4 Curent Format AP Practice Tests
with
Answers
Explanations
Scoring Guidelines
AP Scoring Worksheet
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Question Sets
(4 Total Practice Tests)
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Brief
Simple
Easy-to-understand
Answer Explanations
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Days 1, 3 and 5
60 Points
Practice Exam 1
Multiple Choice Answer Sheet
Day 1
C O
D O
E
OA OB O
A O
B O
C O
D O
E
2. O
A O
B O
C O
D O
E
3. O
A O
B O
C O
D O
E
4. O
A O
B O
C O
D O
E
5. O
A O
B O
C O
D O
E
6. O
A O
B O
C O
D O
E
7. O
A O
B O
C O
D O
E
8. O
A O
B O
C O
D O
E
9. O
A O
B O
C O
D O
E
10. O
A O
B O
C O
D O
E
11. O
A O
B O
C O
D O
E
12. O
A O
B O
C O
D O
E
13. O
A O
B O
C O
D O
E
14. O
A O
B O
C O
D O
E
15. O
A O
B O
C O
D O
E
16. O
A O
B O
C O
D O
E
17. O
A O
B O
C O
D O
E
18. O
A O
B O
C O
D O
E
19. O
A O
B O
C O
D O
E
20. O
1.
Day 3
C O
D O
OA OB O
A O
B O
C O
D O
E
2.
22. O
A O
B O
C O
D O
E
3.
23. O
A O
B O
C O
D O
E
24. O
4.
A O
B O
C O
D O
E
5.
25. O
A O
B O
C O
D O
E
6.
26. O
A O
B O
C O
D O
E
7.
27. O
8.
A O
B O
C O
D O
E
28. O
9.
A O
B O
C O
D O
E
29. O
A O
B O
C O
D O
E
30. O
10.
A O
B O
C O
D O
E
11.
31. O
A O
B O
C O
D O
E
12.
32. O
A O
B O
C O
D O
E
13.
33. O
A O
B O
C O
D O
E
14.
34. O
A O
B O
C O
D O
E
15.
35. O
A O
B O
C O
D O
E
16.
36. O
A O
B O
C O
D O
E
17.
37. O
A O
B O
C O
D O
E
18.
38. O
A O
B O
C O
D O
E
19.
39. O
A O
B O
C O
D O
E
20. O
21.
1.
Copyright © 2018 E3 Scholastic Publishing. All Rights Reserved
Day 5
A O
B O
C O
D O
E
O
A O
B O
C O
D O
E
2. O
41.
A O
B O
C O
D O
E
3. O
42.
A O
B O
C O
D O
E
43.
4. O
A O
B O
C O
D O
E
5. O
44.
A O
B O
C O
D O
E
6. O
45.
A O
B O
C O
D O
E
7. O
46.
8. O
A O
B O
C O
D O
E
47.
9. O
A O
B O
C O
D O
E
48.
A O
B O
C O
D O
E
10. O
49.
A O
B O
C O
D O
E
11. O
50.
A O
B O
C O
D O
E
12. O
51.
A O
B O
C O
D O
E
13. O
52.
A O
B O
C O
D O
E
14. O
53.
15. O
A O
B O
C O
D O
E
54.
A O
B O
C O
D O
E
16. O
55.
17. O
A O
B O
C O
D O
E
56.
A O
B O
C O
D O
E
18. O
57.
A O
B O
C O
D O
E
19. O
58.
A O
B O
C O
D O
E
20. O
59.
40.
1.
1
Scoring Practice Exam 1:
Days 1 to 6 Questions
Section I - Multiple Choices: 75 Points (50% of Score)
Section II - Free Responses: 75 Points (50 % of Score)
Total Composite Points:
150 Points
Practice Exam Scoring Worksheet:
Section I: Multiple-Choice Score
Add up Days 1, 3 and 5 points = __________ x 1.25 = _______
Section II: Long Response Score
Add up Days 2, 4 and 6 Points
= _________ x 1.75 = _______
Section II: Short Response Score
Add up Days 2, 4 and 6 Points
= _________ x 1.375 = _______
Your Total Composite Points
= _______
(out of 150)
Your AP Practice Exam Score
=
_______
(out of 5)
AP score conversion chart:
Composite
score range
100
2
150
Your Practice Exam
AP Score
5
81 99
4
62 80
3
49 61
2
0 48
1
E3Chemistry.com
Day 1
20 Points
Multiple Choice Questions
Section 1 Practice
Start: Answer all questions on this day before stopping.
Directions: YOU MAY NOT USE YOUR CALCULATOR FOR THIS DAY.
YOU MAY USE THE PERIODIC CHART AND EQUATIONS TABLE.
Each of the questions or incomplete statements below is followed by four
suggested answers or completions. Select the answer that is best in each
case and then fill in the corresponding circle on the answer sheet.
Note: For all questions, assume that the temperature is 298 K, the
pressure is 1.00 atmosphere, and solutions are aqueous unless otherwise
specified.
1. A student carries out an experiment to collect carbon dioxide by reacting
20 g of calcium carbonate with 1M solution of hydrochloric acid. In the first
trial the student collected approximately 5 L of carbon dioxide. In the
second trial the student needed to collect more carbon dioxide from the
same amount of calcium carbonate. Which of the following actions by the
