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# OCR Chemistry F322~ 4th June 2013~ AS Chemistry watch

• View Poll Results: How did you find the exam
Hard A= 72
42
15.22%
It was okay A= 76
99
35.87%
Easy A= 79
75
27.17%
Very Easy A=82
60
21.74%

1. (Original post by lucy3003)
I'm not sure but I think its because hexane is a component of crude oil which is a liquid?
Hmm possibly! Just a bit irritating losing a mark over something that small
2. (Original post by lucy3003)
Hey could someone possibly explain how to do june 10 question 6 (c)(i)?
You know that the number of moles is 2x10^5 and the enthalpy change is -909kJmol^-1, so you can work out the energy by multiplying them together. (Because if you think about it, kJmol^-1 is worked out from dividing energy by number of moles.) Then, because there are 4 moles of NO in the equation, just divide the answer you get by 4.
3. (Original post by needtosucceed=))
standard conditions are 25 degrees right, so if it boils 65, this temperature and above its a gas. it'll melt at -95 and anything above that. ie any value between -95 and 64 (as it boils at 65) therefore at 25degrees it would be melted and therefore a liquid. make sense?
Ohhhhhhh, yes that does make sense! I was thinking of something like that but apparently I messed up the simplest of chemistry -.-
Well thank you!
4. (Original post by Jimmy20002012)
Biology and IT

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are you ocr for biology? good luck!
5. can someone please list all the conditions needed for this exam and their relevant reactions. e.g Ni catalyst, 150 oC:- for hydrogenation of alkenes . thanks
6. (Original post by needtosucceed=))
are you ocr for biology? good luck!
Naa AQA, thanks, good luck with your exams as well

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7. (Original post by Jonma)
You know that the number of moles is 2x10^5 and the enthalpy change is -909kJmol^-1, so you can work out the energy by multiplying them together. (Because if you think about it, kJmol^-1 is worked out from dividing energy by number of moles.) Then, because there are 4 moles of NO in the equation, just divide the answer you get by 4.
Ohh ok thank you!! can I ask though, how do you know to divide by 4 at the end and not multiply by 4? I suppose I never really understood the difference between the mol you get by working stuff out and the mols of something as in stoich, coefficients?!
8. (Original post by X44)
can someone please list all the conditions needed for this exam and their relevant reactions. e.g Ni catalyst, 150 oC:- for hydrogenation of alkenes . thanks
Hydration alkenes - alcohols :- 300 degrees, 60 atm (or just high temperature and pressure), phosphoric acid catalyst
Addition polymerisation :- ziegler natta catalyst
Cracking :- heat
Elimination/dehydration of alcohols - alkenes: - acid catalyst (hot conc. sulphuric)
Esterification :- acid catalyst and reflux (conc. sulphuric)
Radical sub :- UV light for homolytic fission
Nucleophilic sub halogenoalkanes - alcohols :- Hot aq. alkali
Complete combusion :- sufficient oxygen
Oxidation 1 alcohol - aldehyde :- heat under distillation
Oxidation 1 alcohol - carboxylic acid :- heat under reflux
Oxidation 2 alcohol - ketone:- heat under reflux

I think that's it!

I might have missed some stuff out because I wrote it all quite quickly, I'll have a look at my notes now.
9. (Original post by Anonymous1717)
Ikr, makes no sense...

Suitability is only really to do with Atom Economy (Slight trick question if you will)

But if you get all the points for Atom Economy for the marks then what you have written for Percentage Yield will be "IGNORED" so no marks will be lost. But you are right, very silly question wording

So I always right everything possible for the marks unless it say "give x examples"
10. (Original post by WaseemB)
Suitability is only really to do with Atom Economy (Slight trick question if you will)

But if you get all the points for Atom Economy for the marks then what you have written for Percentage Yield will be "IGNORED" so no marks will be lost. But you are right, very silly question wording

So I always right everything possible for the marks unless it say "give x examples"
GCSE seems to suggest otherwise: http://www.bbc.co.uk/schools/gcsebit...nomyrev1.shtml
But after some research it does seem as if atom economy is more important.

The question's extremely misleading since it asks you to comment on it and then doesn't credit ANY mention of it...
11. Would something like propan-1,2,3-triol be primary or secondary?
12. (Original post by lucy3003)
Ohh ok thank you!! can I ask though, how do you know to divide by 4 at the end and not multiply by 4? I suppose I never really understood the difference between the mol you get by working stuff out and the mols of something as in stoich, coefficients?!
Well, the number of moles they give you in a question will (should) always correspond to the molar quantities in the equation. So in this case there are 4 moles of NO, therefore you should divide.
13. 4eii on May 2012?
Hydration alkenes - alcohols :- 300 degrees, 60 atm (or just high temperature and pressure), phosphoric acid catalyst
Addition polymerisation :- ziegler natta catalyst
Cracking :- heat
Elimination/dehydration of alcohols - alkenes: - acid catalyst (hot conc. sulphuric)
Esterification :- acid catalyst and reflux (conc. sulphuric)
Radical sub :- UV light for homolytic fission
Nucleophilic sub halogenoalkanes - alcohols :- Hot aq. alkali
Complete combusion :- sufficient oxygen
Oxidation 1 alcohol - aldehyde :- heat under distillation
Oxidation 1 alcohol - carboxylic acid :- heat under reflux
Oxidation 2 alcohol - ketone:- heat under reflux

I think that's it!

I might have missed some stuff out because I wrote it all quite quickly, I'll have a look at my notes now.
thanks a lot if you do find anymore please post it on here
15. (Original post by Anonymous1717)
4eii on May 2012?
the numbers become positive as they have given you the reaction values and you want formation vaules of each equation

then you just do 193-111= +82 (dont forget the + sign)
16. (Original post by .raiden.)
You can see in the equation when 4 mol of NO is made 909KJ is released. You want to find the energy released for 2.5x10^5 mol so start by finding energy released by 1 mol.
4 mol -> 909KJ
so divide both sides by 4
1mol -> 227.25KJ
Now to find energy released when 2.5x10^5 multiply both sides by 2.5x10^5.

Hope that helped
So when it says deltaH=xxx per mol on the side of a reaction, what it the delta H per mol for? How do you know which one of the reactants it is for? Or is it for each mol of each reactant?

Thanks!
17. (Original post by needtosucceed=))
if you've got 3 of them correct theres only 1 box left so just use the value you havent used yet?
Oh lol, of course. How silly of me. I thought that box required you to do a calculation
18. (Original post by Polypocus)
Hi guys, there a question on the June 2009 Paper namely, 2c (ii) where it asks us to write the state symbols for an equation.
The equation is 6C + 7H2 --> C6H14 (Hexane)
Another piece of information we were given was that Hexane melts at -95 degrees Celsius and boils at 65 degrees Celsius

Now Carbon is a solid in its standard state and hydrogen is a gas but how exactly are we supposed to know that Hexane is a liquid?
Is that just something we should know or am I missing something :s
Thanks
Generally, if it contains 5 or more carbons, it is a liquid. Less than 5, it's a gas.
19. (Original post by Gotzz)
Oh lol, of course. How silly of me. I thought that box required you to do a calculation
the 4 values in the table each go in one box the final part of the question does require a calculation however
20. Are people ready for this paper?

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