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# A2 OCR Chemistry B F334 - June 2015 Watch

1. hey guys here are some useful notes for this exam hope they help

https://496a684af17e03aa0a7c6602d8e9...%20A-Level.pdf
2. Jan 2012 f334 Q)4cii)

http://www.ocr.org.uk/qualifications...ers-h035-h435/

How are the units worked out..its come a couple of times and im not sure how they work the rate/or even k unit.
3. (Original post by suyoof123)
Jan 2012 f334 Q)4cii)

http://www.ocr.org.uk/qualifications...ers-h035-h435/

How are the units worked out..its come a couple of times and im not sure how they work the rate/or even k unit.
It took me a while but my teacher explained that you have to use the table. So for example if i want to find the order of lets say BrO3-AQ i look at that and pick two values that are different for example 0.01 to 0.02 concentration. However the other two concentrations of Br- aq and H+ aq have to stay constant so i picked the second table and third table for Br03- aq where that doubles and the other two stay constant. then the rate is 4 to 8 which means that when i double the concentration of BrO3- aq is dobule rate doubles so it is first order. Then you repeat the same thing with the other two concentration following this method.

Also for the unit you will need rate=K [A][B] but you need there orders in this case which i cant explain by writing XD You will have to square [moldm-3]^2 and re arrange the equation for K but remmeber some concentrations are second order which means it is to the power 2 and if it is first to the power 1 if it is zero order it is irrelevant. Then you use your mathematical skill to cancle units and get the right unit which differs for different equations of k
It took me a while but my teacher explained that you have to use the table. So for example if i want to find the order of lets say BrO3-AQ i look at that and pick two values that are different for example 0.01 to 0.02 concentration. However the other two concentrations of Br- aq and H+ aq have to stay constant so i picked the second table and third table for Br03- aq where that doubles and the other two stay constant. then the rate is 4 to 8 which means that when i double the concentration of BrO3- aq is dobule rate doubles so it is first order. Then you repeat the same thing with the other two concentration following this method.

Also for the unit you will need rate=K [A][B] but you need there orders in this case which i cant explain by writing XD You will have to square [moldm-3]^2 and re arrange the equation for K but remmeber some concentrations are second order which means it is to the power 2 and if it is first to the power 1 if it is zero order it is irrelevant. Then you use your mathematical skill to cancle units and get the right unit which differs for different equations of k
I understand how to find the rate from the tables by comparing rates to find if 1st/2nd order. Thanks for clearing that up.

Its the unitsI cant figure when mol dm-3 divided by 4 * mold dm-3 comes to mol-3 and dm+9.
5. I think we may get a question on DNA, codons, anti-codons ect. Haven't had that for a while.
6. (Original post by suyoof123)
I understand how to find the rate from the tables by comparing rates to find if 1st/2nd order. Thanks for clearing that up.

Its the unitsI cant figure when mol dm-3 divided by 4 * mold dm-3 comes to mol-3 and dm+9.
Basically you know the rate equation so rate=k [BrO3-aq] [Br-aq]^2 [H+aq]

So k= rate/ all of the reactants

k= 4.5x10-6moldm-3s-1/ moldm-3 x (moldm-3)^2 x moldm-3
Ssee the highlighted part is expand it mol^2dm-6
so then cancel down some parts

k=s-1/moldm-3xmol^2dm-6
k= mol-^3dm9 s-1
7. (Original post by Welbeck)
I think we may get a question on DNA, codons, anti-codons ect. Haven't had that for a while.
Yeah i hardly see a lot of thread of life questions to do with DNA so i will look over them.
8. Do we need to know how to calculate Rf values?
Basically you know the rate equation so rate=k [BrO3-aq] [Br-aq]^2 [H+aq]

So k= rate/ all of the reactants

k= 4.5x10-6moldm-3s-1/ moldm-3 x (moldm-3)^2 x moldm-3
Ssee the highlighted part is expand it mol^2dm-6
so then cancel down some parts

k=s-1/moldm-3xmol^2dm-6
k= mol-^3dm9 s-1

So if you have moldm-3 divided by lets say 2 * moldm-3 then that becomes mol-1dm3?
10. In the revision guide it says that base pairing in DNA allows for replication, why is this?

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11. (Original post by suyoof123)

So if you have moldm-3 divided by lets say 2 * moldm-3 then that becomes mol-1dm3?
if it is moldm-3 divided by moldm-3 becomes 0
If it is moldm-3 divided by (moldm-3)^2 to the power 2 then it becomes moldm-3. Power rules hope that helps.
12. How do you know whether in E cells you use a platinum electrode instead of the electrode corresponding to the metal in the solution?
13. Does anyone have a list of all the reactions we need to know?
How do you know whether in E cells you use a platinum electrode instead of the electrode corresponding to the metal in the solution?
Platinum/graphite is used for a standard hydrogen E cell
15. I have a STRONG feeling this test is gonna be very biology oriented
16. (Original post by thedontom)
I have a STRONG feeling this test is gonna be very biology oriented
I hope so! But why do you think that?

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17. (Original post by Welbeck)
Platinum/graphite is used for a standard hydrogen E cell
Thank you! So for iron you would use a iron electrode and for copper you would use a copper electrode, etc x

(Original post by thedontom)
I have a STRONG feeling this test is gonna be very biology oriented
What bio stuff do we need to know? Is that like the DNA bit, e.g. codons, anticodons, ethics, translation, transcription, etc? x
if it is moldm-3 divided by moldm-3 becomes 0
If it is moldm-3 divided by (moldm-3)^2 to the power 2 then it becomes moldm-3. Power rules hope that helps.
Wouldn't it be mol^-1dm^3 ???
What bio stuff do we need to know? Is that like the DNA bit, e.g. codons, anticodons, ethics, translation, transcription, etc? x
Yeah all that stuff... :/ I bet there's a 6 marker on translation or something stupid -.-
20. (Original post by thedontom)
Wouldn't it be mol^-1dm^3 ???
Yeah sorry my bad inverse of my answer XD silly mistake. Yeah exactly what you said

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