HSC Chemistry: Suggested Answers and Discussion

[Understanding Plants Photosympathise]

http://s.quickmeme.com/img/9c/9c87cde393444a6e42fb713612172c2718347379f5ae69a7b89bc0512f986774.jpg

Bruh, this meme was made like 3 years ago...

[Aliceyyy98]

Question!

"Justify the need for research into biopolymers"

Did we have to talk about disadvantages of petroleum hence the need for biopolymers??

[gabe.ralph]

I do 4U yet I accidentally calculated 8.4 L/g in Q25 in place of g/L

[jakesilove]

Question!

"Justify the need for research into biopolymers"

Did we have to talk about disadvantages of petroleum hence the need for biopolymers??

Not necessarily! It could certainly help your argument, but it definitely wasn't required to answer the question

[jakesilove]

I do 4U yet I accidentally calculated 8.4 L/g in Q25 in place of g/L

Aha, well I guess either way you said it was toxic

[WLalex]

Is Jake doing shipwrecks?

loved this exam!!!

[RuiAce]

Question!

"Justify the need for research into biopolymers"

Did we have to talk about disadvantages of petroleum hence the need for biopolymers??

Well you could've and I'd say doing so would be good (because it allows us to compare between the two). As for whether or not explicitly that was necessary, well that depends on how BOSTES marks it, cause I'd say no as the question can be argued in a multitude of ways.

I do 4U yet I accidentally calculated 8.4 L/g in Q25 in place of g/L

Haha whoops

But then again the diff. between a 4U and a general student doing maths isn't as big as a general and a non-math student doing science calculations I reckon.

[RuiAce]

Is Jake doing shipwrecks?

loved this exam!!!

He probably will but give him time. He has a big exam tomorrow and still kindly took time out to do the short response section

[Nick-Bradbury]

Will there be any answers uploaded for shipwrecks, corrosion, and conservation option?

 

**EDIT** Sorry, just saw last answer! TAKE YOUR TIME!! WE APPRECIATE YOUR WORK!!

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!
__________________________

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. B <- Simply compute the E^o valuse for all four 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.
____________________________________________________________________________________________

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
 

[RuiAce]

Will there be any answers uploaded for shipwrecks, corrosion, and conservation option?

 

See the post above you

[WLalex]

He probably will but give him time. He has a big exam tomorrow and still kindly took time out to do the short response section

Of course!! no rush its much appreciated

[RuiAce]

Aaaaaaaaaand I see shipwrecks

[goandgoo]

Isn't Q16 A? Although B has a higher Eo potential, the cell is represented as Q | Q2+ || R2+ | R, however aluminium is 3+ so therefore it must be A

[jakesilove]

Isn't Q16 A? Although B has a higher Eo potential, the cell is represented as Q | Q2+ || R2+ | R, however aluminium is 3+ so therefore it must be A

Sounds legit to me

[RuiAce]

Isn't Q16 A? Although B has a higher Eo potential, the cell is represented as Q | Q2+ || R2+ | R, however aluminium is 3+ so therefore it must be A

Nice find. Wow.