1999 AP Chemistry MC Exam - (Discussion of answers)
1. C
This is the definition of ionization energy. Notice that I.E. is not the energy to make any type of ion. Its
only the energy required to lose electrons to become a cation. Your CP teacher probably didn’t tell you that the
measured number for I.E. is for atoms in the gas phase, but now you know! So what kind of energy is required
to create anions? (To gain an electron). ELECTRON AFFINITY. (This is another definition you should know.)
2. E
First you need to know that table salt (and all ionic compounds) form regular repeating rows of + and –
ions when they are solids (not dissolve in water). This is called a “lattice”. Undoubtably the word is familiar
from math. Here is what one looks like:
3.
4.
5.
6.
7.
8.
9.
10.
The lattice energy refers to how strongly the solid salt is held together. To sublime into the gas state, you would
have to provide enough energy to overcome the lattice energy. Even if you are just trying to dissolve the salt in
water, you still have to overcome the lattice energy.
B
This is a question, we will specifically discuss in thermodynamics next semester. Skip it for now.
A
Recall that there is a high energy “bump” on an energy diagram between the reactants and products in a
reaction. Reactions with high activation energy are slower than those with lower activation energy. (Review
Kinetics) The “transition state” is at the top of the energy bump and is the moment when the two chemicals are
reacting. Remember that we used the analogy of a relay race. The “transition state” is a picture of the moment
when both runners are holding the baton.
D
Noble gases are unreactive, so look for the electron configuration of a noble gas. Answer D is the
electron configuration of neon. Notice that it has a full valence shell. (The 2s and 2p orbitals are full and
combine to have 8 electrons.)
A
The Aufbau principle says that we have to fill orbitals from the lowest energy to the highest. Answer “A”
does not have a full 1s orbital and so we can’t put electrons in the 2s orbital. This must be an excited state then.
The electron in the 1s orbital was “excited” by absorbing energy and jumping to a higher energy state. Excited
state electron configurations don’t exist for very long.
C
Only those electrons in the highest principle energy level (n) are valence electrons. The energy level is
the number in front of the orbital. The 1s orbital is in energy level 1, so this is an inner energy level. The
outermost energy level in this example is 2. So answer “C” has 2 electrons in the “2s” and 2 electrons in the
“2p” for a total of 4 valence electrons. To see if you understand the concept, notice that answer “E” has only 2
valence electrons.
E
Transition metals are in the “d-block” are of the chart. (The middle of the periodic table.) E is the only
one that has d orbitals so it must be the answer. You have to be cautious though. This electron configuration
[Ar]4s23d104p2 has d orbitals but is not a transition metal element. (It is Germanium) If you don’t
remember electron configurations, look in 5 steps to a 5.
C
NaCl is a salt that has no affect on pH. HCl however is a strong acid and so the pH will be well below 7.
A and B can’t be the answer because those are buffers and involve weak acids and bases. D can’t be the answer
because they are both bases and will have a pH well above 7.
E
NH3 (ammonia) is a weak base and HC2H3O2 (acetic acid) is a weak acid so you would expect them to
mix to make a neutral solution. You might wonder if they aren’t exactly equally weak, wouldn’t that make the
11.
12.
13.
14.
solution not exactly neutral. Yes. (It turns out they have the exact same Ka and Kb values). But when compared
to the other choices, this would be closest to 7.
A
The only two buffer solutions (where you have a weak acid or base and its conjugate) are A and B. (C is
an acid and its conjugate, but HCl is a strong acid and so can’t be a buffer.) A involves NH3, which is a weak
base, so this will be the buffer greater than 8.
B
Same logic as in part 11. When choosing between A and B, B contains H2PO4 which is a weak acid.
Therefore this buffer will be acidic and have a pH less than 7.
A
Read the explanation to question 2 first. Now that we know that a lattice is the form that salts in their
solid form take, all we need to know is that CsCl is the only salt (ionic compound) in questions 13-16. Since it has
ions and is ionic, it will form a lattice. “Electrostatic forces” is just the attraction between + and – ions.
