HSC Chemistry Question Thread

[J.B]

We can rule out B and C because they both have low turbidity which is not likely due to having passed through farmland and a village. We can rule out D because of it's low total dissolved solids - since X is close to the ocean (estuarine), there would be high TDS due to all the minerals dissolved in the ocean.
This leaves out A by elimination- high turbidity and high TDS.
Although someone else's opinion on the pH value would be great- since I don't really understand why the pH is lower at A (since farmland can sometimes have fertiliser run-off that makes water more alkaline).

Thank you

I was also wondering if someone could show me how to do this? (I can do a, but not b and I don't have the answers)

a)
A Hydrochloric acid solution was standardised by titrating 10ml of it against a 0.106 mol/L sodium hydroxide solution. The average titration volume was 35.6ml. Calculate the concentration of the HCl.

b)
A sample of Magnesium Oxide was found to be contaminated with the sodium chloride. Magnesium oxide is not very soluble in water but can be dissolved in an excess of the standardisted HCl

In order to determine the purity of the magnesium oxide, 3.86g of the sample was dissolved in 500ml of Hcl and then 50ml of the resulting solution was titrated against the NaOH solution of known concentration. The average titration volume was found to be 10.4ml.

Calculate the percentage of magnesium oxide in the concentration sample.

Thanks

[toasted]

Thank you

I was also wondering if someone could show me how to do this? (I can do a, but not b and I don't have the answers)

a)
A Hydrochloric acid solution was standardised by titrating 10ml of it against a 0.106 mol/L sodium hydroxide solution. The average titration volume was 35.6ml. Calculate the concentration of the HCl.

b)
A sample of Magnesium Oxide was found to be contaminated with the sodium chloride. Magnesium oxide is not very soluble in water but can be dissolved in an excess of the standardisted HCl

In order to determine the purity of the magnesium oxide, 3.86g of the sample was dissolved in 500ml of Hcl and then 50ml of the resulting solution was titrated against the NaOH solution of known concentration. The average titration volume was found to be 10.4ml.

Calculate the percentage of magnesium oxide in the concentration sample.

Thanks

This is an incredibly hard back titration question with a lot of nuances, i could be way off but here's how i would go about it.

a)
HCl + NaOH ---> NaCl+ H2O
n(HCl)=n(NaOH)
n(NaOH)=0.0356 x 0.106
=0.0037736
n(HCl)=0.0037736
concentration(HCl)= 0.0037736/0.01\
=0.37736 mol/L
b)
always start with the titration in these questions bc we essentially have to find the moles left over and subtract it from initial to find the moles used in the reaction
NaOH+HCl--->NaCl+H2O
n(NaOH)=n(HCl)
=0.0104x0.106 (using n=CV)
=0.0011024
n(HCl) in the 500ml = n(HCl) in the 50ml x 10 (C1V1=C2V2) (kinda)
= 0.0011024 x 10
=0.011024
n(HCl) initial = 0.5 x 0.37736 (because its our standardised hcl solution) (n=CV)
=0.18868
n(HCl) used to react with MgO = 0.18868-0.011024
=0.177656
MgO + 2HCl---> MgCl2 + H2O
n(MgO) =1/2 n(HCl)
=1/2 (0.177656)
=0.088828
mass(MgO)=0.088828 x (24.31+16) (mass=nxmm)
=3.58g
% mass = 3.58/3.86 x 100
=92.76%
=93% (2 s.f.)

I'm not 100% sure this is right but i feel like this is how you would approach the question. Also, if you want some practise on these types of questions try 2016 hsc 29b)
Hope that helps! 

[J.B]

This is an incredibly hard back titration question with a lot of nuances, i could be way off but here's how i would go about it.

a)
HCl + NaOH ---> NaCl+ H2O
n(HCl)=n(NaOH)
n(NaOH)=0.0356 x 0.106
=0.0037736
n(HCl)=0.0037736
concentration(HCl)= 0.0037736/0.01\
=0.37736 mol/L
b)
always start with the titration in these questions bc we essentially have to find the moles left over and subtract it from initial to find the moles used in the reaction
NaOH+HCl--->NaCl+H2O
n(NaOH)=n(HCl)
=0.0104x0.106 (using n=CV)
=0.0011024
n(HCl) in the 500ml = n(HCl) in the 50ml x 10 (C1V1=C2V2) (kinda)
= 0.0011024 x 10
=0.011024
n(HCl) initial = 0.5 x 0.37736 (because its our standardised hcl solution) (n=CV)
=0.18868
n(HCl) used to react with MgO = 0.18868-0.011024
=0.177656
MgO + 2HCl---> MgCl2 + H2O
n(MgO) =1/2 n(HCl)
=1/2 (0.177656)
=0.088828
mass(MgO)=0.088828 x (24.31+16) (mass=nxmm)
=3.58g
% mass = 3.58/3.86 x 100
=92.76%
=93% (2 s.f.)

I'm not 100% sure this is right but i feel like this is how you would approach the question. Also, if you want some practise on these types of questions try 2016 hsc 29b)
Hope that helps! 

Thank you so so much!

[winstondarmawan]

Hey,
In regards to significant figures, I have been taught it is best to round off the the lowest significant figure that is in the question. However, in some molar calculations involving volume, would values such as 25 degrees have any bearing on the value of SF that we should be rounding off to?