student is least likely to lead to an increase in the volume of carbon dioxide
collected?
(A) grinding the calcium carbonate to a fine powder
(B) raising the temperature
(C) raising the atmospheric pressure
(D) raising the concentration of hydrochloric acid
2. A 55.0 gram sample of metal is heated to 100.0 C and then placed into a
calorimeter containing 125.0 grams of water at 24.0 C. If 3.8 kJ of heat is
transferred from the metal to the water, what is the final temperature of
the water? (The specific heat capacity for water is 4.2 J/(g oC).)
(A) 31.2oC
(B) 76.0oC
(C) 40.5oC
(D) 52.8oC
3. A compound contains only the elements carbon, hydrogen, and oxygen.
Upon combustion of 16.5 g sample of the compound, 33.0 g of CO 2 and
13.5 g H2O are produced. A possible empirical formula of the compound is
(A) CH2O
(B) C2H2O
(C) C3H4O
(D) C2H4O
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3
Day 1
Continue
Questions 4 and 5 refer to the information and diagram below.
Br2(g)
+
I2(g)
2IBr(g)
The reversible reaction between bromine gas and iodine gas is
represented by the the equation above.
10 mol of IBr is placed in a rigid 2.0-liter flask and the reaction above
is allowed
to proceed
Substance
S contains
oneat a constant temperature until equilibrium is
reached.
[I2] = 2 M
10 mol
IBr
Initial Mixture
2.0 L
rigid flask
[Br2] = ?
[IBr] = ?
Equilibrium Mixture
4. What is the equilibrium constant for the reaction at the temperature in
which the reaction took place?
(A) 0.25
(B) 0.5
(C) 5.0
(D) 8.0
5. If the initial mixture was placed in a rigid 1.0-liter flask instead of the
2.0-liter flask, and the reaction is allowed to proceed at the same
temperature, the reaction will
(A) reach equilibrium quicker, and the concentration of I 2 will still be
2 M at the equilibrium point.
(B) reach equilibrium quicker, and the concentration of I 2 will be less than
2 M at the equilibrium point.
(C) reach equilibrium quicker, and the concentration of I2 will be greater than
2 M at the equilibrium point.
(D) reach equilibrium slower, and the concentration of I 2 will be less than
2 M at the equilibrium point.
4
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Day 1
Continue
6. 100 grams of O2(g) and 100 grams of He(g) are in separate containers
of equal volume. If both gases are at 50°C, which one of the following
statements is true?
8. Which particle diagram represents a mixture of elements and a
(A) Both gases would have the same pressure.
compound?
(B) The average kinetic energy of the O2 molecules is greater than that of
the He molecules.
(C) There are equal numbers of He molecules and O2 molecules.
(D) The pressure of the He(g) would be greater than that of the O2(g).
Questions 7 and 8 refer to the titration curve below.
9. What is the total amount of heat absorbed by 100.0 grams of water
when the temperature of the water is increased from 30.0oC to 45oC
(1) 418 J
(3) 12 500 J
7.(2)Which
does
6270 Jone of the following combinations
(4) 18
000the
J titration curve
represent?
(A) Addition of a strong base to a weak acid
(B) Addition of a strong acid to a strong base
(C) Addition of a strong acid to a weak base
(D) Addition of a weak acid to a strong base
8. What is the pH of the solution at the point of maximum buffering?
(A) 10.0
(B) 9.3
(C) 5.3
(D) 1.8
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Day 1
Continue
Questions 9 - 12 refer to the information below.
NaOCl is completely dissociated in water to form Na+(aq) and OCl (aq).
In solution, OCl hydrolyzes according to the equation
OCl (aq) + H2O(l)
HOCl(aq) + OH (aq)
Equation 1
9. 100 mL of pure water at constant temperature is added to a 100 mL
solution of 0.10 M NaOCl. When the solution reaches equilibrium again, the
(A) [H+] has decreased.
(B) pH of the solution has decreased.