B
Your CP teacher may have spent very little time on metallic bonding so lets review it here. This question
refers to Au(s), which is solid metallic gold. In metallic bonding, the valence electrons can travel throughout the
metal and not just stay attached to the metal atom they came from. Therefore, since electrons don’t stay with
their “local” atom, they are called “delocalized”. You might also refer to a travelling salesman as delocalized!
15. D
CO2 is a small covalent molecule so you can rule out all answer choices other than C and D. To find out if
you have single or multiple bonds, you need to draw the Lewis structure for CO2. When you do that, you will
find that CO2 has double bonds. If you don’t remember how to draw Lewis structures, review 5 steps to a 5.
16. C
CH4 is a small covalent molecule. You need to draw the Lewis structure and you will find that it has only
single bonds.
17. E
This question probes your knowledge of the elements. At first it may appear to be a tough question, but
let’s eliminate some elements that are obviously not gases. A, B, and D are all metals and metals are never
gases at room temperature (298K). Some of you may be surprised that uranium is a metal, but remember, the
only non-metals are those on the upper right of the table. So that leaves us with (C) bromine and (E) fluorine.
At this point you will either remember that bromine and mercury are the only two liquid elements, or you will
notice the periodic trend that gases tend to be on top of the column and become solids near the bottom. (F and
Cl are gases, Br is a liquid, I is a solid.)
18. A
The question checks to see if you know that the only strong bases are hydroxide bases. Li reacts as
follows: Li + H2O  LiOH + H2 LiOH is a strong base. If you do the same thing with Ni, you get Ni(OH)2 and that
is not a strong base. Remember that only the group one metals and the bottom of group two form strong
bases. This might be a good time to remember the reactivity trend for Alkali metals. Remember that as you go
down the group, the metals become more reactive. Cesium would be much more reactive than Li and would
also produce a strong base.
19. C
This question is discussing kinetics. The spark makes the reaction go faster. The spark is only giving the
reaction enough energy to overcome the energy of activation, as described in the correct answer “C”. It is not
changing the energy of the reactants or products. Answers “A” and “D” are discussing kinetics, but they are
describing what catalysts do to speed up a reaction. The spark is not a catalyst. Remember that catalyst actually
bind the chemicals in a way that allows them to react fast. “B” may very well speed up the reaction, but the
spark cannot increase the concentration of anything.
20. A
When you see a stoichiometry question without a reaction, it is best to draw the reaction. The wording
suggests that Au2S3 and H2 are the reactants. This suggests single displacement (compound + element):
Au2S3 + 3H2  2Au + 3H2S
(Don’t forget to check charges on the HS product, then balance)
Now we can solve the stoichiometry:
0.0500 mol Au2S3
2 mol Au
196.97 g Au
1 mol Au2S3
1 mol Au
=
Since we don’t have a calculator: 0.0500 x 2 = 0.100 , then we get to practice our decimal pushing: 0.100 x
196.97 = 19.697. So the answer is “A”.
21. B
The AP test believes that knowing the color of elements when burned in flame is important. This seems
like an impossible task, but there are only really 4 that show up on the test with any regularity. So let’s
memorize them now: Strontium (Sr) = red Copper (Cu) = green potassium (K) = purple sodium (Na) =
yellow.
22. We learn thermodynamics next semester. Leave it for now.
23. E
This is a gas law question. Start thinking about what we learned in gas laws. Charles liked to balloon
and he related volume and temperature. Expect the answer to have something to do with this idea. “A” is
unreasonable because since the balloon is open to the air, the pressure inside and outside must be the same.
Plus Charles law assumes constant pressure. “B”- Who knows if convection currents are reasonable but it
doesn’t sound like a chemistry concept. “C” - If anything the hotter air inside the balloon is pushing out more,
but as we said earlier, pressure is the same inside and outside the balloon because of the hole in the bottom.
“D”- Diffusion is not important because diffusion is how fast one gas spreads throughout another one. The gas
inside and outside the balloon is the same. “E”- is correct because the expansion of the volume of gas inside the
balloon has cause some to escape out the bottom. This makes the balloon less concentrated on the inside. The
lower density causes it to rise.