Hello!
Can someone please explain the first limitation:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22883625_1443073785806709_699782289_o.jpg?oh=e373a670b3d43488ce77d26f619ea8a4&oe=59F6FCF8
TIA.

Bump :-)

Hi, can someone please help me with Q20 of the 2016 paper? I don't understand how AAS works in general... thanks!
http://educationstandards.nsw.edu.au/wps/wcm/connect/e76232cd-1d84-4d3e-92ce-ea41546810f6/2016-hsc-chemistry.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e76232cd-1d84-4d3e-92ce-ea41546810f6-lDY2cMn

Hello!
So this is a bait question because they gave you lots of unnecessary information that you do not need.
The only piece of theory that you need to know that relates to this question is that the wavelength that is emitted from the lamp must be the same that is detected, as you are measuring the ABSORPTION of the atomised sample.
So, the answer is A.

Mod Edit: Merged posts.

[RuiAce]

Hey,
In regards to significant figures, I have been taught it is best to round off the the lowest significant figure that is in the question. However, in some molar calculations involving volume, would values such as 25 degrees have any bearing on the value of SF that we should be rounding off to?

The 25 degrees does not influence your rounding.

However, the 24.79 L mol^-1 bit certainly does and you are restricted to at most 4SF as a result of it.

[kauac]

What are the differences in definitions of 'ionise' and 'dissociate'? And when it is appropriate to use each, when referring to acids and solutions?

[toasted]

Hey!

On average how many calculation questions (marks) are allocated per hsc chemistry exam?

[J.B]

What are the differences in definitions of 'ionise' and 'dissociate'? And when it is appropriate to use each, when referring to acids and solutions?

Don't quote me but I believe acids ionise in aqueous solution, whereas salts dissociate into their ions in aqueous solution?

[liya1234]

If oxidation always occurs at the anode and reduction at the cathode in both galvanic and electrolytic cells, how come the anode is negatively charged (and the cathode positively charged) in galvanic cells while in electrolytic cells the anode is positively charged (and the cathode is negatively charged)??

Hey!

On average how many calculation questions (marks) are allocated per hsc chemistry exam?

In the recent ones that I've gone through there's usually about 9/10 marks for calculations in total, not including the option (i.e. in multiple choice and core short answer only)

Mod Edit: Merged posts.

[mylinh-nguyen]

please help I don't understand this question

[roygbivmagic]

Hi, can you please explain how to do this question?
'A student prepared a standard solution of Na2CO3 for titration. The student filled the 250mL volumetric flask to the level shown (below the graduation mark) and recorded its concentration as 0.20molL-1. 25mL of this solution was transferred to a conical flask and titrated with an unknown HCl acid solution. The volume of acid delivered for neutralisation was 21.30mL. Identify the mistake the student made and explain the effect this would have had on the calculated concentration of HCl.'
I understand that the mistake was that they filled the volumetric flask to below the graduation mark, but what would the effect be?

[kauac]

Hi, can you please explain how to do this question?
'A student prepared a standard solution of Na2CO3 for titration. The student filled the 250mL volumetric flask to the level shown (below the graduation mark) and recorded its concentration as 0.20molL-1. 25mL of this solution was transferred to a conical flask and titrated with an unknown HCl acid solution. The volume of acid delivered for neutralisation was 21.30mL. Identify the mistake the student made and explain the effect this would have had on the calculated concentration of HCl.'
I understand that the mistake was that they filled the volumetric flask to below the graduation mark, but what would the effect be?

Hey... I just did a very similar question in a past paper today!

When filling a solution into the volumetric flask, the bottom of the meniscus should line up with the calibration/graduation line. Hence if the solution was filled just to below the calibration mark, then the volume would have been less than what it was recorded to be. So this would have increased the number of moles of sodium carbonate, and therefore the number of moles of HCl. So HCl would end up having a calculated concentration higher than its actual concentration.

[MisterNeo]

If oxidation always occurs at the anode and reduction at the cathode in both galvanic and electrolytic cells, how come the anode is negatively charged (and the cathode positively charged) in galvanic cells while in electrolytic cells the anode is positively charged (and the cathode is negatively charged)??

It's got to do with the flow of electrons.
In a galvanic cell, electrons want to flow from anode to cathode, which means that the cathode is positive and "attracts" the electrons.
In an electrolytic cell, electrons don't want to go from anode to cathode, which means that the cathode is negative and "repels" the electrons, hence an external power source is used to overcome this. You don't need to know why they are, just accept that it is.

please help I don't understand this question

Bump :-)

Arrhenius couldn't explain why acids/bases weren't acidic or basic in non-water solvents even though he said they release H⁺ and OH^- ions in solution. He also couldn't explain why ammonia was basic because of the role of water in the basic nature of ammonia.

[ACCC]

Hi guys,
I was doing practice paper 2015 and I got stuck on question 20. I don't know what I'm doing wrong! Any help would be great! Thank you!

[Natasha.97]

Hi guys,
I was doing practice paper 2015 and I got stuck on question 20. I don't know what I'm doing wrong! Any help would be great! Thank you!

Hi!





Hope this helps