PreAP Chemistry Spring Review
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Spring Review 2005
PreAP CHEMISTRY I
(Items in italics are the actual objective…4 questions per objective)
Chapter 11 – The Mole – TEKS Obj. 2C
1. Calculate and evaluate the chemical quantity of the mole in common compounds in terms
of molar mass and number of representative particles. (2C)

p347: 94, 104, 105, 121, 158
94. 0.200 mol Au x 0.75 = 0.150 mol Au
0.150 mol Au x 6.02 x 1023 atoms Au
1 mol Au
22
= 9.03 x 10 atoms Au
104. 10.0g C x
1 mol C x 6.02 x 1023 atoms of C
12.01 g C
1 mol C
= 5.01 x 1023 atoms C
10.0g Ca x 1 mol Ca x 6.02 x 1023 atoms Ca
40.08 g Ca
1 mol Ca
= 1.50 x 1023 atoms Ca
10.0 g C contains more atoms
105. One mole of any substance contains 6.02 x 1023 representative particles. Thus, 10.0
moles of carbon and 10.0 moles of calcium contain the same number or atoms.
10.0 mol x 6.02 x 1023 atoms
1 mol
= 6.02 x 1024 atoms
121. Silver acetate:
1 mol Ag x 107.87 g Ag
1 mol Ag
= 107.87 g Ag
2 mol C
x
12.01 g C
1 mol C
= 24.02 g C
3 mol H
x
1.008 g H
1 mol H
= 3.024 g H
2 mol O
x
16.00 g O
1 mol O
= 32.00 g O
Molar mass
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= 166.91g/mol
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2.50 mol AgC2H3O2 x 166.91 g AgC2H3O2
1 mol AgC2H3O2
417 g AgC2H3O2
2.50 mol AgC2H3O2 x 6.02 x 1023 formula units
1 mol AgC2H3O2
= 1.51 x 1024 formula units
Glucose:
6 mol C x 12.01 g C
1 mol C
= 72.06 g C
12 mol H x 1.008 g H
1 mol H
= 12.01 g H
6 mol O x 16.00 g O
1 mol O
= 96.00 g O
Molar mass
= 180.16 g/mol
324.0 g glucose x
1 mol glucose
180.16 g glucose
=1.798 mol glucose x 6.02 x 1023 molecules
1 mol glucose
= 1.082 x 1024 molecules
Benzene:
Molar mass
= 78.11 g/mol
5.65 x 1021 molecules benzene x
1 mol benzene
6.02 x 1023 molecules benzene
= 9.39 x 10-3 mol benzene x 78.11 g benzene
I mol benzene
= 0.733 g benzene
Lead(II) sulfide:
1 mol Pb x 207.2 g Pb = 207.2 g Pb
1 mol Pb
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1 mol S x 32.07 g S
1 mol S
Molar mass
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= 32.07 g S
= 239.3 g/mol
100.0 g PbS x 1 mol PbS = 0.4178 mol PbS
239.3 g PbS
0.4178 mol PbS x 6.02 x 1023 formula units
1 mol PbS
= 2.516 x 1023 formula units
158. a. 17.63 g CO2
1 mol C x 12.01 g C = 12.01 g C
1 mol C
2 mol O x 16.00 g O = 32.00 g O
1 mol O
Molar mass
17.64 g CO2 x
= 44.01 g/mol
1 mol CO2 x
44.01 g CO2
2 mol O
1 mol CO2
x
6.02 x 1023 atoms O
1 mol O
= 4.82 x 1023 atoms O
b. 3.21 x 1022 molecules CH3OH x
c. 0.250 mol C6H12O6 x
1 atom O
= 3.21 x 1022 atoms O
1 molecule CH3OH
6 mol O
x
1 mol C6H12O6
6.02 x 1023 atoms O
1 mol O
= 9.0 x 1023 atoms O

An individual bag of M&M’s contains 54 pieces of candy. How many bags of M&M does it
take to have a mole of candy pieces?
1 bag
x 6.02 x 1023 pieces M&M
54 pieces
1 mol M&M
0.11 x 1023 bags/mol
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
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If each piece of candy weighed 0.15 grams, how much would one mole of candy weigh?
0.15 g Candy
1 piece
x
6.02 x 1023 pieces of M&M
1 mol M&M
=9.0 x 1022 g/mol
Chapter 12 – Stoichiometry – TEKS Obj. 2C, 11C
2. Use stoichiometric relationships to calculate reactants and/or products involved in a
chemical reaction (2C, 11C) including percent yield and identifying the limiting reactant.

P381: 65, 75, 81, 86, 92, 98
65. 523 g C3H7COOC2H5
75. 0.587 g Na3Ag(S2O3)2
81. 10Cl2(g) + P4(s)  4PCl5(s)
16.0 g Cl2
x
1 mol Cl2
= 0.226 molCl2
70.90 g Cl2
23.0 g P4 x 1 mol P4
= 0.185 moles
123.88g P4
According to the balanced equation, Cl2 reacts with P4 in a ten to one ratio.
0.226 mol Cl2 x
1 mol P4
10 mol Cl2
= 0.00226 mol P4 needed
Cl2 is limiting reactant; P4 is in excess.
86. a. 103 g CO2
b. 94.7%
92. a. 2NH3(g) + 3CuO(s)  3Cu(s) + 3H2O(l) + N2(g)
b. 2.35 mol NH3
1.01 mol CuO
CuO is the limiting reactant.
c. 9.43 g N2
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98. Equation: 8Al + 3NH4ClO4  4Al2O3 + 3NH4Cl
6.0 x 105 kg NH4ClO4 x 1000g NH4ClO4 x
1 kg NH4ClO4
= 5.11 x 106 mol NH4ClO4 x
1 mol NH4 ClO4
117.50 g NH4ClO4
4 mol Al2O3
3 mol NH4ClO4
= 6.81 x 106 mol Al2O3 x 101.96 g Al2O3
1 mol Al2O3
= 6.94 x 108 g Al2O3 x 1 kg Al2O3
1000 g Al2O3
= 6.94 x 105 kg Al2O3
= 6.56 x 105 kg Al2O3 x 100
6.94 x 105 kg Al2O3
= 94.4% yield Al2O3
Chapter 13 – States of Matter – TEKS Obj. 1B, 4B, 5A, 5C, IPC-7A, 8D
3. Relate the kinetic molecular theory with temperature and pressure when describing the
three states of matter (4B)
 Fill in the following chart with as much detail as possible.
Particle
motion

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no effect
Volume is
determined
by…
Density
Effect of
I.M.F.
limit motion
of particles
to vibrations
around fixed
locations
slow
may
slightly
Density;
Limited
LIQUIDS
compress
container
range of
somewhat
motion
fast
as pressure
Pressure and
very weak,
GASES
increases,
temperature;
allows gas
volume
container
particles to
decreases
move freely
Describe what occurs from a pressure, temperature and volume perspective when a
soda can is heated with just a small amount of water & then inverted in a bath of ice
water.
as temperature increases, the pressure and volume increases
as temperature decreases, the pressure and volume also decreases
SOLIDS
vibrational
How Pressure
effects
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4. Explain how dispersion forces, dipole-dipole forces, and hydrogen bonding
(intermolecular forces) affect the three states of matter using phase changes (5A) and
heating curves (5C).
 Draw & label a heating curve – include what state of matter is present at each segment,
as well as what process is occurring at each segment. Include temperature information
for water.
Liquid
Melting
Solid
 Fill in the following chart with as much detail as you can
Dispersion
Type of
molecule
affected
Nonpolar
Weak
Effect on
Likely state of
vapor pressure matter at room
T
Lots of VP
gas
Dipole-Dipole
Polar
Medium
Medium
Gas/liquid
Hydrogen
Bonding
H of one molc w/
NOF of other
Strong (of IMF)
Not much VP
Liquid/solid
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Relative
strength
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 Explain how the size of the molecule can affect the strength of the intermolecular
force.
Size – the bigger the molecule, the stronger the IMF (more electrons to be unevenly
distributed causing greater partial charges
Investigate and identify properties of liquids including viscosity and buoyancy (IPC – 7A)
Explain how water is treated as well as the importance of conserving water and practical
ways to do so. (1B)
 List and explain some impacts on our world as we know it (climate, world distribution of
water, etc) of water’s high heat capacity.
Impacts – climate: tropical area’s that have high humidity don’t have much temperature
fluctuation due to water’s high heat capacity – it holds the energy without changing temperature
significantly. Therefore, arid areas are the opposite. Deserts, etc are very hot in the day since
there is nothing else to absorb the energy & cold at night since there is nothing retaining it.
 Distribution: If the heat capacity were lower, the oceans would evaporate significantly more;
rain less; ice caps would not stay solid at such high temps – melt/flood, etc
 Describe viscosity – what it is; what determines the viscosity of a substance, how it relates to
density; effect temperature changes have on it.
resistance to flow; determined by IMF; higher temps lessen the viscosity. Ex – syrup has a high
viscosity (does not flow much); but hot syrup has lower viscosity – does flow better.
 What is unique about water when it solidifies? Explain why this happens.
 Unlike other substances, when water solidifies, it expands. Most solids contract. This is
due to the extreme partial charges on the water molecules. When the molecules slow
down & begin to get closer together, the partial charges are actually strong enough to
cause some repulsion & push the molecules farther apart.
5.
a.
Chapter 14 – Gases– TEKS Obj. 2C, 7A, 7B
6. Describe interrelationships among pressure, volume, temperature, and numbers of
particles of gases. (7A)
a. Apply stoichiometric principles to calculate the amount of gas produced in a chemical
reaction. (2C)
b. Apply stoichiometric principles to calculate amount of gas produced at STP. (2C)
 How does atmospheric pressure change at higher altitudes? What impact does that
have on cooking (boiling), packing for trips to the mountains, etc?
Atmospheric pressure is lower in the mountains. If pressure is lower, molecules are held
together less, so it takes less energy to break them apart, so the boiling point is lower. ALSO,
as P , T. When going to lower altitude areas (mountains, airplanes), things tend to expand.
The air trapped inside anything (shampoo bottle for ex) is not under as much pressure as it
was, so it is allowed to expand & sometimes even explodes!
 What is the one unit that must ALWAYS be converted? How do you make that
conversion?
One unit ALWAYS converted is CELCIUS to KELVIN. Add 273 to the C
 p 430:22 (& then what if conditions were changed to STP?) P449: 101, 103, 104;
Page 430: No: 22
T1 = 5.0oC + 273 = 278 K
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T2 = 2.09oC + 273 + 275 K
V2 = P1T2V1 = (1.30 atm) (275 K) (46.0 mL)
P2T1
(1.52 atm) (278 K)
= 39 mL
Page 449:
101. N2 + 3H2  2NH3
13.7 L H2 x (2 L NH3)/ (3 L H2) = 9.13 L NH3
103. 2KClO3  2KCl + 3O2
Molar mass = 39.10 g/mol + 35.45 g/mol + (3 x 16.00 g/mol) = 122.55 g/mol
n KClO3 = 20.8 g KClO3 x (1 mol KClO3/122.55 g KClO3) = 0.170 mol KClO3
nO2 = 0.170 mol KClO3 x (3 mol O2)/(2 mol KClO3) = 0.255 mol O2
V = 0.255 mol x (22.4L/1 mol) = 5.70 L O2
104. a. 2CO(g) + 2NO(g)  N2(g) + 2CO2(g)
b. 1 :1
c. nCO =
=
nN2 =
=
VN2 =
=
42.7 g CO x (1 mol CO )/(28.01 g CO)
1.52 mol CO
1.52 mol CO x ( 1 mol N2)/(2 mol CO)
0.762 mol N2
(0.762 mol) x (22.4 L/mol)
17.1 L N2
Explain the relationships between the 4 variables – ex: if T goes up, what happens to P?
What if T & P both go up, what happens to V? For each statement, what assumption has
to be made? If 3 of the 4 variables do not change, what can make the 4th one change?

7.
P & V are inverse – one goes up; other goes down. All others are direct – one goes upothers go up as well. If T & P go up proportionally, V will not change (effects will negate
the other). For the statement to be true, the other 2 variables must not change.
Likewise, if 3 do not change, the 4th won’t either. (P, V, T, n)
Calculate changes in temperature, pressure, volume, and moles of a gas following by
applying the Ideal/Universal gas law: (7B)
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a. Given experimental data, determine if the data is consistent with the Ideal/Universal
gas law.
 P 449: 100
T = -15oC + 273 + 258 K
n = PV/RT =
(1.00atm)(2.00L)
(0.0821 L.atm/mol.K)(258 K)
= 0.0944 mol
molar mass = (3 x 12.01 g/mol) + (8 x 1.008 g/mol) = 44.09 g/mol
mass = n x M = (0.0944 mol)(44.09 g/mol)
= 4.16 g
 Explain why the word “ideal” is used in this law. Under what conditions will a gas NOT
follow the IDEAL gas law?
Ideal implies not always, which is true with gases. If you cool any gas significantly, then as a
liquid, it will NOT follow the ideal GAS law!  High P or low T will not follow this law.
