2007 Chemistry Exam 2 Written Examination

[Collin Li]

SECTION A - Multiple-choice questions

Question 1
CrO3 => +6
Cu2S => +1 (The sulfide ion is -2)
MnCl2 => +2
K2Cr2O7 => +6
The answer is B.

Question 2
The reaction is a fusion reaction involving species smaller than iron-56, so: I. "The reaction is endothermic" is incorrect.
Energy is released, and according to E=mc^2, mass is lost to compensate for the release in energy, so: II. "The mass of the carbon-12 nucleus is greater than the combined masses of the reactants" is incorrect.
The answer is D.

Question 3
Let a be the proportion of rhenium-187:
185(1-a) + 187a = 186.2
=> 2a = 1.2
=> a = 0.6 = 60%
The answer is C.

Question 4
SO3(g) + H2O(l) --> H2SO4(aq)
The salt will be Na2SO4, while the oxide ion and protons have fun neutralising each other.
The answer is D.

Question 5
Ligands must have a spare electron pair, or be negatively charged, to be attracted to the central metal cation:
Set A contains H+: incorrect
Set B appears correct
Set C contains Na+: incorrect
Set D contains CH4: incorrect (no free electron pair, all are participating in C-H bonds)
The answer is B.

Question 6
Transition metal compounds tend to exhibit colour: barium, aluminium, potassium and sodium are not transition metals, but manganese (Mn) is.
The answer is D.

Question 7
Enzymes are class of proteins that are known for their catalytic ability in biological reactions. Statements I and II are correct. III is incorrect, a catalyst does not affect the equilibrium constant at all.
The answer is A.

Question 8
Antioxidants sacrifically oxidise in reaction with oxidants in defence of foods that would otherwise become rancid. In order to sacrificially oxidise, it must be a good reductant. (answer B or D)
A key feature of antioxidants is the number of OH groups they possess, so antioxidants are also soluble in water.
The answer is D.

Question 9
The compound is not a protein, because there is no nitrogen. This means it is a carbohydrate or fat, which consist of hydrogen only:
=> Hydrogen: 100 - 76.2 - 11.3 = 12.5%
n(C) : n(H) : n(O) = 76.2/12 : 12.5/1 : 11.3/16
= 6.35 : 12.5 : 0.706
= 9 : 18 : 1
It is most likely a fat of the molecular formula: C18H36O2. Carbohydrates have a much higher oxygen content.
The answer is A.

Question 10
In process I, NH3 accepts a proton, it acts as a base.
In process II, NH4+ is oxidised from an oxidation state of -3 to +5.
The answer is D.

Question 11
(+) Half cell I | Half cell II (-) => B2+ is oxidising while A2+ is reducing
(+) Half cell II | Half cell III (-) => C is oxidising while B3+ is reducing
While B3+ is a stronger oxidant than C+, A2+ is a stronger oxidant than B3+ (reduces more readily). Therefore A2+ would be a stronger oxidant than C+ too.
The answer is A.

Question 12
The positive section is the anode, where the oxidation of Cl- occurs. X is chlorine.
The negative section is the cathode, where the reduction of H2O occurs. Y is hydrogen (see electrochemical series).
The answer is C.

Question 13
The porous diaphgram allows the movement of ions (Na+ particularly) so that NaOH can be formed.
The answer is B.

Question 14
The highly concentrated salt solution exceeds the standard conditions of 1M, thus creating non-standard conditions that cause the order of the chlorine and water reaction to swap, allowing chlorine to be produced in preference to oxygen gas.
The answer is B.

Question 15
While generating electricity, M acts as the anode. Therefore, a reduction is occurring in the nickel half-cell, with NiO(OH) being reduced, thus acting as an oxidant.
The answer is A.

Question 16
While being recharged, the positive terminal is the anode where oxidation occurs. In the forward reaction, NiO(OH) acts as the oxidant and is reduced to Ni(OH)2. Therefore in the reverse reaction, Ni(OH)2 must be oxidised instead.
The answer is C.

Question 17
The answer must be C or D, as they are the only options that show oxidation reactions at the anode.
The reductant is methanol, as it becomes oxidised at the anode (as suggested by options C or D, and by commonsense that oxygen is an oxidant).
The answer is C.

Question 18
The calibration factor = energy input / change in temperature. Energy input is Q*V, while the change in temperature is ∆T1. (Typically, we use E = VIt, but since Q = It, we can say E = V*Q)
The answer is D.

Question 19
∆H = - (CF * ∆T2) * 2/n
The factor (2/n) was to adjust the energy output (CF * ∆T2) into per mol. There are 2 moles of butane per mole of reaction, and n moles of butane was combusted, so this is a compensation factor.
VCAA forgot the negative sign!
The answer is B.