24. D
Lets look at what we definitely don’t want to do: “A” – Drying will paper towels would not remove the
acid. “B”- Na2SO4 is not an acid or a base. It is a neutral salt and would have no affect. “C”- NaOH is a strong
base. While it may neutralize the acid, having a strong base is potentially much worse. “E” – Adding an acid to a
hand that already has acid on it, will do nothing to improve the situation. “D” – is the correct answer. This is
baking soda, a weak base.
25. C
This is a cooling curve. We looked only at heating curves, but the principle is the same. The flat part is
the phase change, where you have dynamic equilibrium. Answer “C” is the only answer that discusses the
entire flat part of the graph, and the phase change occurs over the whole flat section. This is why “A”, “B”, “E”
won’t work. Answer “D” includes the flat part but also includes the section “T” where you have only liquid. You
may be wondering about the little dip that occurs at point Q. This is called supercooling. Essentially, the liquid is
cold enough to crystallize into a solid but doesn’t realize it yet. We will discuss supercooling more next semester
in colligative properties.
26. C
This is a straight forward balancing question. Not simple, but straight forward. I started by balancing C,
then H, then S, then O.
27. A
A spectrophotometer is a fancy name for a colorimeter, like we used in our kinetics and equilibrium
units. So we are looking for applications where color can be measured. Now is a good time to learn a
fundamental observation in chemistry: MOST COLORED SOLUTIONS AND SALTS ARE CONTAIN TRANSITION
METALS. Notice, we are not talking flame tests where you are burning them, we are just talking about taking a
salt and dissolving it in water. So in option “I” of this question, it seems likely that we will be able to measure
the concentration of Cu(NO3)2 because it has the ion of a transition metal (Cu+2). In part “II” – KMnO4 does not
contain a transition metal (K is an alkali metal) so this will not be a colorful solution. Plus they are talking about
conductivity (electricity) instead of colorimetry. Part “III” none of the three ions are transition metals, so they
are not likely to have color, so we won’t be able to use a spectrophotometer (colorimeter). So the answer is A.
28. B
Now that we know what lattice energy is, (see question 2), this question asks what factors cause salts to
have high lattice energy. Salts with high lattice energy, like MgO in this example, have high melting points. This
is kind of like intermolecular forces in covalent compounds, high IMF’s have high boiling points. So what causes
salts to want to stick together in a lattice? Ions that have higher charge will have stronger lattice energy. Mg+2
is more attractive to negative ions than Na+, so part “I” is true. Obviously, more negatively charged ions are
more attractive anions, so O-2 is more attractive than F-, so part “II” is true. Part “III” is not true because of the
periodic table size trend. As you go to the right and up in the P.T., elements get smaller. Therefore O cannot be
smaller than F. “B” is the correct answer.
29. E
This is a ketone. It has a carbonyl with carbons on both sides. It is not an aldehyde. Aldehydes have
one side that has only a hydrogen. As for the other answers: “A” -An organic acid would have the COOH
functional group in it. “B” – an alcohol would have an –OH functional group. “C” – An ether has no carbonyl.
Just O in the middle of a carbon chain.
30. D
This is a redox question. While we will do redox with electrochemistry next semester, you learned to
assign oxidation numbers last year in CP. Remember that oxidation numbers are like charges, and in redox
reactions, they change when electrons are exchanged between compounds. There are rules to assigning
oxidation numbers that are too much to put here, so look them up in 5 steps to a 5. Here are the oxidation
numbers for each atom:
H2Se: H is almost always +1, so Se must be -2.
O2F2: F is -1, so O is +1 in this example:
SeF6: F is -1, so Se is +6
HF:
H is +1, so F is -1
O2: You might think that O is -2, but since this is oxygen in its element form, the oxidation number is zero.
Looking at how the oxidation numbers change, “D” must be the correct answer because all of the others are
untrue. One last note: A “disproportionation” is when one compound or element is being both oxidized and
reduced in one single reaction.
