equation:
Complete Ionic (Total) Equation:
Net Ionic Equation:
Spectator Ion:
NOTE: THIS IS WHERE MOST PEOPLE HAVE TROUBLE ON THE EXAM.
PLAN ON EXTRA STUDY TIME.
Study Guide Example – not covered in class:
Write balanced net ionic equations for the reactions that occur when each of the
following are mixed. Assume all reactions are in aqueous solution. It is possible that
no reaction will occur.
Answers
a. H2SO4 + BaCl2
a. SO42– + Ba2+

BaSO4
b. AgNO3 + K2CO3
b. 2 Ag+ + CO32–

Ag2CO3
+
c. ZnS + HCl
c. ZnS + 2 H

Zn2+ + H2S
3–
+
d. Li3PO4 + HBr
d. PO4 + 3 H

H3PO4
e. RbCl + NH4NO3
e. RbCl + NH4NO3 
No Reaction
f. Sr(OH)2 + HI
f. OH– + H+

H2O
g. NaOH + HC2H3O2
g. OH– + HC2H3O2 
H2O + C2H3O2–
h. Cr(OH)3 + HClO3
h. Cr(OH)3 + 3 H+ 
Cr3+ + 3 H2O
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4.4 Oxidation-Reduction (Redox) Reactions
Oxidation-Reduction (Redox):
Oxidation:
Reduction:
Example:
2 Mg + O2 
2 MgO
Reducing Agent:
Oxidizing Agent:
This is very useful with the activity series.
Study Guide Example – not covered in class:
Based on the Activity Series, write balanced molecular and net ionic equations for
each of the following.
Answers:
a. Hydrobromic acid with iron
a. 2 HBr + Fe 
FeBr2 + H2
+
2 H + Fe 
Fe2+ + H2
b. Sulfuric acid with nickel
b. H2SO4 + Ni  NiSO4 + H2
2 H+ + Ni 
Ni2+ + H2
c. Hydrochloric acid with chromium
c. 6 HCl + 2 Cr  2 CrCl3 + 3 H2
6 H+ + 2 Cr  2 Cr3+ + 3 H2
d. Cu + CoCl2
d. Cu + CoCl2  No Reaction
e. Pb + AgNO3
e. Pb + 2 AgNO3  Pb(NO3)2 + 2 Ag
Pb + 2 Ag+ 
Pb2+ + 2 Ag
f. H2 + Hg(NO3)2
f. H2 + Hg(NO3)2  2 HNO3 + Hg
H2 + Hg2+ 
2 H+ + Hg
g. H2 + MnCl2
g. H2 + MnCl2  No Reaction
h. Li + H2O
h. Li + 2 H2O 
2 LiOH + H2
Li + 2 H2O 
2 Li+ + 2 OH– + H2
i. Zn + HC2H3O2
i. Zn+2HC2H3O2  Zn(C2H3O2)2 + H2
Zn+2HC2H3O2  Zn2++2C2H3O2– + H2
j. Sn + CuSO4
j. Sn + CuSO4  SnSO4 + Cu
Sn + Cu2+ 
Sn2+ + Cu
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4.5 Concentrations of Solutions
Concentration may be expressed in many ways – one of which is molarity.
(Recall 1.00 L = 1.00 dm3)
Molarity = M =
Note: Many students get into trouble because they incorrectly use m and the abbreviation
for moles. The correct abbreviation is mol. The abbreviation m means molality,
which appears in chapter 13. The use of an incorrect abbreviation has resulted in
students making very significant factor label mistakes, and losing several points on an
exam.
Example: A solution of NaCl contains 39.12 g of this compound in 100.0 mL of
solution. Calculate the molarity of NaCl.
Example: A maximum of 85.4 g of Na2SO4 will dissolve in 200. mL of water. It
completely separates into sodium ions and sulfate ions. What is the
concentration of sodium ions in the solution?
Example: How many moles of ammonium ions are in 0.100 L of a 0.20 M
ammonium sulfate solution?
Example: How many mL of 0.20 M lithium carbonate solution are necessary to
supply 1.0 mole of lithium ions?
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Example: How would you prepare 650.0 mL of a 0.1000 M ammonium sulfate
solution?
Sometimes it is necessary to change the concentration of a solution. The
concentration may be changed by changing the amount of solute or solvent.
M1V1 = M2V2
BE CAREFUL OF WHEN TO APPLY THIS.
Example: How much 16.0 M NaOH solution is necessary to prepare 1.000 L of
1.000 M NaOH?
Example: To 350. mL of a 0.250 M sodium chloride solution is added 450. mL of
water. What is the final concentration?
4.6 Solution Stoichiometry and Chemical Analysis
Quantitative Analysis:
Two types discussed here.
1.
2.
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Example: How many grams of KCl must be added to precipitate all the mercury(I)
ions in 0.10L of a 0.020 M mercury(I) nitrate solution? The mercury will precipitate
as mercury(I) chloride.
Example: The sulfur in a 0.3054 g sample containing the mineral chalcopyrite
(CuFeS2) was converted to sulfate. The precipitation of this sulfate with barium gave
0.6525 g of barium sulfate. What was the percentage of CuFeS2 in the sample?
Titration:
Titrant:
Equivalence Point
End Point:
Indicator:
Any type of reaction may be used.
A neutralization reaction
Acid +
Base 
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The net ionic equation:
H+(aq) + OH–(aq)  H2O(l)
Example: Using 100.0 mL of 0.100 M acid plus x mL of 0.100 M RbOH, what will
be the concentration of acid after each of the following?
(a) The acid is HNO3 and 100.0 mL of RbOH is used.
(b) The acid is HNO3 and 75.0 mL of RbOH is used.
(c) The acid is H2SO4 and 100.0 mL of RbOH is used.
(a)
(b)
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(c)
Example: A Na2CO3 solution was reacted with an HCl solution. The sodium
carbonate solution contained 0.5015 g of Na2CO3 in 100.0 mL of water. A total of
48.47mL of acid was necessary for the titration. What was the original concentration
of the acid? The reaction was:
Na2CO3 + 2 HCl 
2 NaCl + H2O + CO2
Example: If 35.00 mL of a 0.1500 M KOH solution is required to titrate 40.00 mL
of a phosphoric acid solution, what is the concentration of the acid? The reaction
is:
2 KOH + H3PO4

