HSC Chemistry Question Thread

[bananna]

So for this question, we're trying to convert heat of combustion from kJ/g to kJ/mol. Molar mass is in g/mol. Looking at the units, if we times heat of combustion in kJ/g by molar mass in mol/g, we'll get kJ/mol, which is what we want!

Does that make sense? So to find heat of combustion in kJ/mol, you would multiply the values in the table by their respective molar masses

thank you!

[bananna]

hi!

can someone help me solve this q from hsc 2012

thank you!

[kiwiberry]

hi!

can someone help me solve this q from hsc 2012

thank you!

13) First of all, remember that the oxidising agent does the oxidising, so is being reduced. This means that the oxidising agent has to be on the reactants side, so this gets rid of B. Also note that a substance is being reduced when its oxidation state reduces.
Let's look at the oxidation states of the others
A) H⁺: +1 at first, and stays at +1 in water
C) Fe²⁺: +2 at first, +3 after. Iron is being oxidised, so the answer isn't c)
D) Cr₂O₇^2-: Looking at Cr, +6 at first (let me know if you don't understand how I got this), +3 after! So it has been reduced, therefore it is the oxidising agent and the answer is D

14) So the oxidation and reduction eqns are:
Oxidation: 6Fe²⁺ <--> 6Fe³⁺+6e^-
E_oxidation= -0.77 V (remember to change the sign for oxidation when using the values on the data sheet)
Reduction: Cr₂O₇^2-+14H⁺+6e^- <--> 2Cr³⁺ + 7H₂O
E_reduction= 1.36V
Therefore E_total= 1.36 - 0.77 = 0.59 V! So the answer is A

[J.B]

Hey! This is just a stupid Chemistry thing that you'll have to memorise; Citric acid is triprotic. If you look at the structural formula, it sort of makes sense (there are three 'loose' Hydrogen). But, that doesn't matter; just memorise the point!

Ok thank you so much! and apart from
sulfuric acid - diprotic
Hydrochloric acid, acetic acid - monoprotic
Citric, Phosphoric acid - triprotic.
Are there any other ones that I will need to know?
Thank you.

[legorgo18]

Hey how would you approach this question

Calculate the volume of a 0.025 mol/L solution of KOH required to neutralise a 250 Ml of the 0.100 mol/L solution of HCl (ph of 2)

I just wrote the equation and was thinking, moles? c = n/v? degree of ionisation?

Please dont solve it for me, want to have some playtime

[RuiAce]

Hey how would you approach this question

Calculate the volume of a 0.025 mol/L solution of KOH required to neutralise a 250 Ml of the 0.100 mol/L solution of HCl (ph of 2)

I just wrote the equation and was thinking, moles? c = n/v? degree of ionisation?

Please dont solve it for me, want to have some playtime

The equation is HCl + KOH -> KCl + H2O

1 mole of HCl reacts with 1 mole of KOH, So the more ratio of HCl and KOH is 1:1, so for perfect neutralisation you need the same amount of moles of each. You need to adapt to scenarios where you compare the moles of two different substances.

(Remark: The degree of ionisation of both substances is 100% here.)

[J.B]

Hi,
I was just wondering, as 1M of HCL has a ph of 0, what does say 10M of HCL have as a pH?

[kiwiberry]

Hi,
I was just wondering, as 1M of HCL has a ph of 0, what does say 10M of HCL have as a pH?

It would have a pH of -1 in theory! The pH scale is not restricted to 0-14, it can go negative and above 14 as well

[Zainbow]

Hello 

this answer the question below is A.

Can you please explain why it's that answer? And what do they mean by 'acidified' when the potassium dichromate doesn't have H?

[RuiAce]

Hello 

this answer the question below is A.

Can you please explain why it's that answer? And what do they mean by 'acidified' when the potassium dichromate doesn't have H?

Something being 'acidified' just means that it got mixed with an acid. Hence, H⁺ ions are involved in the entire mixture, not the compound being referred to.

The relevant half-equations from your table of standard reduction potentials are thus:
SO₂^4- + 4 H⁺ + 2 e^- ⇌ SO_2(aq) + 2 H₂O_(l)
1/2 Cr₂O₇^2- + 7 H⁺ + 3 e^- ⇌ Cr³⁺ + 7/2 H₂O_(l)

I will leave the rest as your exercise. Come back if you are still stuck - the process should be standard from here.

[Zainbow]

Something being 'acidified' just means that it got mixed with an acid. Hence, H⁺ ions are involved in the entire mixture, not the compound being referred to.

The relevant half-equations from your table of standard reduction potentials are thus:
SO₂^4- + 4 H⁺ + 2 e^- ⇌ SO_2(aq) + 2 H₂O_(l)
1/2 Cr₂O₇^2- + 7 H⁺ + 3 e^- ⇌ Cr³⁺ + 7/2 H₂O_(l)

I will leave the rest as your exercise. Come back if you are still stuck - the process should be standard from here.

thank you, I think I've got it now

[anotherworld2b]

Thank  you for your help kiwiberry
I was also wondering what would happen if  you add water to a reaction?
Would it always decrease the concentration of reactants so it'll favour the side that produces more molecules/particles? Or will there be situations where nothing happens to the equilibrium position? I also wanted confirm that adding a inert gas (noble gas) to a reaction will not change the equilibrium position
For a rate vs time equilibrium graph if you add heat would the graph dramatically increase(vertical line?)

[strawberriesarekewl]

Hey guys

In HSC chemistry what are the hardest parts in production of materials, Acidic environment, Chemical management and monitoring and Industrial chemistry (option topic)

[RuiAce]

Hey guys

In HSC chemistry what are the hardest parts in production of materials, Acidic environment, Chemical management and monitoring and Industrial chemistry (option topic)

Hardest is biased towards each person.

What I felt:

PoM: (roting = rote learning)
- Roting all the stuff on nuclear chemistry involving commercial radioisotopes and blah
- Roting all the elctrochemistry
- Ethanol bullshitting (although this got better)

AE:
- Whole ton of experiments to memorise

CMM:
- The entire water section. I seriously hated it.
- Ways of measuring ozone concentrations in the atmosphere
- AAS

Industrial chem:
Tbh, the whole thing was hard. But the calculations in this section are intense (equilibrium constant stuff) and the Solvay process is really hard to know off by heart. So are the industrially used electrolytic cells for NaOH

[strawberriesarekewl]

Cheers rui
How did you do about approaching these parts? Did you watch videos on them or do specific practice questions? (Sorry I'm on mi phone at the moment using the atarnotes app an I can't quote you)