Question 20
For a fixed amount of electrical output, we would require the highest amount of chemical energy in coal and oxygen, as this is the first step in the process, and a lot of the energy is lost as heat.
The key is identifying which steps are the earliest in the power production process. The generation of steam occurs next, and there will be more energy in the steam then there will be from the turbine spinning, because not all of the thermal energy of steam will be transferred to the turbine. (I < III < II)
The answer is C.

Since this is the last MC question, I'll digress: this question takes a while to digest. Here's an alternative, more risky route for those who couldn't comprehend it. From an earlier post:

Heh, answer is C without even reading the question. Options A and C indicate II is certainly the last one. Options B and C indicate I is certainly the first one. C is the intersection. (edit: you can even reinforce this further by using this reasoning on the middle one, it's certainly III by options C and D, C is the ultimate intersection!)

An illustrative example (this is a fitting final MC, because it is confusing):
Suppose there is a 1000 units of chemical energy. 400 units of this becomes thermal energy, while 600 units is lost as heat. Then, 300 units of the remaining 400 units is converted into the mechanical energy of the turbine. 100 units must be lost from heat. We are producing a fixed amount of electricity (300 units), where the required units of input vary (the earlier steps require more input to compensate for the certain loss of energy from heat).

END OF SECTION A

...beginning of dinner, be back soon.

[bilgia]

damn three wrong for me...things looking not so good....just waiting for section B now..crickey!
thanks coblin

[randomnezz]

nice work typing it up although i'll try to avoid the answers ...i don't want to get depress cause i still got 1 more exam left 2 do :/

[Collin Li]

SECTION B - Short answer questions

Question 1
i. Mg
ii. U
iii. Al
iv. B
v. O
vi. Sn

Question 2
a)
Any two of these three features:
- Elements with similar chemical properties were placed in the same "group."
- There were spaces where Mendeleev predicted "missing elements."
- The table was ordered by increasing atomic mass.

b)
i. Mass spectrometer
ii. Protons: 113, Electrons: 113, Neutrons: 171
iii. Group: III, Period 7
iv. Tl
v. Fr is in group I, so it has a weaker nuclear core charge than Uut, which means the electrons in Uut are more tightly bound to the nucleus, hence the atomic radius of Uut is smaller than Fr.
vi. Al is in period 3 and since it has the same nuclear core charge as Uut, it has a stronger attraction on the fewer number of electrons it has compared to the electrons on Uut, hence electrons on Uut are easier to ionise.
vii.

Code: [Select]

283         4     279
   Uut --->  He +    Rg
113         2     111

Question 3
a)
i. [He]2s22p2
ii. [Ar]3d64s2
iii. [Ar]3d6
(These should be superscript instead)

b)
Group: VI, Period 5

Question 4
a)
i. C6H12O6
ii. C12H22O11 + 12O2 ---> 12CO2 + 11H2O
iii. ΔH = -2816*2 = -5632 kJ/mol. 1 mole of maltose can hydrolyse into 2 moles of glucose.

b)
i. C15H29COOH => 30 hydrogen atoms
ii.

Code: [Select]

 H
  |
H-C-O-H
  |
H-C-O-H
  |
H-C-O-H
  |
  H

c)
i. CH3OH + C17H35COOH ---> H2O + C17H35COOCH3
ii. C17H35[COO]CH3 (in square brackets)

d)
Lecithin has a non-polar half, and polar half, which allows lecithin to interact with both water and oil, and can create oil in water suspensions, and vice versa.

Question 5
a)
i.

Code: [Select]

   CH3
    |
H2N-C-COOH
    |
    H
ii.

Code: [Select]

  H
   |
HO-C=O
   |
 H-C-H
   |
   N-H
   |
 O=C
   |
 H-C-CH3
   |
   NH2

b)
i. Circle any one of the many:

Code: [Select]

  |
   N-H
   |
 O=C
   |
ii. Z2 (-CH2CH2CH2CH2NH2)
iii. Z1 (-CH2SH)
iv. Enzymes are sensitive to pH and temperature. Under different pH conditions, it's shape can change and it can denature.
v. Urea:

Code: [Select]

H   H
 \ /
  N
  |
  C=O
  |
  N
 / \
H   H

Question 6
a)
i. E = (100-18.5)[550*4.18 + 150*0.900] = 198*10^3 J = 198 kJ
ii. n(C2H5OH) = 198/1364 = 0.145 mol
=> m(C2H5OH) = 0.145 * 46.0 = 6.69 g
iii. mass = (1/0.35) * 6.69 = 19.1 g

b)
n(C2H5OH) = 10.0 / 46.0 = 0.217 mol => E = 0.217 * 1364 = 296 kJ
n(C4H10) = 6.00 / 58.0 = 0.103 mol
=> ΔH = -296*2 / 0.103 = -5730 kJ/mol
(2 moles of butane per mole of reaction)