K2HPO4 + 2 H2O
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NOTE: See the PROBLEM SOLVING section at the very end of this guide.
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
END EXAM 2 MATERIAL
Note: After the exam, you should move immediately to Oxidation Numbers and
Nomenclature III on the website.
Reminder: It is exceedingly important that, while it is fresh in your mind, to make sure that
you have the correct solution to every problem on the exam (and re-test). Not doing so will
have a strong negative influence on your grade on the final exam.
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Reminder: Start the Oxidation Number and Nomenclature III sections on the website.
Reminder: If your grade on the test is not what you want, you should immediately refer to
the syllabus and/or see the instructor.
CHAPTER 5
Note: Many students consider these end-of-chapter problems to be very difficult to interpret.
However, most of them are easy to solve. Factor label will be exceedingly
important. This chapter REQUIRES the extensive utilization of factor label, if you
have postponed learning factor label you need to catch up.
Note: Other than, the equations given in class do not waste your time on any other
equations given in the textbook. It is more important to learn the necessary material
than to waste time on unnecessary equations. Of course, if you have plenty of extra
time to memorize extraneous material, you can use these equations (though probably
not on an exam).
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
Thermochemistry:
Thermodynamics:
5.1-5.2 The Nature of Energy/The First Law of Thermodynamics
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Energy:
Potential Energy:
Kinetic Energy:
Energy – Joule = J =
– calories = cal
1 cal = 4.184 J (exactly)
Heat:
Law of Conservation of Energy (First Law of Thermodynamics):
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System:
Surroundings:
State:
State Functions –
Example: T, P
Energy in the system: Internal Energy = E
Changes in E can be measured = E (final – initial)
E =
Work done by the system –
Work done on the system +
There are many types of work
5.3 Enthalpy
Enthalpy:
Exothermic:
Endothermic:
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5.4 Enthalpies of Reaction
Chemical equations may be written with energy changes included. These equations
specifically deal with moles (never molecules), and are called Thermochemical
Equations.
Thermochemical Equation:
NOTE: They are the only equations that allow fractions.
5.5 Calorimetry
Calorimetry:
Need: Energy
Temperature change (T)
Conversion factor:
Specific Heat:
The units tell exactly
how to solve the problem.
(J/g°C or cal/g°C)
Heat Capacity:
(J/°C or cal/°C)
Molar Heat Capacity:
(J/mole°C or cal/mole°C)
Heat of Reaction:
For problems factor label to get the proper units
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(Mass is needed)
Example: The ignition of a 1.5886 g sample of glucose (C6H12O6) in a bomb
calorimeter resulted in a temperature increase of 3.682°C. The heat capacity of the
calorimeter was 3.562kJ/°C, and the calorimeter contained 1.000 kg of water. Find
the molar heat of reaction for:
C6H12O6(s) + 6 O2(g) 
6 CO2(g) + 6 H2O(l)
NOTE: Watch the signs; they depend on definitions and not on mathematics.
5.6 Hess’s Law
Thermochemical equations may be combined to produce new thermochemical
equations.
Hess's Law –
Example: Given the following information:
C(s) + O2(g) 
CO2(g)
H2(g) + (1/2) O2(g) 
H2O(l)
C2H2(g) + (5/2) O2(g) 
2 CO2(g)+ H2O(l)
Find the enthalpy change for:
2 C(s) + H2(g) 
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–393.5 kJ
–285.8 kJ
–1299.8 kJ
C2H2(g)
5.7 Enthalpies of Formation
Reactions may be carried out in many different ways.
Standard Heat of Reaction:
H°
(Note ° sign)
The heat of reaction is measured under Standard Conditions.
NOT STP
Standard Conditions:
Standard State:
Standard Heat (Enthalpy) of Formation:
H°f
NOTE:
° = standard conditions
f = formation (only one-way to write the equation.)
For an element, it is 0.000
Hrxn = H°f(products) – H°f(reactants)
Example: Calculate Hrxn for:
6 H2O(g) + 4 NO(g) 
5 O2(g) + 4 NH3(g)
NOTE: A common mistake is not to subtract all the reactants from all the products.
5.8
Read
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
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CHAPTER 6
6.1 The Wave Nature of Light
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Electromagnetic Radiation (Electromagnetic Energy):
It is possible to treat light as waves.
 = c
 = Wavelength:
 = Frequency:
c = the Speed of Light
6.2-6.3 Quantized Energy and Photons/Line Spectra and the Bohr Atom
Photon:
The energy of a photon may be calculated:
E = h
h=
Bohr Theory of the atom:
Energy terms – not a physical three-dimensional arrangement.
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Ground State:
Excited State:
6.4 The Wave Behavior of Matter
De Broglie first predicted the wavelike properties of matter:
de Broglie:
NOTE: Do not mistake v for .
Electron behavior 
wave behavior 
Schrödinger Equation
Heisenberg Uncertainty Principle:
Only the probability of locating an electron is possible.
Only a region in space: Probability Density
6.5 Quantum Mechanics and Atomic Orbitals
Quantum Mechanics Results from the Schrödinger Equation – Quantum Numbers
Quantum Numbers:
1. Principle Quantum Number = n =
Shell:
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2. Angular Momentum Quantum Number = l =
Subshell:
Orbital:
3. Magnetic Quantum Number = ml =
4. Electron Spin Quantum Number = ms =
6.6 Representations of Orbitals
What does an atomic orbital look like? Only a probability
n 
s Orbitals:
p Orbitals:
d Orbitals:
f Orbitals:
size
l 
shape
ml