31. This is a colligative properties question and we will do this next semester.
32. Also next semester.
33. C
This is a stoichiometry question, and so we need a balanced reaction: Cl- + AgNO3  AgCl + NO3Total moles of Cl-: 0.1 mole from NaCl and 0.20 mol from CaCl2 = 0.30 moles total
0.30 mol Cl-
1 mol AgNO3 =
0.30 mol AgNO3
1mol Cl34. We will study electrochem next semester.
35. We will study electrochem next semester.
36. B
A classic kinetics question because it has a rate table. We need to find the rate in terms of NO and O2.
Let’s start with the most ridiculous answer. “E” is incorrect because we need to remember that products over
reactants is an equilibrium idea, not kinetics. Now to figure out the rate in terms of NO: Compare experiments
#1 and #2. The concentration of NO doubles, and the rate doubles. The reaction is first order in NO. For O2,
compare experiments #2 and #3. As the concentration of O2 increases 4x, the rate increases 16x. This is
second order in O2. The answer is “B”.
37. We will study first versus second ionization energy next semester. Skip this one.
38. B
This is the definition of a Lewis acid. “C” is the definition of a Lewis base. “A” is the definition of a
Bronsted-Lowry base and “D” is the Bronsted-Lowry definition of an acid.
39. C
“C” is on the line between the solid and liquid phase. “Normal” melting point is the melting point at 1
atm or 760 mmHg.
40. D
Molecules that have “large dipole moments” are just polar molecules. Your CP teacher may or may not
have used this phrase specifically. But now you know! So to find which is the most polar, draw the Lewis
structure of each one and learn its shape. You will find that CO and HF are the only polar molecules. HF is the
most polar because it has a bigger difference in electronegativity between its two elements. C and O are right
next to each other and have more similar electronegativities.
41. E
“A” can’t be the answer because the only thing that will change the Keq is to change the temperature.
“B” is difficult to say. There are not equal moles of gas on both sides but you are increasing the moles of O2
when shifting the equilibrium, so it is a tough call. Lets see if there is a better answer. “C” is not correct since
the equilibrium is shifting left, the SO3 will increase. “D’ is not correct because you are injecting O2. While
some of it will shift left, it will not get back to its original amount. “E” is the answer. The shift to the left will
lower the amount of SO2.
42. E
Just a balancing equation problem. A tough one though.
43. This involves colligative properties. Skip it until next semester.
44. C
This is a gas law question. “A” –can’t be true because they specifically said in the problem that it is a
rigid tank. “B” is not true because the pressure will be going up from the addition of gas to a rigid container.
“C” –This is the correct answer. Since the temperature is constant, the speed of the molecules must remain
constant. “D”-If you are adding gas, then the moles of gas can’t remain constant. Silly! “E”-Constant volume
and more gas molecules will actually lead to less space between the gas molecules.
45. D
This is a classic weak acid equilibrium question. It will require our good math with no calculator skills.
Draw an ICE table and an equilibrium expression. When you plug in the equilibrium concentration you get:
5.0 x 10-10 = x2/0.05M
rearrange to
(0.05)( 5.0 x 10-10) = x2
Now we have to multiply 0.05 and 5.0 x 10-10
Well, 5 x 5 = 25, so following that logic we get:
-12
2
25 x 10 = x
5 x 10-6 = x = [H+]
46. C
This is a complex ion question. They are adding “excess” NH3 which is a good ligand to a metal ion
+2
(Cu ). Unfortunately, the question expects you to know the color of the complex ion solution. You will recall
that we learned earlier that transition metals make brightly colored solutions. The other useful information to
know is that complex ions are usually soluble. I know that this is not a satisfying answer. Some AP questions
are very tough. Knowing that Cu and NH3 make a bright blue solution is a piece of info that is good to file in the
back of your mind. Also, remember that we did Fe and SCN in our lab, and that was red.
