HSC Chemistry Question Thread

[iimooncabbageii]

Hi~
Having some trouble with this question

Spoiler

Apparently the answer is B. I do understand why Mn would have a lower oxidation state, but wouldn't it have lost electrons instead of gaining them?

[KT Nyunt]

Hi~
Having some trouble with this question

Spoiler

Apparently the answer is B. I do understand why Mn would have a lower oxidation state, but wouldn't it have lost electrons instead of gaining them?

Hi there! Welcome to ATAR notes. Yes the answer is B, but Mn does gain electrons.

In (MnO₄^-)... Mn has an oxidation state of +7
(this is because the O always has an oxidation state of -2, we times this by 4 so the O₄ has an oxidation state of -8. The molecule as a whole has a valency of -1 so... +7 -8 = -1)

So Mn⁷⁺ goes to Mn²⁺
This means the oxidation state is lowered (7 to 2)
And Mn has gained electrons because it needs to gain negative charges for it's positive valency to decrease.

It might help to think of it this way:
Remember: OIL RIG
Oxidation - It Loses (electrons)
Reduction - It Gains (electrons)

Lowering in oxidation state means Mn has undergone reduction, so it has gained electrons

Hope this helps

[sarra.mourad2000]

I presume you meant "Explain why alkanes and corresponding alkenes have similar PHYSICAL properties, but very different chemical properties" :

When explaining physical properties, you would talk about the intermolecular bonds i.e. the bonds BETWEEN molecules. So, we know that both alkanes and alkenes can only form weak dispersion forces with other molecules, therefore they both are insoluble in water, have a low boiling point and low melting point.

Whereas with explaining chemical properties, you would talk about the intramolecular bonds, i.e. the bonds WITHIN molecules. So, we know that alkanes have single bonds whilst alkenes have double bonds. This means that the alkenes' double is highly reactive and can readily undergo addition reaction.

I hope that help.

Thanks heaps, I find that question so odd for some reason, but that really helped

[deprezj]

Hi Jake,

When using the equation for fermentation of glucose is it enough to just say yeast for the catalyst or do you need to specifically say zymase, and in responses when they ask for the conditions for this process do you label yeast as a catalyst or a condition?

Thank you lifesaver.
Jen

[deprezj]

Hi Jake,

When using the equation for fermentation of glucose is it enough to just say yeast for the catalyst or do you need to specifically say zymase, and in responses when they ask for the conditions for this process do you label yeast as a catalyst or a condition (like temperature)?

Thank you lifesaver.
Jen

[amelia20181]

is it possible to cram the whole syllabus in 2 weeks??

[clovvy]

Hi Jake,

When using the equation for fermentation of glucose is it enough to just say yeast for the catalyst or do you need to specifically say zymase, and in responses when they ask for the conditions for this process do you label yeast as a catalyst or a condition (like temperature)?

Thank you lifesaver.
Jen

Write the equation out and describe the conditions after that....  To be fancy yes feel free to add yeast on the arrow...  However yeast is NOT a catalyst...  A catalyst does not react directly towards the reactants to produce the required products, but merely create the required environment so to speak.... 

[KT Nyunt]

is it possible to cram the whole syllabus in 2 weeks??

I did it once in two days - Although I really don't recommend doing this! This is a wayyy too stressful process that you shouldn't go through. Space yourself out as much as you can, try teaching the content to someone - this is a really fast way to make content stick. Sometimes cramming is even the way to go. It's not going to help you learn and remember in the long run, but it does mean everything will be fresh in your head by the time the exam comes around. You got this! 

[key to success]

Please please help! im incredibly stuck on this attached question.
Any help will be appreciated! 
Thanks.

[KT Nyunt]

Please please help! im incredibly stuck on this attached question.
Any help will be appreciated! 
Thanks.

Hey I think you may have forgotten to attach the question 

[Katie-E]

Hey, I was wondering whether anyone could give me a hand on these two calculations and how to do them. This question and the attached one as well please. Cheers 

6. 14 g of coal undergoes complete combustion in 58 g of oxygen gas.
What volume of gas (measured at 25 C and 100 kPa) is present when the reaction is
complete?
(A) 16 L
(B) 29 L
(C) 45 L
(D) 74 L

[clovvy]

Hey, I was wondering whether anyone could give me a hand on these two calculations and how to do them. This question and the attached one as well please. Cheers 

6. 14 g of coal undergoes complete combustion in 58 g of oxygen gas.
What volume of gas (measured at 25 C and 100 kPa) is present when the reaction is
complete?
(A) 16 L
(B) 29 L
(C) 45 L
(D) 74 L

I will guide you through the first question you ask, I will start off with reaction:

[Katie-E]

Thanks so much. I made a stupid mistake and didnt use 2 oxygens haha. do you have any idea about the screen shot one. Thanks so much btw   

[Dragomistress]

Hey, I am not quite sure how to do this. How do I know?

[Fergus6748]

Hey, I am not quite sure how to do this. How do I know?

So the Zinc electrode is oxidising so is called the reductant, you can see this as the Zinc comes off the electrode and reacts with the chlorine to make the ZnCl2 solution. To answer the question, the Platinum electrode is undergoing reduction.