(C) concentration of HOCl has increased.
(D) value of the equilibrium constant has halved
10. If K1 is the equilibrium constant for the reaction above, then the value of
K2, the equilibrium constant for the reaction
2OCl (aq) + 2H2O(l)
2HOCl(aq) + 2OH (aq)
at the same temperature, is equal to
(A) K1
(B) 2 × K1
(C) 4 × K1
(D) K12
Equation 2
11. The HOCl produced in a solution of NaOCl can react further to produce
small amounts of chlorine, Cl 2(aq), in water according to the equation
HOCl(aq) + H+(aq) + Cl (aq)
Cl2(aq) + H2O(l)
Equation 3
Which of the following, when added to a solution of NaOCl, would not raise
the concentration of Cl2 in the solution?
(A) NaCl
(B) NaOH
(C) H2SO4
(D) HOCl
12. Which correctly identifies the oxidation number of chlorine in HOCl and
OCl- ?
(A) -1 in HOCl and +1 in OCl(B) -1 in HOCl and -1 in OCl(C) +1 in HOCl and +1 in OCl(D) +1 in HOCl and -1 in OCl-
6
E3Chemistry.com
Day 1
Continue
Note: Electrons are negatively charged particles
NO + Cl2
NOCl2
2)13.3 AnNote:
Malleability
and
good
conductor
are physical
propertiesabove.
of metals.
equation for an elementary reaction
is represented
Relate:
Element
29,
Copper,
is
a
metal
What is the rate law and molecularity for this reaction?
(A) rate = k[NO][Cl2]; unimolecular
3) 3 Note: Sodium phosphate
(Na3PO4) is an ionic compound with
2
(B) rate =three
k[NO][Cl
2] ; unimolecular
different
elements.
2
(C)Note:
rateIonic
= k[NO][Cl
2] ; dimolecular
compounds
with 3 or more elements always contain
ionic2and
covalent bonds
(D) rate =both
k[NO][Cl
]; bimolecular
Questions 14 and 15 are based on the information on the table below.
4) 2 Note: Vapor pressure-temperature relationship is on Table H
Use Reference Table H to determine answer.
Reaction
Chemical Process
5) 2 Note: Saturated hydrocarbons are the alkanes.
1 Propane2S(s)
3O2name
(g) ending)
2SO3(g)
Relate:
(with+ ane
is an alkane. +800 kJ/mol
6) 4
2 In nuclear
2SO
2SOONLY
O2(g) nuclei spontaneously
-200 kJ/mol
3(g)
2(g) + unstable
Note:
radioactivity,
decay
by
emitting
a
beta,
an
alpha
or
a
positron
particle.
3
S(s) + O2(g)
SO2(g)
?
14. The chemical process in Reaction 2 is
(A) not thermodynamically favored at any temperature
(B) thermodynamically favored at all temperatures
(C) thermodynamically favored at high temperatures, but not at low
temperatures
(D) thermodynamically favored at low temperatures, but not at high
temperatures
(A) 300 kJ
(B) 500 kJ
(C) 600 kJ
(D) 1000 kJ
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7
Day 1
Continue
Relative Intensity
Questions 16 and 17 are based on the mass spectrometry chart of
zirconium shown below.
100
90
80
70
60
50
40
30
20
10
0
I I
85
I
I
I
I I I I
90
(m/z ratio)
I
I I
95
I
I
I
I
100
16. Analysis of the information from the chart can be used to contradict
an early theory of atoms that
(A) electrons in atoms occupied specific energy levels
(B) all atoms of the same element are identical
(C) the nucleus of all atoms is positively charged
(D) the internal structure of atoms contains embedded positive and
negative charges
17. The presence of five distinctive peaks on the mass spectra chart of
zirconium indicates that
(A) zirconium tends to form a +5 charge
(B) zirconium has five occupied energy levels
(C) zirconium has five stable isotopes
(D) zirconium has five electrons in its valence shell
8
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Day 1
Continue
18 A student performed two titration experiments with two different acids
as shown in the diagram below.
50.0 mL of 0.10 M
weak monoprotic acid
50.0 mL of 0.10 M
strong monoprotic acid
Flask 1
Flask 2
Which one of the following statements is true?
(A) The weak acid will require a greater volume of NaOH solution than the
strong acid to reach the equivalence point.