orientation
(n, l = 0, ml = 0 )
(n, l = 1, ml = –1, 0, 1)
(n, l = 2, ml = –2, –1, 0, 1, 2)
(n, l = 3, ml = –3, –2, –1, 0, 1,2,3)
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
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6.7 Many Electron Atoms
Pauli Exclusion Principle:
Effective Nuclear Charge:
6.8 Electron Configurations
This may be expressed as the Electron Configuration or as an Orbital Diagram.
Electron Configuration:
Orbital Diagram:
NOTE: Many students get into trouble by giving the wrong one of these on an exam (both is
not a substitute).
Aufbau Principle:
Each orbital in a set of orbitals may hold 2 electrons (ms = +1/2 and –1/2)
First

n=1
so: l = 0 ml = 0
ms = +1/2 or –1/2
H
He
n
l
ml
ms
H, He
1
0
0
+1/2
He
1
0
0
–1/2
The Pauli Principle leaves no other choices, so the next electron must
go into the next available orbital.
3 electrons (Li)
May be drawn vertically:
4 electrons (Be)
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5 electrons (B)
6 electrons (C)
Hund's Rule:
7 electrons (N)
8 electrons (O)
9 electrons (F)
10 electrons (Ne)
11 electrons (Na)
Note: H, Li, B, C, N, O, F, and Na are Paramagnetic
Paramagnetic:
He, Be, and Ne are Diamagnetic.
Diamagnetic:
NOTE: Do not use abbreviated electron configurations when the complete one is
needed.
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Ordering:
___ ___ ___ ___ ___
4d
___ ___ ___
4p
___ ___ ___ ___ ___
3d
___ ___ ___
3p
___
4s
___
3s
___ ___ ___
2p
___
2s
___
1s
Predictions are complicated, but:
1s
2s
3s
4s
5s
6s
7s
2p
3p
4p
5p
6p
7p
3d
4d 4f
5d 5f
6d 6f
Also: s 
p
People get in trouble with this because
they either draw it sloppy or they add
impossible orbitals (i.e., 2d).
d
f 
2 electrons
6 electrons
Example: Fe 26e–
Example: Ra 88e–
KNOW:
There are some exceptions:
Cr
Cu
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10 electrons
14 electrons
Example: List the quantum numbers for each of the electrons in the outer shell of
fluorine.
6.9 Electron Configurations and the Periodic Table
NOTE:
H
Li
Na
Valence Shell:
Valence Electrons:
Core Electrons:
Main Group Elements (Representative Elements):
Transition Metals:
Inner Transition Metals:
Lanthanide Series:
Actinide Series:
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Relating to the periodic table:
The Periodic Table gives a lot of important information. Such as:
1. Number of electrons in the outer shell (valence electrons) = Group Number
2. Electron configuration (Position)
3. Group Names:
4. Metals give cations
Nonmetals give anions
Electron configurations of ions
Anions: simply continue to add electrons (aufbau)
Cations: remove electrons from the outer shell (maximum n, not aufbau)
Isoelectronic:
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CHAPTER 7
7.1-7.2 (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Read
7.3 Sizes of Atoms and Ions
Atomic Radius
The downward increase is because there are more shells.
The leftward increase is because there are fewer electrons in the outer shell, so
there is less shielding.
There are some minor variations, especially in the transition elements.
The conversion of an atom to an ion not only changes the number of electrons but
also the size.
Atom 
atomic radius
Ion

