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OCR Salters Chemistry F335 12th June 2013 Exam revision thread

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can someone explain why this paper which is mainly synoptic, is before F334? it makes no sense to me and just makes my revision structure harder to keep on top of.

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Original post by abzy1234

It's stupid isn't it? I haven't heard or read anywhere why F334 is the last exam, and like you said, it makes sense to have the synoptic paper last. It's the same case for OCR A, I don't think it's the same for the other boards.

Oh well, F334 and F335 are quite complementary, as all the organic stuff are built on; though I imagine sections on polymers, complexes, electrode potentials and biochemistry can seem a bit abstract.

Best to plough on with revision regardless :smile:

This is definitely an OCR thing, my biology final synoptic paper is also about a week before the smaller paper. Annoying but at least it gets the longer ones out of the way first!

Reply 61

abzy1234

Original post by Salmonidae

This is definitely an OCR thing, my biology final synoptic paper is also about a week before the smaller paper. Annoying but at least it gets the longer ones out of the way first!

Yah. I guess it also means extra motivation for unit 4 as well; if we do have a horrendous F335 paper, we can at least try and do better in F334 :tongue:

Reply 62

Kreayshawn

Does anyone have the mark scheme for the Jan 2013 paper? Tried to get it off my teacher today but she didn't have it handy

Reply 63

LOLAKALMAO

Does anyone have the f335 mark scheme for January 2013? Need it urgently please!

Reply 64

EatRainbows_x

Jan 2013 mark scheme?

Reply 65

TheNote

anyone know if we need to know what compounds give certain colours?

Reply 66

Salmonidae

Original post by TheNote

anyone know if we need to know what compounds give certain colours?

Yeah, we need to know some indicators e.g a manganate redox titration goes from colourless to pink.

You also need to know some transition metal ion colours detailed in the Steel Story chapter.

Fe²⁺_(aq) = Green solution
Fe(OH)₂_(s) = Green precipitate

Fe³⁺_(aq) = Orange/Brown solution
Fe(OH)₃_(s) = Rust Brown precipitate

Cu²⁺_(aq) = Pale blue solution
Cu(OH)₂_(s) = Blue precipitate

Also the way in which ammonia reacts with copper ions in solution and the colours formed. This is an example of ligand substitution.

With little ammonia:

[Cu(H₂O)₆)]²⁺ + 2NH₃ --> [Cu(H₂O)₄(OH)₂] + 2NH₄⁺
Pale Blue Solution ---> Blue Precipitate

With excess ammonia:

[Cu(H₂O)₆)]²⁺ + 4NH₃ --> [Cu(NH₃)₄]²⁺ + 6H₂O
Pale Blue Solution ---> Intense blue solution

Also to do with colour, we need to be able to detail how to carry out Colorimetry and also explain why we see certain compounds as coloured (links to Colour by Design chapter).

Hope this helps,

S

Reply 67

tsr1

do we have to learn the mass spec procedure for f334? sorry couldn't find the thread for it...

Reply 68

super121

Original post by tsr1

do we have to learn the mass spec procedure for f334? sorry couldn't find the thread for it...

Yes, and there is a thread for F334:
http://www.thestudentroom.co.uk/showthread.php?t=2307436&page=2&p=42576669&highlight=f334#post42576669

Reply 69

tsr1

Original post by super121

Yes, and there is a thread for F334:
http://www.thestudentroom.co.uk/showthread.php?t=2307436&page=2&p=42576669&highlight=f334#post42576669

Thank you

Reply 70

Kreayshawn

Can anyone run me through the calculation on 2)d)iii) and iv) on June 2012? I thought I was good with the pH calculations but I guess I've forgotten how to do some of them

Reply 71

Salmonidae

Original post by Kreayshawn

Can anyone run me through the calculation on 2)d)iii) and iv) on June 2012? I thought I was good with the pH calculations but I guess I've forgotten how to do some of them

2)d)iii)

Ok so what we have to know here is that pK_a is simply the -log of the K_a. Therefore all we have to do is find the inverse log of minus pK_a to get K_a back.

10^(-3.86) = 0.000138

0.000138 is approximately equal to 0.00014, 1.4x10^-4.

2)d)iv)

This question is simply a case of rearranging the equation for acidity constant.

We know the following:

K_a = ([H⁺] x [A^-]) / [HA]

With the numbers we are given we can substitute in to get:

1.4x10^-4 = ([H⁺] x [A^-]) / 0.1

We can also assume that all of the H⁺ is coming from the acid, and therefore for each H⁺ there will also be an A^-. Therefore we can assume both these concentrations are the same leaving us with:

1.4x10^-4 = [H⁺]² / 0.1

With a little rearranging we get:

1.4x10^-4 x 0.1 = [H⁺]²

which gives a [H⁺]² value of 0.000014.

All we need to do then is square root that number and find the minus log to get pH.

sqrt of 0.000014 = 0.00374165738

-log(0.00374165738) = 2.43.

So pH is 2.43!

Hope this helps!

Reply 72

tasniaa

how's everyone feeling about this exam?

Reply 73

thelion0

I hope the grade boundaries don't go up like it did in F332 for us last june

Reply 74

super121

Original post by thelion0

I hope the grade boundaries don't go up like it did in F332 for us last june

I think they will because they went up for F334 in January as well, suggesting we have a smart cohort? :/

Reply 75

thelion0

Original post by super121

I think they will because they went up for F334 in January as well, suggesting we have a smart cohort? :/

January was fine?

Reply 76

super121

Original post by thelion0

January was fine?

No, they went up

June 2012
A: 63
B: 56
C: 50
D: 44
E: 38

Jan 2013:
A: 71
B: 64
C: 57
D: 51
E: 45

Reply 77

thelion0

June 2011:

A - 68
B- 61
C- 54

Three marks difference compared to January 2013. F332 however as I mentioned before really went up with a difference of about 10 marks from the highest its ever been.

Reply 78

calmpeach

Hey guys, hope revision is going well for you.
I've got something that I can't get my head around and it'll be great if could help me :smile:

On p55 in new edition salters revision guide by OCR it says on the top right corner (for visible absorption spectroscopy):
absorption is more intense and the wavelength of lamda-max increases for organic molecules with large delocalised systems.

correct me if I'm wrong, but I thought as the delocalisation increases, the excitation energy of electrons would decrease?
help! :frown: