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Edexcel A-Level Chem Paper 1 Advanced Inorganic and Physical Chemistry [Exam Chat]

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Reply 20
Original post by TypicalNerd
Because I'm in a generous mood, I have made some practice papers for you lot to use in your revision (note that I will NOT be making any more of these, because of how time consuming it is to do so - if you want more practice papers, you may have to make your own and I'd recommend using google docs if that is ever the plan).

Pages 1 - 24 on papers 1 and 2 are the question papers. Anything beyond that is the MS.
Pages 1 - 32 on paper 3 are the question paper and anything beyond that is the MS.

Disclaimer: these are NOT predicted papers or even remote attempts to guess what will actually come up. Anything that is not assessed on these practice papers could come up, so do not make these the entirety of your revision.

Thank you!!
hi what would you guys recommend for me to do to go from roughly a B in paper 2 and 3 to an A* ?
Original post by 3ASTARSINSHALLAH
hi what would you guys recommend for me to do to go from roughly a B in paper 2 and 3 to an A* ?


Here are three ideas (the third should be already pretty obvious):

-Try using quizzes on sporcle.com for inorganic ion colours (paper 3) and organic reactions (papers 2 and 3).

-Try teaching others some of the harder topics.

-Use past papers regularly.
thank you so much what tips would you give to also potentially break through the a to a* boundary for paper 1 as i feel im barely on an A
Original post by TypicalNerd
Here are three ideas (the third should be already pretty obvious):

-Try using quizzes on sporcle.com for inorganic ion colours (paper 3) and organic reactions (papers 2 and 3).

-Try teaching others some of the harder topics.

-Use past papers regularly.


specifically the calculation questions for 12 14 and some equilibrium / topic 5
Original post by 3ASTARSINSHALLAH
specifically the calculation questions for 12 14 and some equilibrium / topic 5

In my experience, it has been a matter of using ICE tables for equilibrium problems and using Red-Ox (i.e electrode potential = reduction electrode potential - oxidation electrode potential) for electrode potentials as well as regular use of past papers to familiarise myself with the methods required.

I have found that model answers were actually the way forward for a number of my classmates when I tried teaching them myself. Davies A level chemistry and Physics and Maths tutor should be the best places to look if you want to try and find some. Some used the calculations in chemistry book by Jim Clark (who also wrote ChemGuide, which has plenty of example calculations too, now that I mention it) and that worked beautifully for them.
Reply 26
those ICE tables confuse the hell out of me
Original post by TypicalNerd
In my experience, it has been a matter of using ICE tables for equilibrium problems and using Red-Ox (i.e electrode potential = reduction electrode potential - oxidation electrode potential) for electrode potentials as well as regular use of past papers to familiarise myself with the methods required.

I have found that model answers were actually the way forward for a number of my classmates when I tried teaching them myself. Davies A level chemistry and Physics and Maths tutor should be the best places to look if you want to try and find some. Some used the calculations in chemistry book by Jim Clark (who also wrote ChemGuide, which has plenty of example calculations too, now that I mention it) and that worked beautifully for them.


thanks so much bro youre a legend !
Anyone got a idea where grade boundaries will be this year, hopefully slightly lower than 2019 if the papers are roughly the same difficulty?
Anyone have the locked specimen papers?
Original post by student_1999
Anyone have the locked specimen papers?


if you get them please send them to me as well.
does anyone know if we have to know ALL of these? some of them like Zn, Ni thought we didn’t have to know😭
3053E1BA-3ED3-4608-84A2-445A47743F35.jpg.jpeg
Original post by hhhhheeeee
does anyone know if we have to know ALL of these? some of them like Zn, Ni thought we didn’t have to know😭
3053E1BA-3ED3-4608-84A2-445A47743F35.jpg.jpeg

Silver, cobalt and nickel are the only ones on there that I don’t believe are actually required. Maybe an old spec resource?
Any help with this question please 😩
Original post by medic talks
Any help with this question please 😩

Lol wtf… I thought weak base calculations were undergrad level. Alas, here’s what you do:

Consider the equilibrium constant Kh, which is the equilibrium constant for the given equation.

So Kh = [CH3COOH][OH^-]/[CH3COO^-]

A value of Kh can be found as follows:

Kh = Kw/Ka

This works because Kw = [H^+][OH^-] and Ka = [H^+][CH3COO^-]/[CH3COOH] in this case.

Using the relationship between pKa and Ka:

Ka = 10^-pKa = 10^-4.76

So Kh = 10^-14/10^-4.76 = 10^-9.24

Similarly to with weak acids, we assume that there is negligible hydrolysis of the CH3COO^- ion and that the concentration of the CH3COOH = OH^-, so we can use [CH3COO^-] = 1.00 mol dm^-3 and [OH^-] = [CH3COOH]

So Kh = [OH^-]^2/[CH3COO^-]

==> 10^-9.24 = [OH^-]^2/1.00
==> 10^-9.24 = [OH^-]^2
==> 10^-4.62 = [OH^-]

Using Kw again:

10^-14 = [H^+] x [10^-4.62]
==> 10^-9.38 = [H^+]

So pH = -log[10^-9.38] = 9.38
Original post by TypicalNerd
Lol wtf… I thought weak base calculations were undergrad level. Alas, here’s what you do:

Consider the equilibrium constant Kh, which is the equilibrium constant for the given equation.

So Kh = [CH3COOH][OH^-]/[CH3COO^-]

A value of Kh can be found as follows:

Kh = Kw/Ka

This works because Kw = [H^+][OH^-] and Ka = [H^+][CH3COO^-]/[CH3COOH] in this case.

Using the relationship between pKa and Ka:

Ka = 10^-pKa = 10^-4.76

So Kh = 10^-14/10^-4.76 = 10^-9.24

Similarly to with weak acids, we assume that there is negligible hydrolysis of the CH3COO^- ion and that the concentration of the CH3COOH = OH^-, so we can use [CH3COO^-] = 1.00 mol dm^-3 and [OH^-] = [CH3COOH]

So Kh = [OH^-]^2/[CH3COO^-]

==> 10^-9.24 = [OH^-]^2/1.00
==> 10^-9.24 = [OH^-]^2
==> 10^-4.62 = [OH^-]

Using Kw again:

10^-14 = [H^+] x [10^-4.62]
==> 10^-9.38 = [H^+]

So pH = -log[10^-9.38] = 9.38


I’ve never seen a question like this or even anything like this on the spec ?
Original post by 3ASTARSINSHALLAH
I’ve never seen a question like this or even anything like this on the spec ?

It isn’t on the spec. As I said, it’s undergraduate level material…
Original post by TypicalNerd
It isn’t on the spec. As I said, it’s undergraduate level material…

Oh ok
Reply 38
how come halides are reducing agents, i thought halogens lose an electron? sorry if this is a bit of a stupid question
Original post by kettle02
how come halides are reducing agents, i thought halogens lose an electron? sorry if this is a bit of a stupid question


Halides (e.g. Cl^-) lose electrons to form halogens (e.g Cl2) when they react with sufficiently strong oxidising agents. Thus they are reducing agents because they reduce the oxidising agent.

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