F334 (19/06/2013 9:00AM) - Salters Chemistry (Chemistry of Materials) Revision Thread

Original post by abzy1234

Thank you so much! You did so well in that paper (even got the rate of disappearance!) :smile:

I can't believe I messed up; looking back at that paper, I'm beginning to think OCR were justified in hiking the grade boundaries :redface:

Cheers again! :biggrin:

No bother, and thanks :tongue:

Can't believe I forgot the bloody units, as you can see by my angry underlining :mad:

It always seems easier in retrospect though... I still think it was a difficult paper compared to previous years. At least now you've already had a go, so you'll most likely do better :biggrin:

Reply 22

abzy1234

Original post by suncake

No bother, and thanks :tongue:

Can't believe I forgot the bloody units, as you can see by my angry underlining :mad:

It always seems easier in retrospect though... I still think it was a difficult paper compared to previous years. At least now you've already had a go, so you'll most likely do better :biggrin:

Aha, how did you work it out? I still have no clue -_-

That's so true! The question on electrode potentials was perhaps the hardest question; honestly, I think it's the exam pressure that gets to most, especially with salters chemistry! :tongue:

Yah hopefully! I think as I'm doing unit 2 and 5 now, so am literally induated with chemistry (fml), it should hopefully boost my knowledge. :biggrin:

How are you finding unit 5? :smile:

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Reply 23

suncake

Original post by abzy1234

Aha, how did you work it out? I still have no clue -_-

That's so true! The question on electrode potentials was perhaps the hardest question; honestly, I think it's the exam pressure that gets to most, especially with salters chemistry! :tongue:

Yah hopefully! I think as I'm doing unit 2 and 5 now, so am literally induated with chemistry (fml), it should hopefully boost my knowledge. :biggrin:

How are you finding unit 5? :smile:

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I looked at it again yesterday, and I'm not entirely sure myself! I think it's because there were two moles of I- in the equation, so it would be used up at double the rate... Or something...

Yeah I agree, well, it was probably alright if you understood it, but I never really learnt that topic properly in the first place :tongue:

Exam pressure's a bitch haha.

Ooh that's a lotta chemistry, fun times :mmm:

It's not too bad I guess, just loads of reaction mechanisms and equations etc. to remember! And the fact that the paper is mainly synoptic is pretty scary!

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Reply 24

abzy1234

Original post by suncake

I looked at it again yesterday, and I'm not entirely sure myself! I think it's because there were two moles of I- in the equation, so it would be used up at double the rate... Or something...

Yeah I agree, well, it was probably alright if you understood it, but I never really learnt that topic properly in the first place :tongue:

Exam pressure's a bitch haha.

Ooh that's a lotta chemistry, fun times :mmm:

It's not too bad I guess, just loads of reaction mechanisms and equations etc. to remember! And the fact that the paper is mainly synoptic is pretty scary!

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That's sorta makes sense! My teachers still have no clue -_-

Yup, especially now, as it's the final throw of the dice tbh! I see you've insured Leicester, I've also insured it too! :tongue:

I'm literally counting down the days when I don't have to do no more chemistry! Gosh that seems so far away :frown:

I am petrified for that paper, seriously! It's more or less 100% synoptic, giving them the license to ask whatever they want! Oh well, best just to prepare fully and be ready for the worst :biggrin:

Reply 25

has anyone done question 5aii on june 2010 paper and understand it? can u please help.. its a calculation question

Reply 26

Original post by tsr1

has anyone done question 5aii on june 2010 paper and understand it? can u please help.. its a calculation question

The electrode one? You just do the more positive electrode minus the more negative one, so 0.80-0.34 = 0.46

Reply 27

Kreayshawn

Original post by tsr1

has anyone done question 5aii on june 2010 paper and understand it? can u please help.. its a calculation question

We were always taught to do this as the more positive Ecell value minus the less positive value, so higher value - lower
They say the copper and silver electrodes, so we are looking at the values for the Cu and Ag half reactions
(+0.80) - (+0.34) = 0.46

Reply 28

Original post by Kreayshawn

We were always taught to do this as the more positive Ecell value minus the less positive value, so higher value - lower
They say the copper and silver electrodes, so we are looking at the values for the Cu and Ag half reactions
(+0.80) - (+0.34) = 0.46

Original post by super121

The electrode one? You just do the more positive electrode minus the more negative one, so 0.80-0.34 = 0.46

sorry... there seems to be confusion, its the one with dilution factor, I have checked the paper twice but it is june 2010 and question 5aii

Reply 29

Original post by tsr1

sorry... there seems to be confusion, its the one with dilution factor, I have checked the paper twice but it is june 2010 and question 5aii

Lol, I thought this was too easy :P
What's the question asking?