ionic radius
In an atom, there is a balance between the nuclear attraction (shielded) and the
electron repulsion.
Cations are ______________ than their parent atoms.
Anions are _______________ than their parent atoms.
The greater the number of electrons lost or gained the greater the change in radii.
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7.4 Ionization Energy
Ionization Energy:
A(g) + energy 
A+(g) + e–
(First) ionization energy
A+(g) + energy 
A2+(g) + e–
Second ionization energy
There are periodic trends
In general, the value is higher for elements to the right and to the top of the
periodic table.
Larger radii (further from the nucleus)
Low ionization energies 
metals (cations)
Specifics:
Filled shells (noble gases)
Filled subshell (IIA)
Half-filled subshell (VA)
high
high
high
Removal of more than one electron:
7.5 Electron Affinities
Electron Affinity:
A(g) + e– 
A–(g) + energy
The energy may be endothermic or exothermic
A–(g) + e– 
A2–(g) + energy
Second electron affinity (always endothermic)
It is difficult to add more electrons than are required to fill the valence shell.
Best for small atoms – electrons enter nearer the nucleus, also for nearly filled shells.
The Noble gases have no affinity.
There are some variations in the trend (c.f. ionization energy)
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7.6-7.8 Chemical Properties
Read
Study Guide Example – not covered in class:
Write balanced chemical equations for each of the following.
a. Sodium is added to water
b. Strontium is added to water
c. Magnesium is heated in nitrogen to produce magnesium nitride
d. Zinc burns in oxygen
e. Lithium reacts with chlorine
f. Hydrogen reacts with rubidium
g. Barium oxide is added to water
h. Iron(II) oxide reacts with hydrochloric acid
i. Selenium trioxide reacts with water
j. Carbon dioxide reacts with potassium hydroxide
k. Tetraphosphorus decaoxide reacts with water
l. Calcium hydride reacts with water
Answers:
a. 2 Na + 2 H2O
b. Sr + 2 H2O
c. 3 Mg + N2
d. 2 Zn + O2
e. 2 Li + Cl2
f. H2 + 2 Rb
g. BaO + H2O
h. FeO + 2 HCl
i. SeO3 + H2O
j. CO2 + 2 KOH
k. P4O10 + 6 H2O
l. CaH2 + 2 H2O












