HSC Chemistry: Suggested Answers and Discussion

[RuiAce]

Hopefully that's the last we ever need to think about biochemical oxygen demand ever...

We'll have our answers up and running shortly! What did you think of the paper? Any curveballs?

Also, how many of us here are glad to have sat their LAST EXAM?

Check out a copy of the exam here!
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Sample solutions

MULTIPLE CHOICE

1. D <- Classic monomer, also given away by the Cl atom
2. D <- Sodium isn't classified as a heavy metal. It is also found everywhere in the salty sea water.
3. C <- Can be drawn, or use the formula C_nH_2n+1O₂
4. C <- The benzene rings add rigidity. Then obviously plastic bags are not rigid.
5. B <- By elimination: D is wrong because that H is not polar; A and C are wrong because those dipoles are of the same polarity.
6. A <- Most basic would be pairing a strong base with a weak acid
7. B <- Use the table given and find an indicator which distinguishes the two
8. D <- By elimination: C is for reliability, and both A and B would mean you have no experiment anymore
9. D <- Balance it out. 2 protons and 2 neutrons are missing, i.e. a helium nucleus. Which is an alpha particle
10. C <- Cheat way: Decrease the charge and the number of H's by 1. Normal way: Just write out the equation
11. D <- By elimination: Only A and D have the correct alphabetical order. Then, the sum of locants on D is 5, as opposed to 7 for A
12. C <- 0.01 mol L^-1 means that the pH is at 2. So ti bring it to 4, we need to raise the volume 100x
13. A <- Sodium has no special flame, and the precipitate was BaSO₄
14. B <- Forces the pressure to increase, and the reactants have fewer moles of gas
15. A <- By elimination: Only the alkene decolourises the HOBr, so B and D are wrong. Then methanoic acid is going to be much more soluble than butanoic acid as it is smaller
16. A <- B is disqualified, due to the fact the cell must have 2⁺ charges. So simply compute the E^o valuse for the three remaining options and find the highest value
17. D <- This is polypropylene. We need the double bond to be there ready to be opened. (Or alternatively, just find the monomer 1-propene)
18. B <- See below
19. C <- See below
20. A <- The two wavelengths must correspond to each other for AAS analysis.
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18. You can tell by looking it how many moles of HCl are left, but properly done:
n_NaOH = 0.04 * 0.1 = 0.004 mol
n_HCl = 0.06 * 0.1 = 0.006 mol
1:1 mole ratio implies HCl is in excess, and moles of HCl left = 0.002 mol, and the combined volume is now 100mL
[H⁺] = [HCl] = 0.002/0.1 = 0.02 mol
pH = -log₁₀0.2 = 1.6989

19. Ba(NO₃)₂ + Na₂SO₄ -> 2 NaNO₃ + BaSO_4(s)
n_Na2SO4 = CV = 0.2 * 0.2 = 0.04 mol
n_BaSO4 = n_Na2SO4 = 0.04 mol
m_BaSO4 = n*MM = 0.04 * 233.37 = 9.33

CORE SHORT ANSWER

INDUSTRIAL CHEMISTRY

a) (i) A: Superheated water goes in. B: Molten sulfur emulsion comes out

a) (ii) Consider the process and the relevant properties.
Process: Superheated water melts the sulfur. Compressed air pushes the sulfur out of the pipe.
Properties: Sulfur has a low B.P. and can easily be melted.
It is insoluble and thus can be easily collected
Low density is what allows the emulsion to be produced, allowing compressed air to push it out more easily
(Could possibly discuss limited risk of health, however not necessarily required here)

b) (i) Potentially briefly describe the Hoffman Voltameter apparatus and procedures. Safety could include chlorine gas is toxic -> fume cupboard, or just the basic nature of NaOH

b) (ii) Both equations:
2 H₂O + 2 e^- -> H_2(g) 2 OH ^-
2 Cl^- -> Cl_2(g) + 2 e
Tests worth mentioning:
Chlorine - bleaches litmus paper
Hydroxide ions - Add drops of phenolphthalein
Hydrogen gas - using pop test on a sample

c) (i) \(K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \)
Decrease in volume (similar to increase in pressure) causes all concentrations to increase. Because all substances are affected by the same volume change, the concentrations spike upwards, and hence a higher reaction quotient. Then by LCP, equilibrium shifts to side with fewer moles of gas (i.e. reactants) bringing the reaction quotient back down to K.

Decrease in temperature also shifts the equilibrium to the left (as per LCP) due to the nature of the endothermic reaction. However, at a new temperature, a new equilibrium constant will be formed. This equilibrium constant will be lower than the original.

c) (ii) The reaction is straightforward: NH₄HS ⇌ NH₃ + H₂S. But the reactant is a solid.
\(\therefore K = [NH_3][H_2S] \) and also note the 1:1:1 mole ratio
0.0328 mol of NH₃ was produced, meaning that 0.0328 mol of NH₃ was made in the reaction (initial moles = 0)
Hence the moles of H₂S made is also 0.0328 mol
Note that the concentration of NH₄HS will be irrelevant.