Question 7
a)
i. Platinum
ii. temperature: 25?C, pH = 0

b) Cadmium
An increase in pH indicates H+ has been consumed. This means the forward reaction involves the reduction of H+(aq) into H2(g). Thus, the stronger reductant out of H2(g) and Cd(s) is cadmium.

c)
i. n(X2+) = n(initial) - n(final) = 0.100*(1.00 - 0.725) = 0.0275 mol
ii. Q = 2654 => n(e-) = 2654/96500 = 0.0275 mol
=> n(X2+) = n(e-) = 0.0275 mol
=> n(X2+):n(e-) = 1:1
iii. X3+
iv. X2+ ---> X3+ + e-

Question 8
a)
cathode: Cu2+(aq) + 2e- ---> Cu(s)

b)
Q = 1.62*581 = 941 C
n(Cu) = 0.306/63.6 = 4.81*10^-3 mol

c)
n(e-) = 2*n(Cu) = 9.62*10^-3 mol
=> F = Q/n(e-) = 941 / (9.62*10^-3) = 97800 C/mol

d)
Any of the following:
- Only a fraction of the extra copper was weighed (due to flaking).
- Inefficient cell and/or resistance in wires causing a lower n(e-) than expected.

END OF EXAM

[kingmar]

Interesting. I talked about a systematic error with scales.

The third actinide is Uranium, according to my textbook. So we have some problems.

[cara.mel]

6a you missed the energy to heat the aluminium pot
Other then that, thank you ^_^

So pissed off at myself for misreading/interpreting one of the multi choice, especially as I originally wrote the correct answer >_<

[Collin Li]

Interesting. I talked about a systematic error with scales.

The third actinide is Uranium, according to my textbook. So we have some problems.

Sorry, I was looking at the Wikipedia periodic table, which is weird (updated).

[Collin Li]

6a you missed the energy to heat the aluminium pot
Other then that, thank you ^_^

So pissed off at myself for misreading/interpreting one of the multi choice, especially as I originally wrote the correct answer >_<

Yeah, oops. I misread questions when they're on the computer.  :cry:

[Collin Li]

I have the full copy of Section B (thanks to asa.hoshi). I will have the full solutions in a bit. I may need to draw and upload.

[principe]

I have the full copy of Section B (thanks to asa.hoshi). I will have the full solutions in a bit. I may need to draw and upload.

Thanks coblin  

[bilgia]

Interesting. I talked about a systematic error with scales.

The third actinide is Uranium, according to my textbook. So we have some problems.

Sorry, I was looking at the Wikipedia periodic table, which is weird (updated).

what the..is it not uranium??? this is very confusing

[hakeemthedream]

farrrk

i just cant stop making bullshit mistakes

6 marks off - 7 maybe at worst

what study score am i looking at with 5 marks off 1st exam

[bturville]

Interesting. I talked about a systematic error with scales.

The third actinide is Uranium, according to my textbook. So we have some problems.

Sorry, I was looking at the Wikipedia periodic table, which is weird (updated).

what the..is it not uranium??? this is very confusing

actinide series, a series of radioactive metallic elements in Group 3 of the periodic table. Members of the series are often called actinides, although actinium (at. no. 89) is not always considered a member of the series. The series always includes the 14 elements with atomic numbers 90 through 103. The other members are (in order of increasing atomic number) thorium, protactinium, uranium

(The Columbia Electronic Encyclopedia, 6th ed. Copyright ? 2007, Columbia University Press. All rights reserved.)

Wouldn't that make that question ambiguous? I said Uraninum, but lots of periodic tables I've looked up include actinide in the series!?


AHHHH! RIPPED OFF!

[Collin Li]

Interesting. I talked about a systematic error with scales.

The third actinide is Uranium, according to my textbook. So we have some problems.

Sorry, I was looking at the Wikipedia periodic table, which is weird (updated).

what the..is it not uranium??? this is very confusing

actinide series, a series of radioactive metallic elements in Group 3 of the periodic table. Members of the series are often called actinides, although actinium (at. no. 89) is not always considered a member of the series. The series always includes the 14 elements with atomic numbers 90 through 103. The other members are (in order of increasing atomic number) thorium, protactinium, uranium

(The Columbia Electronic Encyclopedia, 6th ed. Copyright ? 2007, Columbia University Press. All rights reserved.)

Wouldn't that make that question ambiguous? I said Uraninum, but lots of periodic tables I've looked up include actinide in the series!?


AHHHH! RIPPED OFF!

Nah, you'd use the VCAA Periodic Table, where U was the third member. I was looking at the periodic table off Wikipedia, which seems to believe in the other way. My mistake, sorry (obeys VCAA's Chemistry laws... but NEVER their ideas on ligands!)

I'm almost done, was sidetracked from Heroes (what a GOOD episode!)

[ezflow]

Coblin you are my hero

Can you send me photos of you so I can put them up around my room and worship you?

No, seriously, thanks so much for doing this.