2 NaOH + H2
Sr(OH)2 + H2
Mg3N2
2 ZnO
2 LiCl
2 RbH
Ba(OH)2
FeCl2 + H2O
H2SeO4
K2CO3 + H2O
4 H3PO4
Ca(OH)2 + 2 H2
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
END EXAM 3 MATERIAL
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Reminder: It is exceedingly important that, while it is fresh in your mind, to make sure that
you have the correct solution to every problem on the exam (and re-test). Not doing so will
have a strong negative influence on your grade on the final exam.
Reminder: If your grade on the test is not what you want, you should immediately refer to
the syllabus and/or see the instructor.
Bonding – increased stability exhibited by the Bond Energy
CHAPTER 8
CHANGE ORDER (Cover 8.4 first, and then cover in order)
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
8.4 Bond Polarity and Electronegativities
Ionization energy and electron affinity both deal with electron attraction – sometimes
one or the other is important. More often, a combination is necessary:
Electronegativity
Electronegativity:
The highest electronegativity is _____.
In general, the closer an element is to _____ the higher is its electronegativity.
For two elements the same distance away, the one __________ on the periodic table
has the higher value.
Example:
H—H
Na — Cl
H — Cl
The degree of polarity depends on just how different the electronegativities are:
Identical atoms
Different nonmetals
Metal + nonmetal
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8.1 Lewis Symbols
Lewis Symbol:
Use only the valence electrons
They only work well for representative elements
The number of valence electrons equals the group number (except helium)
The maximum for an element is 8
Example: Se
Most ions have the same number of electrons in their outer shells as a noble gas. For
all noble gases except He this is 8.
The formation of ionic bonds involves a transfer of electrons.
8.2 Ionic Bonding
Example: 2 Na + F2 
2 NaF
Example: Li + S what compound will form?
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Lattice Energy:
It is always endothermic.
The Born-Haber Cycle is a method for determining lattice energies. This is an
application of Hess's Law
8.3 Covalent Bonding
The atomic orbitals may overlap and allow electrons to be shared:
The electrons can more through the overlap region and are shared by both atoms.
The atoms can only get so close 
bond length
Example: Cl2
(Single) Covalent Bond:
Do not use lines for bonds yet
Example: O2
Double (Covalent) Bond:
Bonding Pair:
Nonbonding Pairs (Lone Pairs):
Example: CN–
Triple Bond:
The maximum number of pairs between two atoms is 3.
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Octet Rule:
8.5 Drawing Lewis Structures
Based on the Octet Rule
Example: HBr
Note: If the Lewis structure that you draw does not match one given in class, do not
simply correct your result. You should write the correct answer next to your
result, then when you review you can analyze your mistake and make sure that is
does not happen again. Finding out why you made a mistake is more helpful than
simply correcting you mistakes.
Example: H2S
Example: C + F
Example: NF3
Example: HCN
Example: NH4+
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Example: NO
Example: HNO2
Note: Watching Lewis structures being drawn often leads to overconfidence on the part of
the student. It takes a lot of practice before they become as easy as they appear in class.
Start practicing these as soon, and as often as possible.
Formal charge may be used to check to see if a structure is reasonable.
Formal Charge =
The formal charges always sum to the charge on the molecule or ion.
Elements with a high electronegativity usually do not have a positive formal
charge (and vice versa for low electronegativity).
Formal charges are usually +1, 0 or –1
Adjacent atoms are not normally both + or both –.
8.6 Resonance Structures
Example: SO3
Resonance:
Resonance Hybrid:
Resonance Structures:
Example: NO2–
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8.7 Exceptions to the Octet Rule
There are exceptions to the octet rule.
Less than an octet:
Near He (He structure = 2 electrons)
Many elements with less than 4 valence electrons
Sometimes it is possible to complete the octet through reaction with
other molecules.
Coordinate Covalent Bond:
Odd electron molecules
If there is not an even number of electrons, there must be an odd
electron.
Example: NO2
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More than an octet
This may occur for elements in the third period or below.
It is the most obvious for elements with 5 or more bonds.
Example:
XeF4 (Xe already has an octet, but the fluorines are attached
so it will get more electrons)
The most electronegative element will get its octet.
NOTE: Go to Lewis Structures (see the website).
8.8 Strengths of Covalent Bonds
Bond Energy =
H =  bonds broken –  bonds formed
For a reaction:
The types of bonds are important (single, double or triple)
It takes more energy to break a bond if resonance is present.
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
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CHAPTER 9
9.3 Molecular Shape and Molecular Polarity
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Polar Molecule:
Dipole:
Dipole-Dipole Force:
Dipole Moment:
Ion-Dipole Force:
Instantaneous Dipole:
The instantaneous dipole creates an induced dipole.
This is related to the number of electrons and how tightly they
are held (polarizability).
London Dispersion Force:
van der Waals Force:
If only one covalent bond is present, a molecule is polar if the bond is polar.
If more than one covalent bond is present, a molecule may or may not be
polar. Both the polar covalency and the shape are important considerations.
Example: HCl and H2O
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Kinetic Molecular Theory applied to solids and liquids
Same as for gases, except there are different numbers
There is a competition between the kinetic energy and the intermolecular forces.
K.E. >> Intermolecular forces 
K.E.  Intermolecular forces 
K.E. << Intermolecular forces 
Changes of state
Freezing (Solidification):
Fusion:
Vaporization (Evaporation):
Condensation:
Sublimation:
Deposition:
9.1-9.2 and 9.4-9.5 VSEPR – VB
Structures of molecules
Two basic methods (for representative elements)
Both give the geometry. The geometry is fundamental to many physical and
chemical properties.
A. VSEPR: All electrons in the valence shell of the central atom are
significant as are all electrons donated to the central atom. These are
determined from a correct Lewis structure.
Electrons normally occur in pairs. Bonding pairs (bp) and Lone pairs (lp)
Repulsion ranking:
lp-lp >> lp-bp > bp-bp
The Lewis structure gives the Orbital Geometry (Electron-Pair Geometry).
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Orbital Geometry (Electron-Pair Geometry):
The orbital geometry plus the atoms gives the Molecular Geometry.
Molecular Geometry:
B. VB Theory: The electron orbitals may rearrange to form a more
stable structure. Orbitals overlap and electrons are shared.
In order to make more orbitals available (for more bonds), hybridization
(mixing) may be required.
Set up a table (on a separate page)
e–pairs | lone pairs | orbital
| molecular | hybridization | polar |
|
| geometry | geometry |
|
|
Polar
A diatomic molecule is polar if the two atoms differ in electronegativity, and
is nonpolar if they have the same electronegativity.
For polyatomic molecules, the polarity depends on the electronegativity
differences and the geometry.
Example:
BeI2
Compounds with this type of Lewis structure are linear. This is commonly
observed for a small group of covalent compounds containing the metals Be,
Zn, Cd, and Hg. It is also observed for some compounds containing multiple
bonds.
VSEPR
VB
Hybridization:
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Remember: If the Lewis structure that you draw does not match one given in class, do not
simply correct your result. You should write the correct answer next to your result, then
when you review you can analyze your mistake and make sure that is does not happen again.
Finding out why you made a mistake is more helpful than simply correcting you mistakes.
Example:
BCl3
and
NO2–
VSEPR
VB
The double bond is special – an unhybridized p orbital is required for it. (2 are
required for a triple bond.)
p2 (the other p is not hybridized)
This would give an angle of 90°, which is too small. The observed
angle is near 120°, so sp2.
Example:
CCl4
NCl3
VSEPR
Polarity:
However, if one atom is changed:
CHCl3
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OF2
VB
Note: for a larger central atom (i.e., S) and a smaller substituent (i.e., H) the
opening is not needed.
H2O 105°
Example:
SbF5
H2S 90°
ClF4+
ClF3
VSEPR – Orbital Geometry – trigonal bipyramid
VB
Example:
ClF6+
IF5
VSEPR
VB
NOTE: Go to VSEPR on the website.
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BrF4–
KrF2
9.6 Multiple Bonds
Direct overlap:
H2
HF
F2
All are  bonds
 bond:
Formation of a double bond
CO2
There is no more room between the C and the O's for additional electron pairs.
 Bond:
Delocalized Orbitals
NO2–
Lewis
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9.7 Molecular Orbitals
Molecular Orbital Theory:
Molecular Orbital Energy Level Diagrams –
Example: H2
Bonding Molecular Orbital:
Antibonding Molecular Orbital:
For p orbitals both  and * bonds are possible.
The electrons are indicated as in an orbital diagram ( || ) the orbitals are filled
beginning with the lowest. Hund's rule and the Pauli Exclusion Principle still hold.
H2
__
*
__
1s
__
1s
__

Bond Order =
In general, all s combinations give the same pattern (incomplete as shown):
__
*
__
__
ns
ns
__

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p orbitals are in a set, so they must be dealt with together (incomplete as shown).
__
*
__ __
* *
__ __ __
np
__ __ __
np
__

__ __
 
The patterns for O, F, and Ne are slightly different (incomplete as shown):
__
*
__ __
* *
__ __ __
__ __ __
np
np
__ __
 