So since the moles is known, using C=n/V we have C=0.0109333333333... mol L^-1
\(K=(0.01093333333333...)^2\approx 1.20\times 10^{-4} \) at 3 s.f.

d) Bit annoying but just rote work. Some possible points you could mention are (but not limited to):
Laboratory:
- Reflux may be employed
- Glycerol is just dumped or is kinda stuck inside the soap
- Soap is very impure
- Reaction doesn't really go far and can be completed quite quickly

Industrial
- Glycerol kept and potentially sold
- Soap is quite pure
- Uses much higher temperatures
- Don't need reflux - less volatility
- Fragrances added

In both cases:
- Equation: Fat/Oil + 3 NaOH -> 3 Soap + Glycerol
- Salting out may be employed (however kinda compulsory for the industrial method)
- Soap: always washed
 

SHIPWRECKS, CORROSION AND CONSERVATION

a)
i)

This is an example of an electrolytic cell, in which electrolysis will occur. This is because energy is being supplied to the cell (the 6V battery), thus inducing components to oxidise and reduce where they normally would not.

ii)

We know that electrons are being supplied to the electrode NOT labelled X. Thus, that electrode will cause something to reduce (reduction it gains!). So, whatever is happening at the electrode labelled X will have to be some kind of Oxidation reaction. Now, there are multiple things that could be oxidised; Sodium chloride (specifically, the Chlorine could be losing electrons to form Chlorine gas) or the water. Choose one, and write out the relevant equations!

b)
i)

Describe your procedure, whatever it is, making sure to do multiple tests (to increase reliability). Discuss safety procedures, such as keeping acid away from skin, keeping glassware away from the edge of the bench etc.

ii)

Colder temperatures corrode less, and ocean depths are very cold. Discuss this briefly, and then move onto your other factor (eg. Pressure, acidity, Sulfate, low oxygen) linking this specifically to corrosion of deep-sea wrecks

c)
i)

An example of a passivating metal is Aluminium. It reacts with oxygen to form an oxide layer (write out the equation!) thus physically protecting the metal from corrosion. This is very useful, as even if a layer is scratched away, another will form if the passivating metal permeates the alloy

ii)

Sacrificial anodes set up a galvanic cell in which a more active metal oxidises, forcing the Iron to reduce (thus converting rust back into Iron). Draw an example if possible, and the relevant chemical equations. Explain problems (need to replace etc).

d)

This is a brilliant question; just go through the standard steps. Usually, these are:
Bring the cannon to the surface. Keep submerged in sea water whilst traveling to facility to ensure salt doesn't crystalise.

Leech salts using 2% NaOH until the concentration is low enough. Replace solution, repeat

Can set up an electrolytic cell to reduce the Iron. Use a graphite mesh etc. keeping Hydrogen bubbling to a minimum

Apply a lacquer, and keep in low light conditions

Not too bad of a section! 

[edmododragon]

Surprisingly I found it very easy

[RuiAce]

Surprisingly I found it very easy

I still haven't even seen the paper yet haha you're in before Jake put it up

[edmododragon]

There was a big bunch of hard calculations tho. I feel like those who aren't as good at math would have struggled

[Wasaabi]

I don't do any maths so I really struggled with those calculations, though I thought the long response questions were pretty ok.
Industrial I found really difficult, but multiple choice was fairly simple except for the last one

[RuiAce]

There was a big bunch of hard calculations tho. I feel like those who aren't as good at math would have struggled

Someone was telling me about seashells...?

[edmododragon]

Someone was telling me about seashells...?

The seashells question was one of the most beautiful Chem calculations I've ever done. Back titration, and then percentage mass

[RuiAce]

The seashells question was one of the most beautiful Chem calculations I've ever done. Back titration, and then percentage mass

Oh boy.

The instant you said percentage mass I was like yep they need to be decently good at maths.

Easy, maybe, but just as easily overlooked.

[bohimei]

Boy I'm finally done

[bohimei]

20 should be A right? the wavelengths need to match?

[thomasdo1]

Question 20 is A because elements absorb different wavelengths differently hence 2 of the same wavelengths must be compared to eachother, once initially and once at the detector

[Mal08]

20 is A

In AAS you compare the intensity of the light of the SAME wavelength after is passes through the sample

[RuiAce]

I actually thought it was A as well. I just wasn't sure as to whether or not the intensity or something would change so when someone told me B I was like ok that'll do until corrected.

[RuiAce]

20 should be A right? the wavelengths need to match?

Question 20 is A because elements absorb different wavelengths differently hence 2 of the same wavelengths must be compared to eachother, once initially and once at the detector

20 is A

In AAS you compare the intensity of the light of the SAME wavelength after is passes through the sample

It's A until you use physics apparently and consider my suspicions.

[bohimei]

So whats the verdict? A or B?