HSC Chemistry Question Thread

[RuiAce]

What is it about HF (and F's high electronegativity) that makes it such a weak acid?
Is it because of something that happens in solution? Why is it's degree of ionisation so low if it's looking to give away it's electrons so desperately?

Fluorine's high electronegativity is mostly prelim content. And only barely; most of the reasoning is not explained until university.

The most electronegative elements are fluorine, oxygen and nitrogen. Incidentally, they're both non-metals, take electrons in to fill their outer shell, AND have only two electron shells.

Incidentally, it is this electronegativity that MAKES hydrofluoric acid a weak acid. Whilst in general Cl, Br and I follow similar rules, the electronegativity of F distinguishes it a bit from the other halogens.

That's all you might want to know, and it still isn't a part of the HSC course anyway. They just care that you know that HCl and etc. are strong acids.

[lzxnl]

As a small insight into why HF behaves that way, think about it this way. Acid strength is dependent on the stability of the conjugate base. Now, there is one main issue with the fluoride ion. It's tiny. It has a massive core charge and an extra electron. As a result, it's going to be able to get near other positive charges, like protons.

So we have a tiny proton-seeking ion floating around in solution. How stable is that going to be? What's it going to do if it runs into a proton? It's going to be much more likely than the other halide ions to abstract a proton. The other halides are far larger, so this isn't nearly as much of a problem.

[RangaTurtle]

A quick google search yielded this link here. Check it out!

Thanks a lot for the help

[Paul.I]

Hi l was wondering if anyone could help me with this question?

"Sodium metal is heated in chlorine gas."
a. Use the standard electrode potentials to predict a reaction the could take place. (NaCl is what l have put down)
b. Write an equation to represent the overall oxidation/reduction reaction. (Here l have put the two theoretical half equations but the problem is both substances are oxidizing and therefore l cant work out what is reducing.)
c. Write the name and formula of the oxidant and the reductant in this reaction. (Again same problem with b. l just figgured being that an ionic compound is forming one isn't forcing the other to reduce or oxidize.)

[QC]

Isn't Na oxidising and Cl reducing, Na+ ions are formed which is sodium losing an electron and Cl- ions are formed which is chlorine gaining an electron i.e OILRIG. so the overall reaction would be 2Na_(s)+Cl_2(g)->2NaCl
oxidant is chlorine gas (Cl₂) and reductant is Na metal. I might just be overlooking something?

[kiwiberry]

Hi l was wondering if anyone could help me with this question?

"Sodium metal is heated in chlorine gas."
a. Use the standard electrode potentials to predict a reaction the could take place. (NaCl is what l have put down)
b. Write an equation to represent the overall oxidation/reduction reaction. (Here l have put the two theoretical half equations but the problem is both substances are oxidizing and therefore l cant work out what is reducing.)
c. Write the name and formula of the oxidant and the reductant in this reaction. (Again same problem with b. l just figgured being that an ionic compound is forming one isn't forcing the other to reduce or oxidize.)

So the reaction is 2Na + Cl₂ --> 2NaCl

Na and Cl₂ both have oxidation states of 0. In NaCl, Na⁺ has an oxidation state of +1, and Cl^- has an oxidation state of -1

We can see that the state of sodium increases from 0 to +1, so it has oxidised. Chlorine's oxidation state decreases from 0 to -1, so it has reduced.

[QC]

So the reaction is 2Na + Cl₂ --> 2NaCl

Na and Cl₂ both have oxidation states of 0. In NaCl, Na⁺ has an oxidation state of +1, and Cl^- has an oxidation state of -1

We can see that the state of sodium increases from 0 to +1, so it has oxidised. Chlorine's oxidation state decreases from 0 to -1, so it has reduced.

True, defs a better explanation then mine haha

[Paul.I]

So the reaction is 2Na + Cl₂ --> 2NaCl

Na and Cl₂ both have oxidation states of 0. In NaCl, Na⁺ has an oxidation state of +1, and Cl^- has an oxidation state of -1

We can see that the state of sodium increases from 0 to +1, so it has oxidised. Chlorine's oxidation state decreases from 0 to -1, so it has reduced.