__

He2
ions:
H2+
H2–
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9.8 Period 2 Diatomic Molecules
Li2
Only use those parts of the valence shell that are occupied, but make sure all are used.
Be2
bond order
=
B2
=
C2
=
N2
=
O2
=
F2
=
Ne2
=
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For ions fill the molecular orbitals as expected, then add or subtract the appropriate
number of electrons.
Heteronuclear diatomics
To a certain extent, they are like homonuclear cases.
Example:
CN–
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
END OF EXAM 4 MATERIAL
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Reminder: It is exceedingly important that, while it is fresh in your mind, to make sure that
you have the correct solution to every problem on the exam (and re-test). Not doing so will
have a strong negative influence on your grade on the final exam. The Final is getting
close.
Reminder: If your grade on the test is not what you want, you should immediately refer to
the syllabus and/or see the instructor.
Note: Many people get into trouble on this last material because the final not only contains
this new material, but it also contains a comprehensive portion. The cause of the difficulty is
that the students budget their normal amount of study time here, however this is only
sufficient time to prepare the new material OR review for the comprehensive portion. In the
past a large percentage of the persons taking the test have done well on one portion of the test
and exceedingly poorly on the other. From now until the end of the term your study sessions
should contain at least some review of the old exams.
CHAPTER 11
11.1 (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Read
11.2 Intermolecular Forces
London Dispersion Forces attract nonpolar molecules to each other.
Types of intermolecular forces:
1. "Normal" bonding –
Ionic, Covalent(C, BN, metalloids and SiO2),
and Metallic
2. Ion-Dipole Forces – an intermediate case normally involving two
substances. Also stronger than a simple dipole-dipole
3a. Dipole-Dipole Forces – between polar molecules, not as strong as a
regular bond (only partial charges are involved)
3b. Hydrogen Bond: A special case of dipole-dipole involving H directly
bonded to N, O, or F.
4. Ion-Induced Dipole
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5. London Dispersion Forces – always present
Instantaneous Dipole (Temporary dipole)
The instantaneous dipole creates an induced dipole.
This depends on the number of electrons and how tightly they
are held (polarizability).
These non-bonding intermolecular forces are labeled van der Waal's Forces.
The distance between nonbonded atoms (minimum) is equal to the sum of the
van der Waal's radii.
11.3 Select Properties of Liquids
Surface Tension:
Capillary Action:
Viscosity:
Amorphous Solid (Glass):
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11.4 Phase Changes
The main difference between gases vs. liquids and solids is due to Intermolecular Forces.
Evaporation (Vaporization, Condensation)
Evaporation, Vaporization
Liquid <==================> Gas
Condensation
Melting Point (Freezing, Melting, Solidification)
Freezing
Liquid <=======> Solid
Melting
Solid-Gas (Sublimation, Deposition)
Sublimation
Solid <===========> Gas
Deposition
At the boiling point, a lot more energy is necessary to overcome the intermolecular
forces and to convert the liquid to gas  Molar Heat of Vaporization (energy
required to vaporize a mole of a substance).
Hvap
CH4
9.2 kJ/mole
Et2O
26.0
it is related to the
H2O
40.8
intermolecular forces
The reverse is the Molar Heat of Condensation (energy released when a mole
of a substance condenses) – the same numerical value, but the opposite sign.
For other transitions
Molar Heat of Fusion (Crystallization, Solidification)
Molar Heat of Sublimation (Deposition)
Heat of ....: Energy change when a substance undergoes the indicated change.
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Heating curve:
Cooling follows the same curve in the opposite direction..
Superheating:
Supercooling:
11.5 Vapor Pressure
Vapor Pressure
Evaporation goes for a while, and then it seems to stop.
Vapor Pressure:
Dynamic Equilibrium:
Le Châtelier's Principle:
Volatile:
The vapor pressure is due to molecules escaping from the surface – below the
surface, they are trapped.
When Pvap = Pext
bubbles can form below the surface and not be
crushed.
Boiling Point:
Normal Boiling Point:
Variations in the vapor pressure with temperature may be calculated:
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P
ln (P1 ) =
2
∆Hvap
R
1
[T −
2
1
T1
]
The T's are in Kelvins
R = 8.314 J/mole K = 1.987 cal/mole K
NOTE:
Example:
The normal boiling point of ethyl alcohol is 78.4°C. The heat of vaporization of this
compound is 40.5 kJ / mole. Calculate the vapor pressure of ethyl alcohol at a
temperature of 55.0°C.
11.6 Phase Diagrams
Phase Diagram: A summary of the phase relationships.
P
T
All phase diagrams look basically the same (for one-component)
Triple Point:
Critical Point:
11.7 Liquid Crystals
Read
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Reminder: It is exceedingly important that, while it is fresh in your mind, to make sure that
you have the correct solution to every problem on the exam (and re-test). Not doing so will
have a strong negative influence on your grade on the final exam. The Final is getting
close.
Reminder: If your grade on the test is not what you want, you should immediately refer to
the syllabus and/or see the instructor.
Note: Many people get into trouble on this last material because the final not only contains
this new material, but it also contains a comprehensive portion. The cause of the difficulty is
that the students budget their normal amount of study time here, however this is only
sufficient time to prepare the new material OR review for the comprehensive portion. In the
past a large percentage of the persons taking the test have done well on one portion of the test
and exceedingly poorly on the other. From now until the end of the term, your study sessions
should contain at least some review of the old exams.
CHAPTER 12
12.1 Classification of Solids
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
These attractions are not as strong as “normal” bonding: ionic, covalent, and metallic.
“Normal” bonding occurs in Ionic Solids, Network (Covalent) Solids, and
Metallic Solids. (“Normal” bond is a simplification – it is never an
acceptable answer to any problem on an exam.)
Ionic Solid:
Network (Covalent) Solid:
Metallic Solid:
All solids not held together by “normal” bonds are Molecular Solids,
Molecular Solid:
Any attraction, other than a “normal” bond, is a van der Waal’s Force.
All attractions of any kind are Intermolecular Forces.
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12.2 Structures of Solids
Crystalline Solid (true solid):
Amorphous Solids