(B) The weak acid will require a smaller volume of NaOH solution than the
strong acid to reach the equivalence point.
(C) The weak acid will require the same amount of NaOH solution as the
strong acid to reach the equivalence point.
(D) The equivalence point in a titration of a weak monoprotic acid with
NaOH solution cannot be determined.
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9
Day 1
Continue
19. Consider the equilibrium:
[Co(H2O)6]2+(aq) +
(pink ion)
4Cl-(aq)
[CoCl4]2-(aq) + 6H2O(l).
(blue ion)
The equilibrium constant for the reaction is 1.2 at 350 K and 2.5 at 275 K.
What is true about this reaction if it was conducted in a sealed vessel put
on ice?
(A) It would be endothermic, and the solution would look pink
(B) It would be endothermic, and the solution would look blue
(C) It would be exothermic, and the solution would look blue
(D) It would be exothermic, and the solution would look pink
20. A gravimetric analysis was used to determine the percent by mass of
lithium iodide in a contaminated mixture containing lithium iodide. Which
of the following compounds would be the best substance to react the
contaminated mixture with in order to determine the mass percent of
lithium iodide?
(A) sodium bromide
(B) lead nitrate
(C) ammonium nitrate
(D) sodium chloride
Day 1
STOP. Check your answers and note how many correct points.
10
E3Chemistry.com
Day 1
Answers and Explanations
Answers: Quick Check
1. C
2. A
11. B 12. C
3. D
4. A
5. C
6. D
7. C
8. B
9. B
10. D
13. D
14. B
15. A
16. B
17. C
18. C
19. A
20. B
Answers and Explanations
1. C
The reaction equation between calcium carbonate and hydrochloric
acid is: CaCO3(s) + 2HCl(aq)
CO2(g) + H2O(l) + CaCl2
Since the reactants are not gases, increase in pressure will not
increase the rate in which they react to produce carbon dioxide.
So, the volume of carbon dioxide collected will not increase.
Actions in choices A, B and D will increase the forward reaction,
therefore, increasing the volume of carbon dioxide produced.
2. A
q
= mC
3800 J = (125 g)(4.2 J/(g o
3800 J
----------------- =
525 J C-1
= 7.2oC
Final temperature =
3. D
24.0oC = 31.2oC
Step 1: Determine the amounts of C and H in the original sample.
33 g CO2(1 mol CO2/44.01 g CO2)(1 mol C/1 mol CO2) = 0.75mol C
0.75 mol C (12.01 g/mol) = 9.0 g C
13.5 g H2O(1 mol H2O/18.01 g H2O)(2 mol H/1 mol H2O) = 1.50 mol H
1.5 mol H(1.01 g/ mol) = 1.5 g H
Step 2: Determine mass and moles of O
Mass of O = 16.5 g - (9.0 g + 1.5 g) = 6.0 g O
Moles of O = 6.0 g O (1 mol/16 g) = 0.375 mol O
Step 3: Determine mole ratio in the empirical formula
O = 0.375 mol / 0.375 mol = 1
H = 1.5 mol / 0.375 mol = 4
C2H4O
C = 0.75 mol / 0.375 mol = 2
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11
Day 1
4. A
Answers and Explanations
Step 1: Determine the initial [IBr]
[IBr] = 10 mol/ 2 L = 5 M
Step 2: Use ICE to determine the equilibrium [Br 2] and [IBr]
Br2(g)
+
I2(g)
2IBr(g)
I
0
0
5M
C
0+2M
0+2M
5 M 2(2 M)
E
+2 M
+2 M
1M
Step 3: Calculate the equilibrium constant, K.
K =
5. C
[IBr]2
(1)2
----------------------- = ------------------ = 0.25
[Br2] [I2]
(2)(2)
Decreasing the volume increases the pressure and [IBr].
Because the initial [IBr] is increased, the reaction will proceed
faster and reach equilibrium quicker.
Because the volume of the reaction cylinder is smaller, the molarity
concentration of I2 will be greater than 2 M.
Note: The number of moles of I2 produced will be the same in
both the 2 L cylinder and the 1 L cylinder.
6. D
12
The true statement to this problem is best determined by eliminating
the obvious false statements.
TIP:
(A) False: O2 and He have different molar masses. Equal mass of O2
and He have different numbers of molecules, thus,
different pressures.
(B) False: The temperature is the same, therefore, the average kinetic
