HSC Chemistry Question Thread

[bsdfjnlkasn]

Hey there, was wondering if someone could please explain the following multiple choice questions?

I thought the answer for Q2 was D as chlorine is more electronegative than bromine so get's the lowest number (i.e. is the beginning of the carbon chain)

With 9, could I possibly get an explanation, I just chose D by a process of elimination since I knew the species had to be the same. Isn't PH₃ acting as a base in the forward reaction? Or does conjugate acid/base pair imply the backwards reaction?

For Q16, I got 18.2mL...

Any help would be super appreciated!

[andiyousif1]

yo can someone help me out with this question

[winstondarmawan]

What is the purpose of a pipette in titration?
It was listed as one of the equipment but I don't see how it is necessary..
TIA

[Fahim486]

Hi so I am looking back at Production of Materials and kinda forgot the difference between Hexene and Hex-2-ene and like what the 2 indicates. Also another question, is bromine water used to distinguish between alkanes and alkenes e.g. propane and propene?

Thanks!

[MisterNeo]

Hi so I am looking back at Production of Materials and kinda forgot the difference between Hexene and Hex-2-ene and like what the 2 indicates. Also another question, is bromine water used to distinguish between alkanes and alkenes e.g. propane and propene?

Thanks!

Hi!
The numbers on hexene represent which carbon the double bond is on.
Hex-1-ene is the same as 1-Hexene, and Hex-2-ene is the same as 2-Hexene.
Bromine water is used to distinguish between alkanes and alkenes. The practical experiments usually use cyclohexane and cyclohexane, but any alkane and alkene can be used because something like hexene will still open its double bond to react with bromine.

[kdawgs]

For the dot point "identify that ethylene, because of the high reactivity of it's double bond, is readily transformed into many useful products" if you were to answer a question on this in an HSC exam, do you have to include information regarding the properties of alkanes and alkenes?

I only ask because I've been looking over chemistry study notes and for most of the notes under this dot point, people have included large sums of information relating to the physical and chemical properties of alkanes and alkenes.

[MisterNeo]

For the dot point "identify that ethylene, because of the high reactivity of it's double bond, is readily transformed into many useful products" if you were to answer a question on this in an HSC exam, do you have to include information regarding the properties of alkanes and alkenes?

I only ask because I've been looking over chemistry study notes and for most of the notes under this dot point, people have included large sums of information relating to the physical and chemical properties of alkanes and alkenes.

For this dotpoint, you only need to identify that ethylene is more reactive than its alkane counterpart - ethane, because of the high electron density in the double bond that allows ethylene to undergo addition reactions. You should include the fact that alkenes are unsaturated, and alkanes are saturated (maximum hydrogen bonded). These are the only two properties you would need to explain the reactivity comparison between alkanes and alkenes.
After that, they may ask you to identify properties of products like vinyl chloride and styrene that are formed by ethylene.

[kdawgs]

For production of LDPE and HDPE, is it important to include the Ziegler-Natta process for HDPE to score a band six if this question comes up? Or would we still get top marks for discussing initiation, propagation, termination and just mentioning the different conditions required for each?

[anotherworld2b]

Hi I was wondering if someone could check my answer for part a in this question please
The student prepared 1.00L of a 0.0248mol.L-1 Na2CO3 solution. He titrated three 25mL aliquots of this solution against the HCL and found a average titre of 24.35mL.
a) Calc concentration of the standardized HCL solution

Na2CO3 + 2HCl -> 2NaCl + H20 + CO2
n(Na2CO3) in 1 litre = 1 x 0.0248 = 0.0248 moles
n(HCl) = 2 x 0.0248 moles = 0.0496 moles
c(HCl) = 0.0496 / 24.35 = 0.0020 mol -1

I was hoping to get help get some help with part b please. I am not how to start

The antacid suspensions were thoroughly shaken and 20mL of each transferred to separate 250mL volumetric flasks. Both were made up to the mark with distilled water and shaken vigorously. 10mL aliquots of the distilled suspension were transferred to conical flasks for titration and an appropriate indicator added.

The tire values obtained for the Al(OH)3 suspension.
 
Average Titre volume HCL mL
Trials
1 - 21.94
2 - 22.62
3 - 21.98
4 - 21.94
Average titre - 21.90

b) calculate the concentration, Mol.L-1 of Al(OH)3 in the original Al(OH)3 suspension

[MisterNeo]

Hi I was wondering if someone could check my answer for part a in this question please

Spoiler

The student prepared 1.00L of a 0.0248mol.L-1 Na2CO3 solution. He titrated three 25mL aliquots of this solution against the HCL and found a average titre of 24.35mL.
a) Calc concentration of the standardized HCL solution

Na2CO3 + 2HCl -> 2NaCl + H20 + CO2
n(Na2CO3) in 1 litre = 1 x 0.0248 = 0.0248 moles
n(HCl) = 2 x 0.0248 moles = 0.0496 moles
c(HCl) = 0.0496 / 24.35 = 0.0020 mol -1

I was hoping to get help get some help with part b please. I am not how to start

The antacid suspensions were thoroughly shaken and 20mL of each transferred to separate 250mL volumetric flasks. Both were made up to the mark with distilled water and shaken vigorously. 10mL aliquots of the distilled suspension were transferred to conical flasks for titration and an appropriate indicator added.