Ideal is used because it describes the physical behavior of an ideal gas in terms of pressure,
volume, temperature, and number of moles of gas present – ideal gases take up no space,
have no IMF’s and follow gas laws under all conditions of temperature and pressure; real
gases deviate from ideal gas law under extremely high pressures and low temperatures.
Chapter 16 – Energy and Chemical Change– TEKS Obj. 5B, 15A
8. Distinguish between energy, heat and temperature.
a. Interpret a graph involving heat changes of a system over time.
b. Describe the energy changes that occur in the system and surroundings for an
exothermic and endothermic reaction.
c. Calculate the quantity of heat absorbed or released when a substance changes
temperature.
600
Would this diagram represent an
endothermic or exothermic reaction?
Explain. Exothermic-end product has
less energy
500
Potential Energy
400
How would you calculate H for this
reaction? H = Hproducts - Hreactants= -200kJ
How would you write a
thermochemical equation for this
reaction? RP +200 kJ OR
RP H = -200 kJ
300
200
100
Reaction Progress
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Which is more stable in this diagram
– the reactants or products? Explain.
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Products – less KE, so less entropy( disorder)
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
Explain energy changes involved in changes of state. Ex – is condensing exothermic
or endothermic? Explain what is occurring on a molecular level & what applications each
change has. (Ex: how sweating cools your body down.) Explain for each state change
(melting, freezing, condensing, vaporizing).
Melting, vaporization, and sublimation-endothermic;
condensation, deposition, and freezing-exothermic.
Melting- heat flows from water to ice and disrupts hydrogen bonds holding water molecules
together in ice crystal; ice absorbs enough energy to break hydrogen bonds, move apart
and enter the liquid phase; melting point – temperature at which forces holding crystal lattice
together are broken and it becomes a liquid.
Vaporization – energy distributed unevenly in liquid, so some have the energy required to
overcome forces of attraction holding molecules together, allowing particles to escape from
liquid and enter gas phase; when vaporization occurs on surface only = evaporation,
requires energy – min. energy required to escaped liquid; how body controls water – water
molecules absorb heat energy from skin and evaporate.
 Sublimation – solid changes directly to gas; air fresheners, moth balls, dry ice, freezedried foods
 Condensation – molecules lose energy and form hydrogen bonds to condense; fog and
dew, clouds
 Deposition – gas to solid; snow crystals (H2O vapor to ice crystals)
 Freezing – molecules lose kinetic energy and velocity; when enough energy removed,
molecules fixed into position; freezing point – temp. at which liquid converted into
crystalline solid

P881: 5, 8
5) C = .2139
8) .493 kJ
SOLUTIONS
Enduring Understandings:
 Chemical solutions are an integral component of everyday life.
 Chemical properties of solutions affect their function.
Chapter 15 – Solutions – TEKS Obj. 12A, 12B, 12C, 13A & 13B
(&CH 10)
9. Describe the nature of solutions. Homogeneous mixtures containing two or more
substances called the solute and solvent; solute is dissolved, solvent is dissolving
medium.
a.
Factors that affect solubility (12A & 12B) solubility -maximum amount of solute that
will dissolve in a given amount of solvent at a specified temperature and pressure;
factors affecting- temperature (T increases, solubility increases in solids and liquids,
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NOT gases), pressure (solubility of gaseous solutes; solubility increases as external
pressure increases) SIP = SIP
b.
Colligative properties (13B) colligative properties -vapor pressure lowering (greater
number of solute particles in solvent, lower the resulting vapor pressure); boiling
point elevation (greater number of solute particles, higher boiling point); freezing
point depression (great number of solute particles, lower freezing point
c.
Saturated, unsaturated, and supersaturated solutions (13A) unsaturated -less
dissolved solute than saturated solution; saturated- contains maximum amount of
dissolved solute for given amount of solvent at specific temperature and pressure;
supersaturated- more dissolved solute than saturated solution at same temperature make saturated solution at high temperature and cool slowly ~ solute stays dissolved
at lower temperature -seed crystal allows precipitate
d.
Explain water’s ability to be the universal solvent. (12C) water molecules have very
strong dipoles (polar) so it dissolves most polar solutes as well as many ionic
compounds; bent shape results in dipoles that do not cancel each other out .