47. C
This is an empirical formula question. We need to find the moles of Hf and Cl in the product. Assume a
100 sample. This means you have 62.2 g of Hf and 37.4 g of Cl. Convert these to moles using approximations:
62.2 g Hf 1 mole Hf = about 0.30 mole
37.4 g Cl
1mole Cl
= about 1.0 mole
178.49 g
35.35 g
So our formula should be Hf0.30Cl1.0 but we need whole numbers so divide both by 0.3 and we get Hf1Cl3
48. B
This is a kinetics question. Half life is the time it takes for half of the material to react. It can be helpful
to write out a half life table:
Time
0
? days (1st half life)
? days (2nd half life)
24 days (3rd half life)
Iodine remaining
100 %
50 %
25 %
12.5 %
Iodine reacted
0%
50 %
75 %
87.5 %
Since the half life is always the same amount of time, The first half life must be after 8 days, the second after 16
days. So the half life is 8 each time.
49. E
This question is about solubility rules, acid base chemistry and lab experience. “B”- can’t be the answer
because KNO3 is soluble and there will be nothing to filter. “C” isn’t correct because KNO3 is a neutral salt and
so can’t be reacted with acids or bases. “D”- Electrolysis would have no effect, but if it did, it would change the
KNO3 into K metal which would not be recovering KNO3. “E” would work great since KNO3 does not evaporate.
After all, it is a salt.
50. D
This is a periodic trend question. Atomic radius (size) decreases from left to right, and from bottom to
top.
51. E
This question asks you to remember that Rutherford discovered the atomic nucleus. Any answer that
discusses electrons is not referring to Rutherfords work. This rules out “A” and “B”. Neutrons were discovered
last and Rutherford didn’t even know they existed. This rules out the rest of the answers. Be sure to review
Millikan and Thomson in your 5 steps to a 5.
52. This is a colligative properties question and we will learn about his next semester.
53. A
This appears to be a gas law question, but is really just a stoichiometry question. Since the conditions
are constant, we can assume that atm is directly related to moles. Notice that not all of the “W” has reacted.
They start with 1.2 atm and at the end they still have 1.0 atm. This means that only 0.2 atm of “W” react.
Since the balanced equation is 1:1 for W and Z, then this means 0.20 atm of Z forms.
54. B
This is an equilibrium question. Remember that the only thing that can affect the equilibrium constant
is temperature. This rules out all answers except “A” and “B”. We know from the delta H that this is an
exothermic reaction. (delta H is negative so the energy of the products is lower than the reactants. Draw
yourself an energy diagram to convince yourself of this.) Since the reaction is exothermic, we put the heat term
on the products side. Now if we decrease the temperature, it is like removing heat (which is a product) and the
reaction will shift to the right. This will cause the equilibrium constant (Keq) to increase because it is shifting
towards products. The question wants Keq to decrease so “B” must be the correct answer.
55. D
This is a stoichiometry question. The conversion is as follows:
2.5 mol I2
10 mol HI
=
5.0 moles HI
5 mol I2
56. A
This is a solubility rules question. “A” – PbI2 will definitely precipitate. “B”- ZnI2 will not. “C”- chromate
can only react with Na+ and 1A cations don’t precipitate. “D” – Same thing, sodium sulfate will not precipitate.
“E” – NaOH will not precipitate. So “A” is the answer. And we didn’t even need to know the color of the
precipitate.
57. This is electrochem. We will wait for next semester.
58. C
This question is a gift to those of us that live in Colorado. Clearly the question is going to relate to the
lower pressure. Since high altitude pressures are lower, this rules out “A” and “B”. “C” is correct. The wording
may sound strange, but they are trying to remind you that when a liquid boils, it is because the vapor pressure
of the liquid is the same as the atmospheric pressure. Recall that the vapor pressure goes up when you start to
heat the liquid until it matches the atmospheric pressure, then voila, it boils.
59. C
This is not a stoichiometry question because there is no reaction. Just find the total moles of OH in the
solution, then divide by the total volume:
Moles of OH from KOH: 0.040L x 0.25M = 0.01moles KOH = 0.01 moles of OHMoles of OH from Ba(OH)2: 0.060 L x 0.15M = 0.0090 moles of Ba(OH)2 = 0.018 moles OHTotal moles OH = 0.01 + 0.018 = 0.028 moles
Molarity = 0.028 moles/0.100L = 0.28 M
(Notice we divide by the total volume.)
