HSC Chemistry Question Thread

[bsdfjnlkasn]

Hey could someone please explain to me why the answer for this question is D. I got B and feel really confused... Thank you!

[beau77bro]

Hey could someone please explain to me why the answer for this question is D. I got B and feel really confused... Thank you!

hey bsdfjnlkasn you need to remove the 3rd measurement as it is an outlier, then take the average of 4 left over measurements.

[bsdfjnlkasn]

how should u move the solution being tested in the conical flask? like how do u fill it accurately? like will all conical flasks have measurements all down the side?

You use the pipette to fill the conical flask (same thing as the volumetric flask). These flasks don't have measurements down the side and don't actually need them since we are accurately drawing up solution with pipettes and transferring them this way.

Here's a useful video that will help you visualise the titration a bit better :
https://www.youtube.com/watch?v=sFpFCPTDv2w

[beau77bro]

You use the pipette to fill the conical flask (same thing as the volumetric flask). These flasks don't have measurements down the side and don't actually need them since we are accurately drawing up solution with pipettes and transferring them this way.

Here's a useful video that will help you visualise the titration a bit better :
https://www.youtube.com/watch?v=sFpFCPTDv2w

our schools have very bad pipettes to put it lightly and im attrocious with them. idk how to deal with that fact, how do u even get better at that omg

theres a second question added above btw:
'also how do we deal with diprotic acids when testing?? do we assume they go to completion, how would we assume they didnt? and what indicator? like i really have no clue what to do with that. so an explanation/ steps would be great (with calculations)'

[bsdfjnlkasn]

our schools have very bad pipettes to put it lightly and im attrocious with them. idk how to deal with that fact, how do u even get better at that omg

See if you have the opportunity to practice with them before your exam - I find them tricky too!

[bsdfjnlkasn]

Hi there,

Could someone please give me a guide as to where the 4 marks in this question would be awarded. It's from the 2004 HSC paper and I can't seem to find proper solutions anywhere.

 I've identified the outlier in trial 4 and stated that this led to invalid results as unreliable results were used. Further, the calculated result for [Hg²⁺] would end up being lower since the provided average is lower. I mentioned that regardless of the experiment being repeated, the results weren't consistent and so the overall experiment was made unreliable. Since the experiment wasn't reliable, we can't completely engage the aim which means that our validity is also compromised.

Is there anything else that I should add? How do I make the link between validity and reliability clearer? Have I missed any important information out?

Thank you, any help would be super appreciated

P.S: Here are the guidelines

The better candidates were able to clearly state the relevant points to support their judgement about both validity and reliability. There were a large number of responses that linked both terms together and showed no clear understanding of the difference between them. Most responses identified the outlier but stated incorrectly that including it in the average made the result more valid or more reliable or more accurate.

[beau77bro]

does this seem solid, am i missing any important tricks - appreciate any advice from students and moderators (Especially)

forgot the requirements for the standard - but im not sure it's in the prac so im not gonna focus too much yet

[jakesilove]

All of the above (4 marker, prac write up) looks good. Unfortunately, guys, I've got exams at the moment as well, so probably won't be able to go in depth on your work tonight. That being said, I seriously do think you've covered everything you need to, and wouldn't have added anything else. To all the users posting quite a lot tonight; that's seriously great to see. Even if I am a bit less active, why don't you try helping each other out? I note that this has already occurred a little bit, but I'm sure you all have different perspectives which will be found very useful here. In any case, great job all! I can see you've come very far in the past few days

[beau77bro]

hey so my teacher hasnt exactly told our class anything for any of our assessments minus just like basically first two mods.

which leads to my question - what exactly are processing skills? and what would a processing skills assessment for chem consist of?

[Alalamc]

Hi Guys so I tried doing this question however is it wrong because I treated X2 as X?
Question: Consider the reaction X2(g)+2Y(g) ↔ 2XY(g). 4.0 moles of X2 are mixed with 6.0 mole of Y in a 1.0-L container. When equilibrium is achieved at 30°C, it is found that 4.0 moles of XY are present. What is the equilibrium constant for this reaction at this temperature?
2 Mols XY / 2 mols of Y= 4 Mols XY/ x
x= 2 Mols of Y present (6-2=4 Mols of Y reacted)

2 Mols Y/ 1 Mols X2= 4 Mols/x
x= 2 Mols X2 reacted(4-2=2 Mols present at Eqm)

At eqm in 1 L
[X2]= 4-2= 2M
[Y]= 6-4= 2M
[XY]= 0+4= 4M

K= [XY] / [Y] [X2]
K= [4]^(2) / [2]^(2) times [2]
K = 2

[bluecookie]

Hey uhh, can I've some help with electrolytic cells?

Okay so, in a galvanic cell the chemical reactions between the anode and cathode electrodes with the solutions produce a current which causes oxidation or reduction to occur. But in electrolytic cells there's no anode and cathode, because in the diagram in my book they're just made of carbon, and there for the purposes of conducting electricity. So to my understanding in galvanic cells the negative/positive charges of the electrodes interaction with the negative/positive solutions generate the current and reaction they undergo at each electrode. But what happens in an electrolytic cell where there's no negative/positive electrodes?

Also, in a galvanic cell there is a solid form of both electrodes as well as a liquid form. So as the reduction/oxidation reactions occur, one of the element's electrolytes gains electrons and forms a solid on the solid electrode of itself? Whilst the other electrode loses electrons and dissolves into a liquid into it's own electrolyte solution? I'm confused because in an electrolytic cell there's no solid form of either of the elements used - they're just aqueous ions in a solution. So how does oxidation occur? When there's no solid electrode to give of an electron in order to become a liquid? I'm doing this question in my textbook (which doesn't have answers for this section, probs cause they thought it was too basic lmao) where its asking the reactions that occur at the electrodes and give equations (for an electrolytic cell of MgCl) and I'm confused cause since there's no solid electrode, wouldn't it just be reduction reactions occuring? Because the aqueous ions don't have another electron to give, so the most they can do is gain electrons from the battery at either electrode and become solids? So yeahhh, confused about that :/ And how/where the oxidation reaction occurs in an electrolytic cell.

[johnk21]

Can i please have help with this question??

[beau77bro]

hey guys. what are like important prac techniques, not procedures but techniques when carrying out parts of the practical - any and all appreciated, especially for the first two mod pracs.  what i mean by this is, like reading the meniscus from the bottom at eye level, and placing the solid to be dissolved for a standard in a solution then pouring that into a volumetric flask and rinsing everything used to transfer it then filling to the meniscus inside (again idek if that's right)

ALL PRAC TIPS WANTED -will add them all to this post in edit for reference for other users

- the water goes into the BOTTOM of a condenser and out the TOP.

[johnk21]

The addition of a catalyst as no effect on the yield as it only increases the rate of reaction and the time it takes to reach equilibrium.
An increase in pressure will shift to the right and increase the yield of N2O4. This is because an increase in pressure will shift towards the side with fewer moles (2:1).
The equilibrium forward reaction is exothermic because new bonds are formed from the two NO2 into one N2O4. I'm not familiar with the K-thing but I assume that's important.
 

Thanks so much

[beau77bro]

what is a solid procedure for carrying out the decarbonation practical - im finding it hard to discern between just letting the can sit on the bench and coool - or using a warm bath - in which case how do i warm the water (school doesnt exactly have a good lab facility so a hot plate is very unlikely - pretty much why we havent done any computer based pracs or even physics cathode ray ones)