energy must be the same. Average KE = Temperature
(C) False: Different molar masses, therefore, different numbers of
moles, thus, different numbers of molecules.
(D) True: He has a smaller molar mass, therefore, a 100 gram sample
has greater numbers of molecules than a 100 g sample of
oxygen.
Greater number of molecules (He) = Greater pressure
E3Chemistry.com
Day 1
Answers and Explanations
7. C
The curve starts at a pH above 7 but much lower than 14.
The analyte must be a weak base.
The curve decreases to near pH 1:
The titrant must be a strong acid.
8. B
Maximum buffering pH is at half the equivalent point, where
[Base] = [conjugate acid]. The slope of the curve changes very little.
9. B
The addition of 100 mL of water to 100 mL of 0.10 M NaOCl solution
will instantaneously result in:
The doubling of the volume.
[OCl-], [OH-], and [HOCl] decreased by half.
Kc or product/reactant ratio reduced by half.
To restore equilibrium:
Kc will have to increase above half (Eliminating choice D)
The forward reaction will have to be favored for K c to increase.
[HOCl] and [OH-] will increase above half when the forward reaction is
favored.
At the new equilibrium point:
[OH-] and [HOCl] will be greater than half their original, BUT they will still
be lower than their initial concentrations (Eliminating choice C).
Since the new equilibrium [OH-] is lower than the initial [OH-], that means
the new equilibrium [H+] will be higher than the initial [H+] (Eliminating
choice A).
Higher [H+] means a lower or decrease in pH (choice B)
10. D
K1 = [HOCl][OH-] / [OCl-]
K2 = [HOCl]2[OH-]2 / [OCl-]2 = ([HOCl][OH-] / [OCl-])2 = K12
11. B
[Cl2] will increase with any change that will shift the reaction to the right.
The change that will shift the reaction to the left will then be the correct
choice to this question.
A. Adding NaCl increases [Cl -] and shifts equilibrium right (Eliminating A)
B. Adding NaOH produces OH-, which reacts with H+. This will decrease
[H+], and the reaction will shift left
C. Adding H2SO4(aq) increases [H+] , which will shift the reaction to the
right (Eliminating C)
D. Adding HOCl will shift the reaction to the right; Eliminating D
12. C
In HOCl, H+1 O-2 Cl+1 = 0
In OCl-, O-2 Cl+ = -1
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13
Day 1
Answers and Explanations
13. D Rate = k[reactant]. The [ ] of each reactant is raised to a power equal
to its coefficient in the balanced equation.
Rate = k[NO][Cl2]
Bimolecular because two molecules are involved as reactants
14. B
Reaction 2 has:
reactant to 3 molproducts ).
at all temperatures)
Any reaction with -
15. A Reaction 3 equation = ½ (Reaction 1 + Reaction 2)
200 kJ) = 300 kJ
16. B
identical.
According to the mass spec: Atoms of the same element have different
masses, therefore, are different.
17. C
5 peaks = 5 different mass to charge ratio = 5 stable isotopes
18. C
In a titration process, the volume of NaOH(aq) needed to reach the
equivalence point depends on the number of moles of H + of the acid.
In both Flask 1 and Flask 2, moles of H+ are the same because:
Volume of the acids is the same: 50 mL
Molarity of the acids is the same: 0.10 M
Both acids are monoprotic, so each produces the same moles of H+.
The number of moles of H + available for titration in each flask is:
(Va)(Ma) = (0.050 L)(0.10 mol/L) = 0.0050 mol H +
Therefore, the same volume of 0.20 M NaOH(aq) is required to reach
the equivalence point for both acids.
19. A The forward reaction is endothermic because:
Higher temp = Higher equilibrium constant = more blue product .
The reaction will shift left at a lower temperature (in ice).
Lower temp = smaller equilibrium constant = more pink reactant is made
20. B In gravimetric analysis, a solid substance must be used.
