Chemistry
2008 June
Exam
Answer Key
2008 Chemistry Exam Key
1-B
a straight knowledge question. If you don’t know it review
2-C
the number of moles is equal, so calculate it for NH3, that will then give it for the F2
3-B
P1/T1 = P2/T2. Don’t forget to convert to Kelvins before, then back to Celcius after!
4- D.
A is not correct, since the molecules do not increase their volume, just the space between
5-A.
Remember you are looking for the reverse reaction!
6-A
-mCΔT =mcΔT, so 250(4.19)(Tf-24) = 2000(4.19)(Tf-4)
7-B
The motions of a liquid’s molecules are mostly rotation (I) and vibration (III).
8-C
Photosynthesis (1) absorbs sunlight(energy). Evaporation (4) absorbs heat
9-D
The graph with the higher Ea (graph B) is uncatalyzed. ΔH = 10-40 = -30 kJ/mol
10- A
This question is a bit bogus, since there are arguably other possible answers, but A is best
Why? because B is false -- digging trenches has no effect on oxygen, C is arguably partly true, but
water mostly cools and only slightly blocks off oxygen; D is also questionable-- foam mostly blocks
oxygen and only slightly cools.
11-D
Although A looks to be correct its not! The water is liquid H2O(l), and should not be included!
12-A
At a higher temperature more product is produced, changing Kw
13-B
14.
Don’t worry about this one, electrochemistry is no longer on June exams
Long Answer questions
15.
The students would smell the ammonia (NH3) first. Its molar mass is approximately 17g/mol
(14+1+1+1) and chlorine (Cl2) has a molar mass of about 71 grams per mole (35.4 x2). According to
Graham’s law, the lower the molar mass of a gas’ particles, the faster it will diffuse.
(presenting a formula of Graham’s Law here to help justify your answer would be a good touch!)
16.
T1=23C = 296K
P1= 315 kPa
m1=84g (N2)
M1= 28.0 g/mol (N2=2 x14.0)
n1 =3.00 mol (see calc.)
T2=15C =288K
P2=235 kPa
n2=?
Answer:
n1= m1/M
𝑃1 𝑉1
𝑛1 𝑇1
n2
𝑃2 𝑉2
=
𝑛2 𝑇2
=84.0 g ÷ 28.0 g/mol = 3.00 mol (keep 3SD)
you can eliminate V, it doesn’t change
=n1T1P2 / P1T2
= (3.00mol) ( 297K)(235kPa)/(315kPa·288K)
= 2.308 mol
2.308 mol x 28 g/mol = 64.63 g
= 64.6 g of N2 remain in the tire.
17.
Step 1: Calculation of moles of NaClO3
m= 3.42 g (NaClO3)
stated in the problem
M=106.5 g/mol
M= 23.0g + 35.5 g + 3 (16.0g) from periodic table
n= 0.0321 mol (NaClO3)
n=m/M so... n = 3.42g / 106.5 g/mol = 0.0321mol
Step 2: Calculation of moles of oxygen by stoichiometry:
2 NaClO3 2 NaCl + 3O2
Molar ratio
2:
3
0.0321mol n(O2 )

2
3
n=0.482 mol (O2)
Step 3: Calculation of R
PV=nRT
PV
nT
102.5kPa  1.240 L
R
0.0482mol  347 K
R
Answer :
R=7.599 or rounded to 3 sig. figs: R=7.60 kPa·L/K·mol
18.
Step 1: calculate number of moles of NH3, either by using PV=nRT or by dividing by 22.4 L/mol
3794.6 mol of NH4, there will be half this many moles of ammonium sulphate by stoichiometry,
1897.3 mol of (NH4)2SO4
Mass = n x M = 1897.3 x (2(18)+32.1+4(16))
= 1897.3 x132.1= 250633.3g
Answer:
=251000 g or 251kg
19.
Step 1: Use Hess’ law to find the ΔH for NaOH(aq)+HCl(aq)  NaCl(aq) + H2O(l)
NaOH(s)

NaOH(aq)
ΔH=-44.2kJ
R
NaOH(aq)

NaOH(s)
ΔH=+44.2kJ
NaOH(s) + HCl(aq)

NaCl(aq) + H2O(l)
ΔH=-100.1 kJ
NaOH(aq) + HCl(aq)

