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OCR A level Chemistry Unified Chemistry paper 3 pre-exam discussion

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Original post by SilverWater
My thinking is that you can only determine the oxidation state of Co from the fact that there are 3Cl's present.
As you said the question doesn't state really state anything so I guess you are forced to assume :lol:

The question is quite retarded imo.

How did you find Paper one and paper 2 btw? I need a B too but I think I have missed out :frown:


I think it was an A* qs anyways lol. I havent even done the practise set paper 1 or 2 yet. Going to try do them today :/
Original post by Nettled
I think it was an A* qs anyways lol. I havent even done the practise set paper 1 or 2 yet. Going to try do them today :/


Oh good luck lol- They are good revision imo.

How did you find this years Physical and organic?
Original post by SilverWater
Oh good luck lol- They are good revision imo.

How did you find this years Physical and organic?


I prefered the Organic over Physical even though I made a lot of silly mistakes :/
Original post by Nettled
I prefered the Organic over Physical even though I made a lot of silly mistakes :/


Yeah, I think I did alright in organic but I did really bad on physical. I just made so many mistakes on the easier stuff which is really annoying

I need a B for uni asw well and I think I'm going to have to settle for a C :frown:
Can someone help me with this pls?

Absorbance is proportional to the concentration of I2. Calculate the concentration of I2 at the start of the experiment and after 500s.

Time/s Absorbance
0 0.80
500 0.18
Original post by ReadySalted28
Can someone help me with this pls?

Absorbance is proportional to the concentration of I2. Calculate the concentration of I2 at the start of the experiment and after 500s.

Time/s Absorbance
0 0.80
500 0.18


Looking at the amount of I2 in the solution. There’s 1cm3 in a solution of 100cm3 (49.5+49.5+1). 1cm3 of 1.00mol dm3 of I2

1 mol in 1000cm3
0.001mol in 1cm3

0.001mol of I2 / 100cm3

= 0.001mol/(0.1)dm3 = 0.01
Reply 66
Original post by ReadySalted28
Can someone help me with this pls?

Absorbance is proportional to the concentration of I2. Calculate the concentration of I2 at the start of the experiment and after 500s.

Time/s Absorbance
0 0.80
500 0.18


Absorbance is proportional to concentration. Therefore at Time = 0s, your concentration must be your initial concentration of 1.00moldm-3, which is given in the Q i think. 1.00 must be the conc at time 0 because nothing has yet reacted!! So conc has not changed yet!

Use the relationship between your known absorbance and known conc, to work out the scale factor. Eg, Divide 0.80 by 1.00 which is 0.80.

Then to work out concentration at absorbance of 0.18, divide 0.18 by 0.8 (your scale factor) to give a concentration value of 0.225moldm-3

Hope that makes sense?? xx
(edited 5 years ago)
hi guys can anyone send me the set 1 paper???
Reply 68
Supposedly meant to get an A* in Chemistry, managed to in the mocks somehow but after stupid mistakes in organic you could say the free energy change for that is positive
I can't hold all this information in my brain ahaha. I reckon they'll definitely put tons of practical questions & a tricky NMR question in.
Good luck everyone though :smile:
(edited 5 years ago)
Reply 69
Original post by jeoja
Supposedly meant to get an A* in Chemistry, managed to in the mocks somehow but after stupid mistakes in organic you could say the free energy change for that is positive
I can't hold all this information in my brain ahaha. I reckon they'll definitely put tons of practical questions & a tricky NMR question in.
Good luck everyone though :smile:



You reckon NMR too? but loads came up in organic ahh I'm aiming for an A* too but after organic im not so sure
Reply 70
Have enthalpy changes of neutralisation come up, cos I always get so confused on which of the moles you divide by. Like if it is HCl and NaOH and there is a different molar ratio which of them do you divide by?
Original post by DODGY MO
does any one have the 2017 chemsitry paper A


If you mean the Unified paper then i have a physical copy of the paper. Not sure how that would help though.
Reply 72
Original post by Kiki99
You reckon NMR too? but loads came up in organic ahh I'm aiming for an A* too but after organic im not so sure


Unfortunately :frown: I was hoping we'd got it out the way in the organic but last year there were two questions on NMR and they haven't asked anything about carbon nmr. Also I doubt they'd let us get away with no benzene ring in an nmr question :'(

Yeah in the mocks I was reasonably pleased with my marks and then I just looked at the grade boundaries for A* and didn't realise how close I was ahaha - and these exams have definitely gone worse (especially in terms of stupid mistakes)! So not even really considering getting an A* anymore hahaha.
Has anyone got a leaked paper 3 chemistry?? Would be a blessing in disguise if you bless it to me xx
Original post by rregdfgdfgdf
Has anyone got a leaked paper 3 chemistry?? Would be a blessing in disguise if you bless it to me xx


If you're talking about the 2017 Unified paper, it's on the OCR website.
The specimen and practice papers are somewhere online, try physics and maths tutor

If you're talking about the 2018 paper we're all sitting tomorrow.. idk about that.
Can someone please explain why a solid A formed between Mg and CH3CH(OH)COOH has an Empirical Formula of MgC6H12O6 and not MgC6H10O6?

Surely a salt between a carboxylic acid and a metal has lost those 2 hydrogen atoms as hydrogen gas right?
Original post by ahhhhhhhh1
Can someone please explain why a solid A formed between Mg and CH3CH(OH)COOH has an Empirical Formula of MgC6H12O6 and not MgC6H10O6?

Surely a salt between a carboxylic acid and a metal has lost those 2 hydrogen atoms as hydrogen gas right?


Cabroxylic acids react with metals to form Carboxylate salts and evolve Hydrogen gas, where did you find this question? Maybe I can take a look and then help you, but from what you've told me, I have no idea.

Maybe the hint is in Empirical formula but still, doesn't make sense as even when drawing the empirical formula the molecule doesn't make sense. (unless you break the double carbon bonds within the carboxyl group and use those to bond with Mg, but I've never seen this mechanism before)
I feel so unprepared for this exam because of revising for biology and maths :frown:
Original post by nisha.sri
I feel so unprepared for this exam because of revising for biology and maths :frown:


Don't start over-thinking and panicking, just get started asap.

Considering you've already done papers 1&2, just be sure to go over the minor details such as reagents, catalysts, mechanisms briefly. Shouldn't take you more than 1 hour to remind yourself how they go.

Then, a lot of people are saying practicals are due to come up, so go on YouTube and look at the practicals and pay attention to how they're prepared. Go through the textbook looking at the practical techniques such as recrystallisation etc.. and if you can try to remember the equipment used.

Remind yourself on the identification tests such as for carbonates, sulfates etc.. colour changes.
Just try to briefly skim and/or read every page, shouldn't take long at all.

Literally just go over how to do the calculation steps, reagents, practicals and practical techniques then you're done and can relax or do some questions from previous Unified papers, optional though as long as you've done questions and know the steps in getting an answer.
Original post by Invective
Cabroxylic acids react with metals to form Carboxylate salts and evolve Hydrogen gas, where did you find this question? Maybe I can take a look and then help you, but from what you've told me, I have no idea.

Maybe the hint is in Empirical formula but still, doesn't make sense as even when drawing the empirical formula the molecule doesn't make sense. (unless you break the double carbon bonds within the carboxyl group and use those to bond with Mg, but I've never seen this mechanism before)


Only hint I could find is that it states after the salt is formed all the water is evaporated leaving solid A, so my guess is that when a salt is dehydrated the H of the carboxyl group is reformed OR ocr got it wrong, I think OCR more than likely got it wrong

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