 Use Fig 15-7 (p 458). Label the following as saturated, supersaturated, and
unsaturated.
o KCl solution has 60 grams of solute per 100 g of H2O at 50C
Supersaturated
o NaCl solution has 37 grams of solute per 100 g of H2O at 50C
Saturated
o KCLO3 solution has 10 grams of solute per 100 g of H2O at 50C
Unsaturated
 Water is the universal solvent. Explain why this is possible using water’s characteristics
(heat capacity, polarity, intermolecular forces and their effect on cohesiveness)
 Heat capacity -can dissolve many different types of solutes and large volumes before it
reaches boiling point; polarity- many polar solutes and molecules; intermolecular forces
-hydrogen bonds ~ attraction between dipoles are stronger than IMFs in compound,
separating and dissolving the solute; cohesiveness -increases cohesiveness in different
mixtures because of strong cohesiveness of water.
 Explain the effect of temperature on gas solubility in solutions. What are practical
applications of this? As solution's temperature increases, solubility of gaseous solute
decreases; soda goes flat faster in warm temperatures than cold temperatures.
 What happens to the boiling point/ freezing point of a solution when a molecular
substance is added? When an ionic substance is added? What types of ionic
compounds will have the greatest effect?

When molecular substances are added, there is a slight change in freezing/boiling
points while when ionic substances are added, there are more drastic changes that
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occur; strong electrolytes that break into the most number of ions would have the
greatest effect on the freezing/boiling points of a solution.
Chapter 17 – Reaction Rates – TEKS Obj. 13C, 15B
10. Relate the rate of reaction to temperature, concentration, surface area, and catalyst
(15B) increase temperature = faster reaction; increase concentration = faster reaction;
increase surface area = faster reaction; add catalyst = faster reaction and lower
activation energy
a. Using energy diagrams, recognize activation energy, enthalpy change, and the effect
of a catalyst.
 Refer to the Energy diagram in ch 16 section of the review. How does a catalyst change
the diagram? What would be the activation energy for both the forward and reverse
reaction?
 Catalysts lower the “hill” – the activation energy. This does NOT change the H (products –
reactants). This will effect both the forward & reverse in the same way – speed them
BOTH up.
 If you add a metal piece to excess acid, what of the listed factors below will increase,
decrease, or have no impact on how fast a reaction happens?
a) Lower molarity (concentration) of a solution Decrease rate
b) Higher molarity (concentration) of a solution Increase rate
c) Increase volume of a solution (with the same molarity) no change
d) Decrease volume of a solution (with the same molarity) no change
e) Ribbon shape of the metal piece Increase rate
f) Cube shape of the metal piece Decrease rate
g) Ball shape of the metal piece Decrease rate
h) Increase mass of the metal piece Increase rate
i) Decreased mass of the metal piece Decrease rate
Chapter 18 – Chemical Equilibrium – (Intro a3)
11. Predict changes in an equilibrium system based on temperature, pressure,
concentration variations.
Le Chatelier's Principle: if you stress a system, it will attempt to undo that stress
 changes in concentration -stress on right  shift to left & vice versa
 changes in volume (pressure) -increased pressure shift to side with less moles
(equal moles means no change); decreased pressure  shift to side with more moles
 changes in temperature -heat added  reaction shifts in direction in which heat is
used up (endothermic); heat removed  shifts to exothermic side
a. Using the equilibrium constant expression, predict the shift in equilibrium
A + B  C + D
Keq = [C][D]/[A][B]
 Leave out solids & liquids since their concentrations don’t change; coefficients
become exponents
 P 574: 11, 15
11) a. shift  b. shift  c. shift 
15) no shift
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 What does it mean when the equilibrium constant
a) Has a large value (Keq >1) – products favored
b) Has a small value (Keq <1) – reactants favored
Chapter 19 – Acids and Bases – TEKS Obj. 14A, 14B, 14C, 14D
12. Relate the properties of acids and bases using graphs, charts, and experimental data.
a. Differentiate between acids, bases, and neutral solutions (including household
products) using indicators (14A)
b. Identify acids and bases using pH (14A)
c. Recognize the effects of acids and bases on ecological systems. (14D)
 Complete the chart below
Property
Acids
Taste
sour
Neutral
none
Bases
bitter
Feel
watery
??
slippery
pH range
1-6
7
8-14
H30 + Concentration
over 50%
> 1 x 10-7 M
less than 50%
< 1 x 10-7 M
red
50%
less than 50 %
< 1 x 10-7 M
over 50%
> 1 x 10-7 M
blue
OH- Concentration
Reaction with litmus paper
Reaction with
Phenolphthalein
Examples of household
products
clear
pickles, batteries,
citrus fruits, soda
50%
n/c
WET!
clear
water
Pink (at pH of 8)
soap, cleaners,
bleach, ammonia
 What does the word amphoteric (amphiprotic) mean? substances that can act as both
an acid and a base (water)
 How does pH relate to [H3O+]? pH = -log [H3O+]
 List examples of common household acids & household bases.
