Answers To Difficult Questions (1 Viewer)

[Martyno1]

I wrote free chlorine radical as an initiator because catalysts don't participate in reactions.
For the calculations:
Sulfur dioxide i got somewhere around 2.4L
Chlorine i put down 1.81 x 10^(4) (3s.f)
KOH i put down 6 ml

Did anyone else forget to name and draw structural formula for water

I got 3mL for the KOH
and catalysts DO participate in reactions as long as they are reformed at the end that's what makes them a catalyst, whereas an initiator isn't reformed.

 

[Martyno1]

There is no half-mark in the HSC.
Only 1, 2, 3 and so on ... whole numbers only.

they meant 1 or 2 marks not 1/2 a mark

 

[bmwz4]

lol hopefully we still get our 95+ UAI aims

hope so too!

 

[Shrikar]

hey guys, just confirming for u, 14 was NOT B as i thought, that answer actually relates to Atomic Emission Spectroscopy, not AAS - so i must conceed that it is D

ppm calculations - i got 18050 ppm chloride, but thought that was too much, so i wrote 180 ppm, but my working was correct - so 1/2marks??
The So2 q - i got 232mL i think - not good for me when i look at ur answers :mad1:

Wasn't the anwer for Question 14 C?

 

[joshuajspence]

I got 3mL for the KOH
and catalysts DO participate in reactions as long as they are reformed at the end that's what makes them a catalyst, whereas an initiator isn't reformed.

yeah its true about the catalysts... i cant remmeber where i read it. but either in esterification or hydration, the sulfuric acid actually attaches to form ethyl something or other, but then is reformed at the end

 

[bmwz4]

I got 3mL for the KOH
and catalysts DO participate in reactions as long as they are reformed at the end that's what makes them a catalyst, whereas an initiator isn't reformed.

how did u work out the KOH? sum1 plz provide working out

 

[Martyno1]

Wasn't the anwer for Question 14 C?

That's what I thought ... apparently it's not but oh well what's done is done.

 

[joshuajspence]

Wasn't the anwer for Question 14 C?

they don't use white light for AAS. they use a lamp that emits radiation of a wavelength known to be absorbed by the sample being tested for. it is D

 

[Martyno1]

how did u work out the KOH? sum1 plz provide working out

2C1V1 = C2V2 (because you needed twice as many moles of KOH, so you multiply H2SO4 by 2)
2 x (forget the conc) x (forget the volume) = 0.005 x (volume KOH)
2(conc)(volume)/0.005= 3 mL

 

[joshuajspence]

how did u work out the KOH? sum1 plz provide working out

i cant remember the numbers. however i can apply a common sense approach...

the H2SO4 was originally the same conecntration as the KOH. However it was diluted by a factor of 10x. Therefore only 1.5mL of KOH would be required (as 15mL of H2SO4 was used)

However, since sulfuric acid is diprotic you need twice as much KOH. Therefore the answer is 3mL

 

[Martyno1]

Does anyone remember how many marks that ppm SO2/S question was worth? The molar volume calculation?

 

[sarang1703]

Can anybody plz plz verify my answers???

SO2 qs, i got 2.32 litres...
pHs of H2SO4 i got as 2 and 3
KOH 3ml,
Chlorine conc. 17500 ppm

thanks a million ppl...
cheers

 

[jbai]

c₁ V₁ = c₂ V₂

(NOTE: You need pH 2 for c₂, to convert:
pH 2 = 10^-2 = 0.01)

0.1 mol L^-1 x 0.09 L = 0.01 x V₂
V₂ = 0.1 mol L^-1 x 0.09 L / 0.01 mol L^-1
V₂ = 0.9 L
V₂ = 900 mL

I don't know if I'm right ... or it should be 900 mL - 90 mL = 810 mL

Why is it 0.09L? Didn't the Q. ask "How much water to be added to dilute 10mL of the solution to get pH of [whatever]?" I got 90mL, just my 2 cents

 

[astrolio]

2C1V1 = C2V2 (because you needed twice as many moles of KOH, so you multiply H2SO4 by 2)
2 x (forget the conc) x (forget the volume) = 0.005 x (volume KOH)
2(conc)(volume)/0.005= 3 mL

i got 17.98 mL :S

 

[Martyno1]

Why is it 0.09L? Didn't the Q. ask "How much water to be added to dilute 10mL of the solution to get pH of [whatever]?" I got 90mL, just my 2 cents

No it says 90 mL. (I have the MC question in front of me)

 

[sykoticx]

Can anybody plz plz verify my answers???

SO2 qs, i got 2.32 litres...
pHs of H2SO4 i got as 2 and 3
KOH 3ml,
Chlorine conc. 17500 ppm

thanks a million ppl...
cheers

S02? I got like 1.2L or something...
pH is the same.. red then violet yeah?
KOH i got 3mL as well..
Chlorine conc... it was liek 18050ppm




and as for that dilution question..
it was (d) 900mL

 

[Martyno1]

S02? I got like 1.2L or something...
pH is the same.. red then violet yeah?
KOH i got 3mL as well..
Chlorine conc... it was liek 18050ppm




and as for that dilution question..
it was (d) 900mL

it was (c) 810mL because it says how much must be added

 

[k4t5UM0t0]

Sorry to burst the bubble but I think the dilution question was c)

It asks how much needs to be ADDED which is 810ml since we already have 90ml
Not the final volume which was 900ml.

 

[Forbidden.]

No it says 90 mL. (I have the MC question in front of me)

At least some people didn't make their papers undergo combustion.

 

[tmde1st]

Can anybody plz plz verify my answers???

SO2 qs, i got 2.32 litres...
pHs of H2SO4 i got as 2 and 3
KOH 3ml,
Chlorine conc. 17500 ppm

thanks a million ppl...
cheers

i got the same for the SO2, the H2SO4, and the KOH, and i cant remeber about the chlorine. i think i screwed that up. but now its over, so i have a couple of days to study for maths/bio sweet.