supercooled liquids (undercooled liquids)
Crystalline solids

ordered array = Lattice
Lattice:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Unit Cell:
There are only 7 different types of lattices with only 14 types of unit cells
(Bravais Lattices) needed to describe every solid.
A unit cell is like a "brick." How do you describe a brick?
Six things are necessary:
Length of sides = a, b, and c
Angles between = , , and 
Cubic
Tetragonal
Orthorhombic
Monoclinic
Triclinic
Hexagonal
Rhombohedral
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Three of the Bravais lattices are cubic:
Simple (Primitive)
Body-Centered
Face-centered
Closest Packed Structure:
12.3/12.5/12.6/12.7 Types of Solids
Types of crystals
Ionic Solids
Contain ions 
ionic bonds
Covalent Solids (Network Solids)
Carbon Allotropes – graphite
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diamond
Silicon dioxide
Metallic Solids
Metallic bonds
Bonding – Band Theory a modification of molecular orbital theory
Molecular Solids/Atomic Solids
Contain individual molecules and in some cases only atoms
Amorphous "Solids"
Glass is a common example, so are some plastics.
Summary:
Property
Ionic
Molecular
m. p.
Solubility
in H2O(polar)
in nonpolar
Conducts electricity
Hard and Brittle
Bonding
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Covalent
Metallic
12.4 (12.7) Metallic Bonding
Metal:
Band Theory:
Bonding – Band Theory a modification of molecular orbital theory
Li [He]2s1
Some network solids may be semiconductors. Band theory still applies. With
covalent bonds, there is a definite separation of bonding and antibonding
molecular orbitals.
Conduction Band
Band Gap:
Valence Band:
Most semiconductors are doped to improve their conductivity.
n-type –
p-type –
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Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
12.8-12.9
Read
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CHAPTER 13
Reminder: Many people get in trouble on this last material because the final not only
contains new material, but it also contains a comprehensive portion. The cause of the
difficulty is that the students budget their normal amount of study time here, however this is
only sufficient time to prepare the new material OR review for the comprehensive portion.
In the past a large percentage of the persons taking the test have done well on one part of the
test and exceedingly poorly on the other. From now until the end of the term, your study
sessions should contain at least some review of the old exams. (This is why you went over
the exams and re-tests as soon as they were returned, and while they were fresh in your mind.
If you did not do this, you have to catch up now.)
Solutions are homogeneous mixtures.
The solvent may be a solid, liquid or a gas. The solute may be a solid, liquid or a gas.
13.1 The Solution Process
(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)
Examples of various solutions
Example: Water plus a polar solute
Solvent-solvent
Solute-solute
Solvent-solute
Example: Water plus nonpolar solute
Solvent-solvent
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Solvent-solute
Example: Water plus a gas
Solvent-solvent
Solute-solute
Solvent-solute
Example: Water plus an ionic compound
Solvent-solvent
Solute-solute
Solvent-solute
Example: Nonpolar plus nonpolar
Solvent-solvent
Solute-solute
Polar substances will dissolve in each other.
Nonpolar substances will dissolve in each other.
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Solvent-solute
Two considerations
Heat of Solution:
Disorder (entropy) –
13.2-13.3 Saturated Solutions and Solubility
Solutions will only hold so much:
Saturated – Unsaturated – Supersaturated
Solute-Solvent Interactions
"Like dissolves like"
Temperature effects
Pressure effects
Henry's Law:
13.4 Concentrations
Molarity =
Mole fraction = X =
Weight Percent (Mass Percent) =
Volume Percent =
Molality = m =
Reminder: Many students get into trouble because they incorrectly use “m” and the
abbreviation for moles. The correct abbreviation is “mol.” The abbreviation “m”
means molality. The use of an incorrect abbreviation has resulted in students
making very significant factor label mistakes, and losing several points on an exam.
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Example: At room temperature, 13.20 g of ammonia will dissolve in 100.0 mL of
ethyl alcohol (C2H5OH). Calculate the mass percent, m, and X of ammonia. (The
density of ethyl alcohol is 0.7893 g/cm3.)
Example: A solution contains 85.4 g of sodium sulfate dissolved in 200.0 g of water
at 100°C. What is the total molality of the ions in this solution?
13.5 Colligative Properties
Colligative Properties:
Vapor pressure of solutions
Two cases:
Ptotal < P°pure
1. Volatile solvent with a nonvolatile solute
P = XP°solvent
Raoult's Law:
2. Both solvent (A) and solute (B) are volatile.
Ptotal = Psolvent + Psolute
Ptotal = XAP°A + XBP°B
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Still Raoult's Law
Ideal Solution:
PB°
PA°
0
XB
1
Boiling point elevation and Freezing point depression
Boiling Point Elevation:
Freezing Point Depression:
T = Kbm or T = Kfm
The T's refer to how much the boiling point in increased or to how much the
freezing point is lowered.
van't Hoff Factor (i):
Given the following setup:
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Osmosis:
Osmotic Pressure =  =
For electrolytes, the van't Hoff factor is needed:
=
Reverse Osmosis –
Example: A 225.0 gram sample of benzene, C6H6, has a 10.0-gram sample of lauryl
alcohol, extracted from coconut oil, dissolved in it. A 1.22°C depression of the
freezing point of the benzene occurred. Determine the molar mass of lauryl alcohol.
The value of Kf for benzene is 5.08°C/m.
13.6 Colloids
There are also systems intermediate between homogeneous mixtures and
heterogeneous mixtures
Colloid (Colloidal Dispersion):
Tyndall Effect:
NOTE: See the PROBLEM SOLVING section at the very end of this guide.
Do not forget the number of assigned problems has been trimmed to the absolute minimum
for the exams. Unfortunately, there is no way to assign fewer problems and assure good
performance on the exams.
END OF 133 MATERIALS
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PROBLEM SOLVING
This is a problem-solving course. You must be able to solve many types of problems.
Nearly every student, who has trouble with problem solving, has trouble because that student
makes the class more difficult than it is. Even students who do not have trouble with
problem solving make the class more difficult. A common example of making the class
more difficult concerns conversions. You are required to learn one English-SI length
conversion. Many students go ahead and learn two or three conversions instead. The
additional conversions may make one or two problems, out of all the problems done during
the term, easier. Therefore, for the sake of maybe one or two problems these students have
doubled or tripled the work required. In addition, the more conversions you memorize the
greater the chance of becoming confused. A very common mistake on exams is to find
mixed conversions. A mixed conversion is occurs when a student combines two or more
conversions. Now, simply learning an extra length conversion is not going to take that much
time, however, students do not stop there; they learn extra items every time. Not all these
extra items make you a better student; they just clog up your mind. Another example occurs
in the Gas Law chapter. Every problem assigned, presented in class, or given on an exam
requires only four equations. However, students insist on learning the 15-20 equations given
in most textbooks; again, there are serious problems with wasting time on the 11-16 extra
equations. The following discussion emphasizes many points that will help simplify this
class.
One of two methods (or a combination of the two) will solve the problems in this course.
The first method is the “plug-in” method, and the other is Factor Label. The plug-in method
concerns the rearrangement of one of a small number of equations given you, and then
numbers given to you are “plugged-in.” The Factor Label method involves tracking the units
so that all unwanted units cancel to leave only the desired unit(s). Finally, some problems
may involve a plug-in followed by one or more Factor Label steps. Many students make this
last case difficult by treating this type of problem as two or more different problems.
The plug-in method requires you to learn one or more of the following equations:
Exam 1 (Chapters 1-3)
NONE
Exam 2 (Chapters 4-10):
PV = nRT
M1V1 = M2V2
P1 V1
RateA
RateB
P2 V2
=n
n1 T1
P
ln (P1 ) =
2
h
∆Hvap
P = P°solvent V = nRT
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Ptotal = Pa + Pb + Pc + ....
2 T2
Exam 3 (Chapters 5-7): E = q + w
 = mv
2
E = 1/2 mv
Exam 4 (Chapters 8-9):
NONE
Exam 5 (Chapters 11-13):
MW
= √MWB
R
1
[T −
2
 = c
1
T1
hc

