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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

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Original post by four4
When making up this years grade boundaries for f324, does OCR have to take last years grade boundaries into consideration? Can take they jump straight 8 marks and go from 42 from last year to 50 this year for an A?

Just a little curious since thats not happened in the past, the biggest difference between 2 years has been 5 marks...


No, because the whole point of an exam is to be a distinctly-unique entity from the one preceding it. The only influence last year has is in how OCR choose to write their paper, etc.

Everyone's grade is marked raw, so out of 60 for F324. If we plot everyone's results, we get a bell curve, where some people do horribly, some people do amazingly, and most do decently.

Imagine we have 100000 students who sat F324. We take the top 10000 scores (top 10%) and we find out what the lowest member of this group got. Say they got 56. That means 56 is the A* boundary. Obviously candidate 10001 may also have 56, so they'll also get an A*. Then we look at the next 10000. Let's say the lowest member gets 50 - an A become an 50. Candidates 20001-20054 may all also have 50, so they'll all get A's. We repeat this until we account for all grade boundaries and all students.

^ This is how grade boundaries are decided. No outside influences, just cold hard data. That way, we have a consistent rubric to match marks against (by taking 10% samples) and usually little variation in the percentage of people attaining a certain grade; 8.39% of students last year got an A* compared to 7.67% in 2014. Everybody is happy c:


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I've been hardcore prepping for this paper and I woke up at 6, just to realise it's tomorrow and today is maths. FML
Original post by Dinasaurus
I've been hardcore prepping for this paper and I woke up at 6, just to realise it's tomorrow and today is maths. FML


looooooooooooool.
Original post by RayMasterio
guys can anyone help me with June 2015 question 2d part i, q 4c part ii, and part d iii


2 d (i) N2O +H2 -> N2 + H2O
If you look at step 2 N2O is produced but this isn't in the overall equation so must react again in step 3 to be converted into something that is. In the overall equation there is one mole of N2 and 2 moles of H2O and yet before step 3 you have no moles of N2 and one mole of H2O. Therefore N2O must be combined with something to form N2 and H2O. Hydrogen.
Original post by rishani
Any predictions for tomorrow


Chemistry.


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Original post by GetOverHere
Imagine we have 100000 students who sat F324. We take the top 10000 scores (top 10%) and we find out what the lowest member of this group got. Say they got 56. That means 56 is the A* boundary. Obviously candidate 10001 may also have 56, so they'll also get an A*. Then we look at the next 10000. Let's say the lowest member gets 50 - an A become an 50. Candidates 20001-20054 may all also have 50, so they'll all get A's. We repeat this until we account for all grade boundaries and all students.

^ This is how grade boundaries are decided. No outside influences, just cold hard data. That way, we have a consistent rubric to match marks against (by taking 10% samples) and usually little variation in the percentage of people attaining a certain grade; 8.39% of students last year got an A* compared to 7.67% in 2014. Everybody is happy c:


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This is not correct at all.

The only boundaries which are set by examiners are the Grade A and Grade E boundaries. These are decided by a mix of statistics (similar to what you mentioned), but also a review of candidate scripts from the current year and previous years, to ensure that the standard of work required to achieve each grade stays the same each year.

Once the Grade A and Grade E boundaries are set then the rest of the boundaries are automatically generated such that there is an equal number of marks between each boundary. That is why the difference between A and A* is always equal to the difference between A and B.
Original post by rishani
Any predictions for tomorrow


hardest paper in the world
Would anyone mind helping me with question 3 (c) (ii) on the June 2012 paper? I seem to be getting the wrong number of moles for butanoic acid and I don't understand how or why the concentrations are calculated
Please someone explain the questions on ratios with buffers I don't get what the rationing actually means what does HCO3- : H2COH being 6.6:1 actually means?
(edited 7 years ago)
Yeah the mark scheme for that paper seems wrong idk why I couldn't get the moles either :/

Original post by Lucy985
Would anyone mind helping me with question 3 (c) (ii) on the June 2012 paper? I seem to be getting the wrong number of moles for butanoic acid and I don't understand how or why the concentrations are calculated
image.jpg