The tire values obtained for the Al(OH)3 suspension.
 
Average Titre volume HCL mL
Trials
1 - 21.94
2 - 22.62
3 - 21.98
4 - 21.94
Average titre - 21.90

b) calculate the concentration, Mol.L-1 of Al(OH)3 in the original Al(OH)3 suspension

For Part A, the volume is supposed to be in litres (0.02435L) so your answer is 1000x too small.
For Part B, you exclude Titre-2 because it's an outlier and you calculate a new average using the 3 consistent results (21.95).

The vigorous shaking is due to aluminium hydroxide being only very slightly soluble.
So we work off the 21.95mL average for HCl used and the concentration obtained in Part A.
-Moles of HCl used is 2.037x0.02195=0.045mol.
-Moles of aluminium hydroxide=0.015mol.
The amount of each aluminium sample is 10mL so we divide 0.015 by 0.01L which is 1.5mol/L in the diluted sample.
The diluted sample is 250mL so we divide the 1.5 by 4 to get 0.375 moles.
These moles were from 20mL of the original sample, so divide 0.375 by 0.02L to get 18.75mol/L.
Annnd that's your original volume.
I think I did that right, do you have answers? 
Hope this helps

[anotherworld2b]

ah so
part a) would be
c(HCl) = 0.0496 / 0.02435 = 2.0369 mol L-1

I get it now thanks for your help
I was also hoping to get some help with these last parts for the question please because I don't have any answers available

f) Mg(OH)2 diluted suspension, found the mass of Mg(OH)2 in the 250mL diluted suspension to be 1.13g.Determine the concentration of Mg(OH)2 in the original undiluted suspension and express your answer in mol.L-1

g) Which of the preparations would be more effective (neutralise more HCl) for a given volume

For Part A, the volume is supposed to be in litres (0.02435L) so your answer is 1000x too small.
For Part B, you exclude Titre-2 because it's an outlier and you calculate a new average using the 3 consistent results (21.95).

The vigorous shaking is due to aluminium hydroxide being only very slightly soluble.
So we work off the 21.95mL average for HCl used and the concentration obtained in Part A.
-Moles of HCl used is 2.037x0.02195=0.045mol.
-Moles of aluminium hydroxide=0.015mol.
The amount of each aluminium sample is 10mL so we divide 0.015 by 0.01L which is 1.5mol/L in the diluted sample.
The diluted sample is 250mL so we divide the 1.5 by 4 to get 0.375 moles.
These moles were from 20mL of the original sample, so divide 0.375 by 0.02L to get 18.75mol/L.
Annnd that's your original volume.
I think I did that right, do you have answers? 
Hope this helps

[phoebegresham]

I am going through the Jakes Chem notes that I purchased at one of the lectures. For Equilibria, it says that for endothermic reactions, an increase in temperature will cause the equilibrium to shift to the right, which I totally understand. However in the next syllabus point, it says that the CO2(g) -><- CO2(aq) is endothermic, however that an increase in temperature would cause a shift to the left. Isn't this contradicting the earlier statement? Thanks

[jakesilove]

I am going through the Jakes Chem notes that I purchased at one of the lectures. For Equilibria, it says that for endothermic reactions, an increase in temperature will cause the equilibrium to shift to the right, which I totally understand. However in the next syllabus point, it says that the CO2(g) -><- CO2(aq) is endothermic, however that an increase in temperature would cause a shift to the left. Isn't this contradicting the earlier statement? Thanks

Hey! You're 100% right, there is a typo there in some of the notes (thought we had sorted it out, sorry about that). The reaction, when shown in that order, is EXOTHERMIC. However, if you put the reaction the other way (ie. aqueous turning into gas) then the reaction is ENDOTHERMIC. Thus, in that case, when you increase the temperature the aqueous carbon dioxide turns into gaseous carbon dioxide, and escapes the solution. In that order, the reaction moves to the right. The issue here is the ambiguity of the whole 'moving right' vs 'moving left' thing, given that equilibrium reactions can technically be written either way. Anyway, I'm sorry for the mistake (totally my fault) but glad that you picked it up and understand it well enough to have come to your own conclusion.

The easiest way to think about this is that when you leave a fizzy coke in the sun, open, the soft drink will go flat quicker than if you had left the same coke in the fridge

[phoebegresham]

Thanks heaps Jake, no worries

[Fahim486]

Hi, I'm confused as to how you would approach this question on addition polymers. Thanks!