]
= iMRT
101
E = h =
T = i Kbm or T = i Kfm
As may be seen for the above list, this entire course consists of 16 equations. Some of these
equations are simple variations of others on the list. There is one additional equation, which,
if you need it, will be given to you in the problem. You only need these equations. The most
equations that you need to learn for any exam is five. Do not be foolish and try to learn
more. You would be surprised to see how many equations, besides those listed above;
people try to learn for Exam 1.
Every problem not requiring one of the above equations requires the Factor Label method.
This method may appear in addition to one or more of the above equations. Factor Label
involves SI prefixes, specifically designated conversions and constants, and definitions. If
you want to avoid making this course more difficult, stay very strictly within each of these
categories.
The SI prefixes and their numerical values are in a table in Chapter 1. Learn these exactly as
they appear in this table. Thus if you learn that c = 0.01 then by a simple substitution: 1 cm =
0.01 m. Many people learn this relationship and then waste their time learning that 100 cm =
1 m. You must learn the table for the Exams; why learn extra superfluous conversions? In
addition, during a normal term 60-70% of the students in the class will make at least one
mistake on the exam using the second type of conversion.
You will need to learn a minimum number of other conversions and constants. In some
cases, it will be your option, such as with the English-SI length and mass conversions, while
in other cases there are very specific conversions. The constants have different numerical
values depending on the units specified; learn only the ones given in class do not waste your
time on others.
Finally, definitions are very important to the Factor Label method. There are two ways of
setting up a definition for Factor Label. The first type uses very specific units, for example, J
= kg m2 / s2 or Pa = kg / m s2. The second type uses classes of measurements, for example,
speed = distance / time or density = mass / volume. The second type means that speed, for
example, may be expressed several ways; some of these would be miles / hour, kilometers /
hour, inches / second, and meters / minute. All that is necessary in this type of definition is
any measurement unit that fits the classification. Many people first see this type of definition
as if it were a mathematical relation; however, the sooner these people leave this limitation
behind the better they will be. The elementary level treatment of, for example, the density
relation as a mathematical relation retards many students. The sooner you go beyond this the
easier other problems will be. The advantage of using density as an equation makes a few
introductory problems easier, and impedes your progress towards excellence in the Factor
Label method. Do not let anything slow your work on understanding the Factor Label
method – you do not want to be 4-5 weeks behind when you get to Chapter 6, because you
will not get pass this chapter without Factor Label. Force yourself to use the Factor Label
method even when you want to fall back on elementary methods.
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