Could some explain why the cell potential increases when water is added?
Just figured it out - they seem to jump a step in the mark scheme -_-
They went straight to finding the excesss moles and didn't include the original moles at first

Original post by tcameron
Yeah the mark scheme for that paper seems wrong idk why I couldn't get the moles either :/


Original post by Lucy985
Would anyone mind helping me with question 3 (c) (ii) on the June 2012 paper? I seem to be getting the wrong number of moles for butanoic acid and I don't understand how or why the concentrations are calculated
Original post by GetOverHere
No, because the whole point of an exam is to be a distinctly-unique entity from the one preceding it. The only influence last year has is in how OCR choose to write their paper, etc.

Everyone's grade is marked raw, so out of 60 for F324. If we plot everyone's results, we get a bell curve, where some people do horribly, some people do amazingly, and most do decently.

Imagine we have 100000 students who sat F324. We take the top 10000 scores (top 10%) and we find out what the lowest member of this group got. Say they got 56. That means 56 is the A* boundary. Obviously candidate 10001 may also have 56, so they'll also get an A*. Then we look at the next 10000. Let's say the lowest member gets 50 - an A become an 50. Candidates 20001-20054 may all also have 50, so they'll all get A's. We repeat this until we account for all grade boundaries and all students.

^ This is how grade boundaries are decided. No outside influences, just cold hard data. That way, we have a consistent rubric to match marks against (by taking 10% samples) and usually little variation in the percentage of people attaining a certain grade; 8.39% of students last year got an A* compared to 7.67% in 2014. Everybody is happy c:


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Partly true, but also partly misinformed. Lets say one cohort of students is a lot more intelligent than a previous year, wouldn't that just skew the results and make it unfair? Grade boundaries are decided by many factors and this is one of them...
I did A2 chemistry last year, hence did F324 and F325

How did you guys find this years paper?
Original post by Lucy985
Would anyone mind helping me with question 3 (c) (ii) on the June 2012 paper? I seem to be getting the wrong number of moles for butanoic acid and I don't understand how or why the concentrations are calculated


Ok, so the butanoic acid is mixed with NaOH and an acid-base reaction would obviously occur.
CH3(CH2)2COOH + NaOH --> CH3(CH2)2COO-Na+ + H2O

You work out the number of moles of butanoic acid: n= c x v/1000
n(butanoic acid) = 0.0125mol
n(NaOH) = 0.0025mol

NaOH is the limiting reactant, so all of it is used up and forms 0.0025mol of the salt, CH3(CH2)2COO-Na+

Because butanoic acid is in excess, only some of it is used up: 0.0125 - 0.0025= 0.01mol
You then work out the new concentrations using c=n/v
c(butanoic acid)= 0.01/(100/1000) = 0.1 mol dm^-3
c(CH3(CH2)2COO-)= 0.0025/(100/1000) = 0.025 mol dm^-3

Then you plug these concentrations into the buffer equation and you get a concentration of H+ of 6.04x10^-5.
Then -log that answer and you get a pH of 4.22
Original post by abdulmilad
I did A2 chemistry last year, hence did F324 and F325

How did you guys find this years paper?


F324 easier than last year's. A few questions were quite wordy, and there were a few traps that might have caught people out. No gigantic proton NMR at the end.
Original post by AqsaMx
image.jpg

Could some explain why the cell potential increases when water is added?


Anyone??
http://us20.chatzy.com/11324377761338
testing each other for f325 tomorrow
when water is added it causes the concentration of whatever ions to decrease, this causes the equilibrium to shift to the side where the ions are, in the direction of ion that have dcreased in conc. This results in a more negative / more negative electrode potential value, greater difference between electrode potential values, larger overall se potential value
Original post by AqsaMx
Anyone??


The addition of water means that Cu2+ concentration is decreased. The equilibrium will then shift to the left to restore the balance, leading to more Cu being oxidised to Cu2+. This releases more electrons, resulting in a bigger difference in electrode potential.

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