 Using Figure 19-18 on page 619, predict the pH of solution A when ….
Soln A + bromothymol blue = green color pH = 7.7
Soln A + phenolphthalein = clear pH = 8.4
Soln A + methyl orange
= yellow pH = 4.3
 Discuss some impacts of acid rain.
acidic air pollution -nitrogen oxide  nitric acid  acid rain; increases acidity of some
soils and remove essential nutrients -adverse affects on vegetation; flow into water
sources -harm aquatic life.
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13. Using the properties and behavior of acids and bases, determine strength of acids &
bases using conductivity (14B) and identify the characteristics of neutralization
reactions. (14C) Higher conductivity, stronger acid/base; neutralization reaction -produce
salt and water (salt- cation from base, anion from acid)





What property categorizes acids and bases as strong or weak? complete ionizationstrong; partial ionization- weak
Using figure 19-7 on page 602, explain why the light bulb is brighter in one solution but
not the other? HC1, a strong acid, is completely ionized, therefore resulting in higher
conductivity; the other is a weak acid, partially ionized, resulting in lower conductivity
What is the purpose of a buffer?
Buffers -solutions that resist changes in pH when limited amounts of acid or base are
added; used to maintain pH values within narrow limit
Write a neutralization reaction for Mg(OH)2 and HCl ?
Mg(OH)2 + 2HCl  MgCl2 + 2H20
Chapter 20 and 21– Redox Reactions and Electrochemistry – TEKS Obj. 10A & 10B
14. Identify changes in energy in electrochemical reactions and electrical circuits.
Electrochemical reactions -voltaic cell- chemical energy  electrical energy -electrolytic
cell- performs electrolysis (converts electrical energy to chemical reaction)
a. Identify oxidation-reduction processes. oxidation- loses elections, reduction- gains
electrons
b. Differentiate between the following terms as they relate to corrosion and
electroplating: corrosion- iron oxidized (reducing agent) and water reduced (oxidizing
agent); electroplating- silver is oxidized at anode (reducing agent) and cathode to be
plated is reduced (oxidizing agent)
reduction, oxidation, reducing agent, and oxidizing agent.
 P640: 3; P658: 44, 49, 51
a. oxidizing agent -I; reducing agent -Mg b. oxidizing agent- H; reducing agent -Na
 c. oxidizing agent -Cl; reducing agent -hydrogen sulfide
44.)a. oxidized- Ga; reduced -Br
b. oxidized- Zn; reduced- HCl c. oxidized -Mg; reduced -N2
49.)a, because neither compound was oxidized or reduced
51) a. –3
b –3
c. +2
d. +5
e. +1
f. +3
.
 What causes corrosion? (pg 679)
 What can be done to prevent corrosion from happening?
 In the galvanizing process, why is zinc used to coat iron instead of another metal?

corrosion is caused by redox reaction of metal with substances in the environment
(usually iron and oxygen)

qalvanizing (coat with layer of zinc) or using a sacrificial anode can prevent corrosion
zinc is a self-protecting metal which oxidizes at the surface, but clings tightly to the metal
and seals from further oxidization; also, when cracked, the zinc sets up a voltaic cell so that
further corrosion doesn't occur
Plano ISD
Revised 2/23/05
PreAP Chemistry Spring Review
Page 15 of 15
15. Demonstrate an understanding of lab safety, techniques, and scientific tools. (1A, 2B)
and apply the principles of experimental design in laboratory and field investigations.
 P23: 5-7
o 5. c; 6. a; 7. d (but I don’t like this answer!)
 In a given reaction, how do you decide what the control will be?
 the control is what would occur under normal circumstances, without the independent
variable being altered in any way
 In making a graph from the results of a reaction, how should you decide what
information to put on which axis?
o the variable that is changed is on the x-axis, the variable that changes in result of
the other variable's change is on the y-axis
 The slope needs to give you what you are looking for… For ex: in a graph reflecting
density, mass has to be on the y axis & volume on the x axis or the slope will NOT
give you density!
Plano ISD
Revised 2/23/05