lead nitrate + lithium iodide
lead iodide (solid precipitate).
None of the other substances will produce a precipitate when reacted
with lithium iodide.
14
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15
16
E3Chemistry.com
Day 2
Free Response Questions
Section II Practice
14 Points
START: Answer all questions on this day before stopping.
Directions: Question 1 is a long free-response question that requires
about 20 minutes to answer and is worth 10 points. Question 2 is a
short free-response question that requires about 7 minutes to answer
and is worth 4 points.
Write your response in the space provided following each question.
Examples and equations may be included in your responses where
appropriate. For calculations, clearly show the method used and the
steps involved in arriving at your answers. You must show your work to
receive credit for your answer. Pay attention to significant figures.
Question 1
Long Response
10 points
Sulfur dioxide gas is commonly used as a preservative in wine. An important
source of SO2 is solid sodium metabisulfite (Na2S2O5; molar mass 190 g/mol).
Na2S2O5 reacts readily with acid as follows.
Na2S2O5(s) + 2HCl(aq)
2NaCl(aq) + H2O(l) + 2SO2(g)
(a) Determine the oxidation number of sulfur in sodium metabisulfite.
(b) Calculate the volume, in liters, of SO2 produced at 1.00 atm pressure and
15.0oC when 250 g of Na2S2O5 reacts with excess acid.
(c) The concentration of an aqueous solution of SO2 (solution A) is to be
determined using its reaction with an aqueous solution of triiodide
ions (I3). The relevant half reactions are
SO2(aq) + 2H2O(l)
I3 .(aq) + 2e
4H+(aq) + SO42 (aq) + 2e
3I (aq)
A student adds 50.0 mL of a 0.0125 M solution of I 3 to 50.0 mL of solution A,
providing excess of I3 . The final 100.0 mL solution is called solution B.
(i) Write an overall balanced chemical equation for the reaction that occurs.
(ii) Identify the oxidized substance.
(iii) Calculate the number of moles of I 3 added to solution A.
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17
The student determines the excess I3 remaining in the solution by titration
with a standard solution of sodium thiosulfate (Na2S2O3). The equation for the
reaction is
S2O32 (aq) + I3 (aq) + H2O(l)
3I (aq) + S2O42 (aq) + 2H+(aq)
It was determined that 14.70 mL of a 0.00850 M solution of Na2S2O3 reacts
exactly with all the I3 remaining in solution B.
(d) Calculate the original concentration of SO2 in solution A.
18
E3Chemistry.com
Day 2
Question 1
Space for Work and Answers
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20
E3Chemistry.com
Day 2
Question 2
Continue
Short Response
4 Points
A student carried out four experiments of reacting solid magnesium
carbonate, MgCO3, in a hydrochloric acid solution, HCl(aq). In each
experiment the student used excess amount of the HCl solution. The data
table below shows the result of the experiments.
(a) Is the reaction exothermic or endothermic? Justify your answer.
(b) Considering experiments 1 and 2, explain why the increase in the initial
temperature has raised the reaction rate.
(c) What is the order of reaction with respect to HCl for this experiment.
Justify your answer.
(d) Results from experiment 1 are plotted on the sketch graph below.
On the same axes, sketch the results from experiment 3.
Day 2
STOP. Check your answers and note how many correct points.
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21
Day 2
Question 2
Space for Work and Answers
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22
E3Chemistry.com
Day 2
Question 1
Answers and Scoring Guidelines
Long Response
10 points
Sulfur dioxide gas is commonly used as a preservative in wine. An important
source of SO2 is solid sodium metabisulfite (Na2S2O5; molar mass 190 g/mol).