NaCl(aq) + H2O(l)
ΔH=-55.9 kJ
Step 2: Find the total heat using calorimeter formula (total m=150g + 150g = 300g)
Q=300(4.19)(13)
Q=16341 J or 16.341 kJ
n =Q/ΔH
= 16.341 kJ / -55.9 kJ/mol
Answer:
0.292 mol
20.
M
= (7 * 12.0) + (6 *1.0) + (3*16.0) = 138 g/mol (salicylic acid)
n
= m/M
= 3.84g / 138g/mol
=0.0278 mol
Q
=ΔH·n
3.02 kJ/mol x 0.0278 mol
= 0.08403 kJ
Q
=mcΔT
84=100g (4.19) ΔT
ΔT=84 / (100·4.19)
Answer:
0.200°C
=84 J
21.
-Q(water)
= Q (stones)
-mcΔT
= mcΔT
-m (4.185 J/g°C)(63°C-100°C) = 3000g (0.84 J/g°C)(63°C - 21°C)
-m (-154.845)
=105840
154.845 m
=105840
m
=683.52 g
Assuming that the density of water is 1 g/mL, then you would need to boil 683 mL or 0.683 L of
water.
22.
This will be a Hess’ Law problem. You actually have three equations to use, but two of them are
disguised as graphs.
6 CO2(g) + 6 H2O(l)
Target equation:
C6H12O6(s) + 6 O2(g)
Eq. 1
Reverse Eq.1
6C(s) + 6H2(g) + 3O2(g)  C6H12O6(s)
C6H12O6(s)
6C(s) + 6H2(g) + 3O2(g)
Eq.2
Multiply Eq.2 x 6
C(s) + O2(g)
6C(s) + 6O2(g)
 CO2(g)
 6CO2(g)
ΔH= -394.1 kJ
ΔH= -2364.6 kJ
Eq.3
Reverse and x6
H2O(l)
6H2(g) + 3 O2(g)
 H2(g) + ½ O2(g)
 6 H2O(l)
ΔH= +286.2
ΔH= -1717.2 kJ
Verify target
C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(l)
ΔH=-2807.3 kJ
ΔH= -1274.5 kJ
ΔH= +1274.5 kJ
The heat of combustion of glucose is 2807.3 kJ/mol, and the reaction is exothermic. However,
we have been asked to find the combustion heat from burning 90 grams of glucose, so we have
to find out how many moles this is. n=m/M, and M(glucose) = (6x12.0)+(12x1.0)+(6x16.0) = 180
g/mol. This means that we have 90g / 180 g/mol = 0.5 mol (glucose). Burning a half a mole of
glucose would produce 0.5 x 2807.3 kJ = 1403.7 kJ of heat. In the question the lowest significant
figures of a measurement was the 3 significant figures in 90.0g, so we can round our answer to 3
significant figures.
Answer: Burning 90 g of glucose would produce 1.40 x103 kJ of heat.
23.
State four modifications that would increase the reaction: 4HCl(aq) + O2(g)  H2O(l) + 2Cl2(g)
A. Increasing the temperature would increase the average speed of the molecules and therefore
increase the number of collisions.
B. Increasing the concentration of the HCl would increase the number of HCl particles colliding
in a given amount of space.
C. Increasing the pressure of the oxygen would increase the number of oxygen particles in a
given space, and therefore increase the number of colliding particles.
D. Introducing a catalyst would reduce the activation energy, so that more of the collisions
would be effective.
24. 2Ag+2HNO3H2 + 2AgNO3
Step 1: Find the increase in H2 mass between 5 and 20 sec. from the graph.
Δm = 0.12g – 0.08g = 0.04g (H2)
Step 2: convert this to moles by dividing by M(H2)
Δn= 0.04g / 2.0 g/mol = 0.02 mol(H2)
Step 3: Find the rate of H2 production by dividing by the time (Δt= 25s – 5s = 20s)
R(H2) = Δn/Δt = 0.02 mol/ 20s = +0.001 mol/s (positive because H2 is increasing)
Step 4: From the stoichiometry of the equation, you know that Ag disappears at twice the rate
that H2 appears.
R(Ag) = -2 R(H2) = -2 x 0.001 = -0.002 mol/s
Step 5: Convert this into g/s by multiplying by M(Ag), 107.9 g/mol
0.002 mol/s x 107.9 g/mol = 0.2158 g/s
Step 6: round to 2 sig figs to match data in the problem  0.22 g/s
25.
The initial and final concentration of HCl can be determined from the pH, since HCl is a strong
acid and dissociates completely. The [H+] will equal the HCl(aq) concentration.
Initial [HCl] = 10-pH = 10-1.00 = 0.100 mol/L