Thank you very much

[Paul.I]

Can someone please help me with B and C. I understand A and D fine, l am just confused in which part of the solution is reducing.
For C l believe it is the Chlorine but l am not sure.

[shreya_ajoshi]

Hi
How do you do this?

The heat of combustion of 1-propanol is 2016 kJ mol-1.
What is the numerical value of the heat of combustion in kJ g–1?
The options are:
(A) 33.60
(B) 2016
(C) 3.360 × 104
(D) 1.210 × 105

Thank you

[Shadowxo]

Can someone please help me with B and C. I understand A and D fine, l am just confused in which part of the solution is reducing.
For C l believe it is the Chlorine but l am not sure.

b) Well you have Zinc nitrate, Zinc, and MnO4 with HCl, and K+ and NO3
To start with, K+ (aq) is fairly unreactive as it's down the bottom of the table - it would only react with Li(s) so it will not be appearing in this reaction, it's just there to balance out the charges. Platinum is unreactive so is ignored. NO3 is unreactive so is ignored.
As the equation \(\ce{MnO4^-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}\) is above \(\ce{Zn^2+(aq) + 2e- -> Zn(s)}\), we know that Zn(s) is oxidised to Zn²⁺, and the MnO4- along with the H+ from HCl is reduced to Mn²⁺ and H2O.
From you have your two equations, their standard reduction potentials, and which is oxidising and reducing
Does this help?

[Shadowxo]

Hi
How do you do this?

The heat of combustion of 1-propanol is 2016 kJ mol-1.
What is the numerical value of the heat of combustion in kJ g–1?
The options are:
(A) 33.60
(B) 2016
(C) 3.360 × 104
(D) 1.210 × 105

Thank you

Well first you need to find out how many grams in a mol of propanol
Propanol = C3H8O
M(Propanol) = 3*12.0 + 8*1.0 + 16.0 = 60.0gmol-1
So there are 60.0g per mol of propanol
So heat of combustion is 2016kJ per mol or 2016 kJ per 60.0g = 33.6kJ per gram

[jakesilove]

Hi
How do you do this?

The heat of combustion of 1-propanol is 2016 kJ mol-1.
What is the numerical value of the heat of combustion in kJ g–1?
The options are:
(A) 33.60
(B) 2016
(C) 3.360 × 104
(D) 1.210 × 105

Thank you

Hey! So, we have the quantity



If we want to turn this into a kJ/g, we need to multiply by moles, and divide by grams. I can tell by looking at the units! What do we know that has both grams and moles? Well, the molar mass of Propanol is



Remember, we need to MULTIPLY by moles, so let's divide our original number by the molar mass



So, our answer is A.


To be honest, that's the ONLY answer it could have been. The value for kJ/g should be LESS than the value for kJ/mol, as there are 60g in a mole. Therefore, the answer had to be less than 2016.

[Paul.I]

b) Well you have Zinc nitrate, Zinc, and MnO4 with HCl, and K+ and NO3
To start with, K+ (aq) is fairly unreactive as it's down the bottom of the table - it would only react with Li(s) so it will not be appearing in this reaction, it's just there to balance out the charges. Platinum is unreactive so is ignored. NO3 is unreactive so is ignored.
As the equation \(\ce{MnO4^-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}\) is above \(\ce{Zn^2+(aq) + 2e- -> Zn(s)}\), we know that Zn(s) is oxidised to Zn²⁺, and the MnO4- along with the H+ from HCl is reduced to Mn²⁺ and H2O.
From you have your two equations, their standard reduction potentials, and which is oxidising and reducing
Does this help?

This is much appreciated thank you

[shreya_ajoshi]

Hey! So, we have the quantity



If we want to turn this into a kJ/g, we need to multiply by moles, and divide by grams. I can tell by looking at the units! What do we know that has both grams and moles? Well, the molar mass of Propanol is



Remember, we need to MULTIPLY by moles, so let's divide our original number by the molar mass



So, our answer is A.


To be honest, that's the ONLY answer it could have been. The value for kJ/g should be LESS than the value for kJ/mol, as there are 60g in a mole. Therefore, the answer had to be less than 2016.

THANK YOUUUUU!!! Makes much more sense now