60. A
0.03 mol of NH4NO3 will produce 0.03 moles N2O and 0.06 mole H2O. Since both products are gases,
this will mean 0.09 moles of gases. Now use PV=nRT to convert moles to pressure:
61.
62.
63.
64.
65.
66.
67.
68.
69.
P = nRT/V
P = (0.09)(0.082)(400)/1
Now we have to estimate the answer. Lets adjust the
numbers to something like: P = (0.1)(0.1)(400)/1 = This is about 4 atm so I will guess “A”
This is thermodynamics. Next semester.
A
This is an equilibrium question. The important thing to note here is that if the equilibrium constant is
large, the equilibrium lies far to the products side. “A” is the correct answer because if CN- is a stronger base, it
will be very affective at taking the proton from acetic acid. C2H3O2- is a weaker base and so it can’t take the
proton from HCN as well. “B” is incorrect because if HCN was a stronger acid, it would be successful in pushing
the reaction to the reactants side. “C” – both are bases so they cannot be conjugates of each other. “D” While Keq can change with changing temperature, we cannot know the change without knowing if the reaction
is endothermic or exothermic.
B
This is a kinetics question. “A”- We know that this can’t be correct. A plot of [x] versus time would be a
straight line if it was zero order. The others are hard to find without graphing the 1/[x] and ln[x] plots. The path
to the answer is to notice the little “dots” on the graph for the data points. The half life of “x” is 1 hour. Notice
that concentration of x consistently gets cut in half each 1 day. This is indicative of 1st order kinetics. 2nd order
reactions do not decay with this consistency.
A
Pinhole leaks are examples of “effusion” Remember that smaller gases will effuse faster. This means
the pressures of the small gases will be less after time than the larger gases.
A
This is a solubility rules/Ksp type question. “B””D”and “E” are all soluble compounds without any help
according to solubility rules and so they can’t be the answer. “A” is the answer because adding acid will lower
the lower the concentration of OH in the Ksp reaction causing a shift to the right: Mg(OH)2  Mg+2 + OHThis is thermodynamics. Next semester.
E
This is a Ksp question. Remember that molar solubility refers to how much dissolves. Draw an ICE table
and an equilibrium expression for this reaction: Ag2CrO4  2Ag+ + CrO4-2
You will get this expression when you plug in the equilibrium values:
8 x 10-12 = 4x3 so then 2 x 10-12 = x3
The question does not require you to take the cubed root.
D
Remember that in phase changes, covalent bonds are not broken, only intermolecular forces.
“A””B””C”and “D” are all phase changes so you can rules these out. “D” – diamond forms covalent networks so
covalent bonds must be broken to allow this to happen.
C
This is a precipitation reaction. Draw out the balanced equation. Find the moles of each reactant, then
do the stoichiometry to find how much of the BaCl2 reacts. The amount that doesn’t react will tell you how
70.
71.
72.
73.
74.
75.
many moles of Ba+2 ions are in solution. Those that do react are not in solution because the BaSO4 precipitates
(so you can ignore them.)
A
It is a good idea to write out the reaction so that you know that Ag3PO4 is precipitating and you know
the balanced equation.
B
This is a solubility rules question. “A” sulfides tend to be insoluble in general, so FeS, CuS, and PbS will
all precipitate. This will not allow you to separate Pb ions. “B’ is the answer. FeCl2 is soluble, CuCl2 is soluble,
PbCl2 is not.
B
This is an error analysis question.
D
This is a dilution question. You can use MV=MV to solve the problem, but notice that they don’t want to
know the final volume, they want to know how much water to add.
A
This is a gas laws question. Recall that kinetic molecular theory says that ideal gases have no volume
and no intermolecular forces (attractive forces). The gas that deviates most will be polar (strong IMF) and will
have high volume (or molecular weight). All of the choices are non-polar except SO2.
This is colligative properties and so we will study this next semester.