Na2S2O5 reacts readily with acid as follows.
Na2S2O5(s) + 2HCl(aq)
2NaCl(aq) + H2O(l) + 2SO2(g)
(a) Determine the oxidation number of sulfur in sodium metabisulfite.
+4
1 point is earned for
a +4.
+1 +4 -2
Na2S2O5
The sum of charges (+2 +8 - 10) = 0
(b) Calculate the volume, in liters, of SO2 produced at 1.00 atm pressure and
15.0oC when 250 g of Na2S2O5 reacts with excess acid.
First: Fine moles of SO2 (nSO2) produced
1 mol Na2S2O5
2 mol SO2
250 g Na2S2O5 x --------------------------- x ---------------------- = 2.63 mol SO2
190 g Na2S2O5 1 mol Na2S2O5
Second: Calculate SO2 volume using the Ideal Gas Equation:
PV = nRT
nRT
(2.63 mol)(8.31 L atm K-1mol-1)(288 K)
V(SO2) = ------- = --------------------------------------------------------------------------P
1.00 atm
V(SO2) = 62.2 L
Copyright © 2018 E3 Scholastic Publishing. All Rights Reserved
1 point is earned
correcting calculating
moles of SO2.
1 point is earned
for a volume
consistent with
moles of SO2
produced.
23
Day 2
Answers and Scoring Guidelines
(c) The concentration of an aqueous solution of SO2 (solution A) is to be
determined using its reaction with an aqueous solution of triiodide
ions (I3). The relevant half reactions are
4H+(aq) + SO42 (aq) + 2e
SO2(aq) + 2H2O(l)
I3 .(aq) + 2e
3I (aq)
(i) Write an overall balanced chemical equation for the reaction that occurs.
SO2 + 2H2O + I3
3I + 4H+ + SO42
When the two half reactions are added, the 2ecancels out, resulting in the above overall equation.
1 point is earned for
the correct equation.
(ii) Identify the oxidized substance and justify your choice.
SO2
The first half-reaction given represents oxidation,
therefore, the substance to left of the arrow is
oxidized by losing electrons.
1 point is earned for
identifying SO2.
OR
The change of SO2 to SO42- represents a gain of
oxygen, which is oxidation. Therefore, SO2 is
oxidized.
1 point is earned for
correct justification
consistent with the
chosen substance.
50.0 mL of a 0.0125 M solution of I 3 is added to 50.0 mL of solution A,
providing excess of I3 . The final 100.0 mL solution is called solution B.
(iii) Calculate the number of moles of I 3 added to solution A.
24
moles of I3
= (0.0125 mol/L)(0.050 L)
moles of I3
= 0.000625 or 6.25 x 10-4 mol
1 point is earned for
correct moles.
E3Chemistry.com
Day 2
Answers and Scoring Guidelines
The excess I3 remaining in the solution is determined by titration with a
standard solution of sodium thiosulfate (Na2S2O3). The equation for the
reaction is
S2O32 (aq) + I3 (aq) + H2O(l)
3I (aq) + S2O42 (aq) + 2H+(aq)
14.70 mL of a 0.00850 M solution of Na2S2O3 reacts exactly with all the I3
remaining in solution B.
(d) Calculate the original concentration of SO2 in solution A.
First: Determine moles of I3 in excess = moles of S2O32 reacted
nI3 (in excess) = (0.00850 mol/L)(0.01470 L)
nI3 (in excess) = 0.000 125 mol
1 point is earned
for correct moles
of I3 in excess.
Second: Determine moles of I3 reacting with SO2
nI3 (reacting) = nI3 (initially)
nI3 (in excess)
nI3 (reacting) = 0.000625 mol - 0.000125 mol
nI3 (reacting) = 0.000500 mol
1 point is earned
for moles of SO2
in Solution A.
Third: Calculate the original concentration of SO2
NOTE: nI3 (reacting) = nSO2 in Solution A = 0.000500 mol
0.000500 mol SO2
Molarity = -------------------------------------- = 0.0100 M SO2
0.0500 L
Copyright © 2018 E3 Scholastic Publishing. All Rights Reserved
1 point is
earned for
concentration
consistent with
moles of SO2.
25
Day 2
Question 2
Answers and Scoring Guidelines
Short Response
4 Points
A student carried out four experiments of reacting solid magnesium
carbonate, MgCO3, in a hydrochloric acid solution, HCl(aq). In each
experiment the student used excess amount of the HCl solution. The data
table below shows the result of the experiments.
(a) Is the reaction exothermic or endothermic? Justify your answer.
Exothermic because the final temperature
is higher than the initial temperature in
each experiment, indicating that heat is
released during the reaction.
1 point is earned for
exothermic with a
correct justification.
(b) Considering experiments 1 and 2, explain why the increase in the initial
temperature has raised the reaction rate.
Increased temperature increases the
number of collisions with energy greater
than the activation energy
Increasing temperature leads to more
frequent effective collisions.
26
1 point is earned for a
correct explanation
referencing more collisions.
E3Chemistry.com
(c) What is the order of reaction with respect to HCl for this experiment.
Justify your answer with specific information from the data table.
Second order, 2nd order.
The rate of CO2 evolution in Experiment 4
quadrupled (is 4 times) that of Experiment 1
when the concentration of HCl was doubled.
1 point is earned for
second order with
correct justification.
Twice the concentration of HCl doubles the
initial rate of CO2 evolution.
(d) The student plotted on the sketch graph below the results from
Experiment 1. On the same axes, sketch the results from Experiment 3.
CO2
evolved
in mL
Time
1 point is earned for a skecth that:
starts at the same point
has a higher gradient or slope
levels out at or very close to twice the
volume of CO2 evolved.
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27
Day 3
Multiple Choice Questions
Section 1 Practice
20 Points
Start: Answer all questions on this day before stopping.
Directions: YOU MAY NOT USE YOUR CALCULATOR FOR THIS DAY.
YOU MAY USE THE PERIODIC CHART AND EQUATIONS TABLE.
Each of the questions or incomplete statements below is followed by
four suggested answers or completions. Select the answer that is best in
each case and then fill in the corresponding circle on the answer sheet.
Note: For all questions, assume that the temperature is 298 K, the
pressure is 1.00 atmosphere, and solutions are aqueous unless
otherwise specified.
1. Gaseous molecules X, Y and Z are placed in a closed container that can
have its volume, temperature, and/or pressure changed at any time.
Rubber lines in and out of the container can be used to add or remove
molecules of the substances. Initially, the container has a volume of
2 liters at 25 degrees Celsius and 1 atmosphere. Which of the following
changes will increase the rate of collisions between molecules of X and Y?
(A)
(B)
(C)
(D)
Changing the volume to 1.5 liters
Changing the temperature to 20 degrees Celsius
Taking away molecules of X
Adding molecules of Z
Molecule
Bond
Bond Length (pm)
NH3
N H
98
PH3
P H
140
2. Based on the information given on the table above, which statement
best explains why P H bond length is longer than N H bond length?
(A) The P atom has a larger atomic radius making the Columbic attractions
weaker and the bond length longer.
(B) The P atom has a larger atomic radius making the Columbic attractions
stronger and the bond length longer.
(C) The P atom has a smaller atomic radius making the Columbic attractions
weaker and the bond length longer.
(D) The P atom has a smaller atomic radius making the Columbic attractions
stronger and the bond length longer.
Copyright © 2018 E3 Scholastic Publishing. All Rights Reserved
29
Reference Materials
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 


























 


 

    
   
  
 ×  


  

 



 
 
 
 
 

 


    

    



 


  
     

 

     

    


 
 

  

302


  

 
   




 

 
 

   


  
  
  
  
 
  
 
 
 
 
  
  
  
     

   
   
   


   
  


  
  
 
 
 
 
 
 
 
  
 
 
  
 


 

  
 
  
  
  
  
  
  
  
  
  
  
  
 



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