This is June 2010 http://www.ocr.org.uk/Images/65142-question-paper-unit-f334-chemistry-of-materials.pdf

(edited 10 years ago)

Reply 30

A solution of hydrogen peroxide s diluted by factor 10. 10.cm3 of this diluted H2O2 reacted with exactly 18.2cm3 of 0.0200 dm-3Mno-2 solution. The equation for the reaction is: 2 MnO4- + 6H + 5 H2O2 ---> 2Mn2+ 8H2O + 5O2... calculate the concentration of undiluted H2O2 solution....... also sorry tsr doest allow me to put spaces?!?!

Reply 31

Original post by super121

Lol, I thought this was too easy :P
What's the question asking?

This is June 2010 http://www.ocr.org.uk/Images/65142-question-paper-unit-f334-chemistry-of-materials.pdf

hey sorry.. it was JAN 2010 paper.. but the question is below:
A solution of hydrogen peroxide s diluted by factor 10. 10.cm3 of this diluted H2O2 reacted with exactly 18.2cm3 of 0.0200 dm-3Mno-2 solution. The equation for the reaction is: 2 MnO4- + 6H + 5 H2O2 ---> 2Mn2+ 8H2O + 5O2... calculate the concentration of undiluted H2O2 solution

Reply 32

Original post by tsr1

hey sorry.. it was JAN 2010 paper.. but the question is below:
A solution of hydrogen peroxide s diluted by factor 10. 10.cm3 of this diluted H2O2 reacted with exactly 18.2cm3 of 0.0200 dm-3Mno-2 solution. The equation for the reaction is: 2 MnO4- + 6H + 5 H2O2 ---> 2Mn2+ 8H2O + 5O2... calculate the concentration of undiluted H2O2 solution

This paper? http://www.ocr.org.uk/Images/65140-question-paper-unit-f334-chemistry-of-materials.pdf

Reply 33

Original post by super121

This paper? http://www.ocr.org.uk/Images/65140-question-paper-unit-f334-chemistry-of-materials.pdf

the question seems to be on none of them :s-smilie:

but I was doing it and I have it printed on me..

Reply 34

Original post by tsr1

the question seems to be on none of them :s-smilie:

but I was doing it and I have it printed on me..

The specimen paper possibly?

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Reply 35

Question_paper_2849-01.pdf194.1KB

Original post by super121

The specimen paper possibly?

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I guess its one of those legacy papers.. lol no wonder why I have 3 of 2010 papers :P attached

(edited 10 years ago)

Reply 36

CureSam

Original post by tsr1

A solution of hydrogen peroxide s diluted by factor 10. 10.cm3 of this diluted H2O2 reacted with exactly 18.2cm3 of 0.0200 dm-3Mno-2 solution. The equation for the reaction is: 2 MnO4- + 6H + 5 H2O2 ---> 2Mn2+ 8H2O + 5O2... calculate the concentration of undiluted H2O2 solution....... also sorry tsr doest allow me to put spaces?!?!

18.2cm^3=0.0182dm^3
0.0182*0.02=3.6*10^-4 moles of MnO4 (vol*conc=moles)
3.6*10^-4*(5/2)=9*10^-4 (moles of H2O2 from the equation)
That number of H2O2 is in 10cm^3, which is 0.01dm^3, giving a concentration of 0.09moles/dm^3, but the solution has been diluted by 10, meaning the original concentration 0.9mol/dm^3.

Hope this helps!

Reply 37

CureSam

Original post by suncake

https://docs.google.com/file/d/1sdHMYjo9qqZslsPAlbdN_DoQTxO0gEJapvUKPDHMPMnjkPKS3egQH2XDf5ja/edit?usp=sharing

Here you go. Yeah, ignore the incorrect scribbles then :wink:

And excuse my awful scanning skills :lol:

You must have got like 90% ums on that?

Reply 38

tasniaa

how's everyone feeling about this exam?

Reply 39

abzy1234