Final [HCl] = 10-pH = 10-2.00= 0.0100 mol/L
Δ[HCl] = 0.100 – 0.010 = 0.090 mol/L. Since the container is 1L, this also means that the change
in the amount of HCl is also 0.090 mol. From the stoichiometry of the equation, the change in
CO2 will be half the change in HCl:
0.090 𝑚𝑜𝑙
2
𝑛
1
= , so n = 0.045 mol (CO2). However, there is one
last step remaining. We need to know the rate of production of the CO2 over the 25 second
interval, so divide: 0.045 mol ÷ 25 s = 0.0018 mol/s, and that is the answer.
26.
2 H2O2(aq)  2 H2O(l) + O2(g) + energy
The first stress is temperature change. Increasing the temperature would favour the
endothermic reaction, as the system tries to absorb energy and reduce the temperature. This means
the reverse reaction would be favoured by an increase in temperature, and the direct (forward) reaction
would be favoured by a decrease in temperature.
The second stress is changing the concentration of hydrogen peroxide. Increasing the
concentration of Hydrogen peroxide would favour the direct (forward) reaction, as the system tries to
reduce the H2O2 concentration. Lowering the concentration of H2O2 would favour the reverse reaction,
as the system tries to increase H2O2 concentration.
The third stress is changing the concentration of oxygen in the reaction chamber, but without
changing the pressure. Higher concentration of O2 would favour the reverse reaction, as the system
would try to use up the O2. Lower concentration of O2 would favour the direct (forward) reaction.
The fourth stress is pressure change. Increasing the pressure by reducing the size of the sealed
container would favour the side of the reaction with the fewest gas molecules, which is the reverse
reaction, as there are no gas molecules on the left side of the equation. Decreasing the pressure by
enlarging the reaction container would favour the forward (direct) reaction.
27.
In a 4L container, 2 SO2(g) + O2(g)  2 SO3(g)
The amounts are given in moles. To change them to concentrations in mol/L you must divide by
4 (the volume of the container) , so 1.20 mol  0.30 mol/l; 1.00 mol  0.25 mol/L; 0.40 mol  0.10
mol/L. Then we put the numbers into an ICE table to find the equilibrium concentrations. The bold
numbers are what you start with, the italics are the ones you would fill in.
Substance
Ratio
I
C
E
SO2
2
0.30 mol/L
-0.20 mol/L
0.10 mol/L
O2
1
0.25 mol/L
-0.10 mol/L
0.15 mol/L
SO3
2
0.0 mol/L
+0.20 mol/L
0.20 mol/L
[𝑆𝑂3 ]2
0.202
𝐾𝑐 =
=
[𝑆𝑂2 ]2 [𝑂2 ] 0.102 × 0.15
Kc=26.6667
The lowest significant digits in the data was 2, so we can round the answer to K c = 27
28. Use the Ka formula. We will assume that because it is a weak acid, the initial concentration 0.30
mol/L will be the same as its equilibrium concentration. We will also assume that it dissociates equally,
so that [H+] = [CH3COO-]
𝐾𝑎 =
[𝐻 + ][𝐶𝐻3 𝐶𝑂𝑂−]
0.30 𝑚𝑜𝑙/𝐿
, substituting in the Ka value, and changing the numerator to [H+]2, we
end up with: 1.8 × 10−5 =
[𝐻+]2
0.30
, or [H+]2 = 5.4 × 10-6 so [H+] = √5.4 × 10−6
[H+] = 0.00232379. to find the pH, find –log(0.002323790) = 2.63380312
Now round our answer to two significant figures, to match the data in the problem
Answer: The pH of the acetic acid solution is pH=2.6
29. Does not need to be done, since electrochemistry is no longer required. For those of you
who are interested, the answers are:
1. Zn + 2Fe3+  Zn2+ + 2Fe2+
2.
3. Fe3+ is the oxidizing agent
4. No, adding more Zn2+ would NOT favour the reduction of Fe3+