Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed) Watch

Poll: How pumped up are you for this exam?-(warning)-(bad jokes arene this poll!)
"Titanium-I'm not going to corrode (even at high temperatures)" (A*) (22)
16.67%
"Benzene's my middle name, give me the paper in a week and I'll ace it!" (A) (27)
20.45%
"Yeah, I'm fairly electrophillic (positively charged) about the exam" (B) (27)
20.45%
"I'm in the middle of the salt bridge, but I will pass-eventually" (C) (21)
15.91%
"I'm feeling rather electroNegative about this exam" (D) (18)
13.64%
"Benzene, what's that?" (E) (6)
4.55%
"Chemistry, what's that?" (F) (11)
8.33%
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JoshL123
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#1141
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#1141
Hi guys, just some quick question. For a mass spectrum. if we were to put down/ give a fragment, would it be incorrect to represent say C02H+ as CH02+ ? Hope that makes sense. If so why? Thanks
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Knoyle quiah
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#1142
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(Original post by James A)
You will never get a question which asks you to chose a starting point on the 6 carbon atoms on the benzene ring.

However when naming a compound that involves a benzene ring only and groups attached, you always start with the group on top as position 1.

However in your question, they would accept all answers, doesn't matter where you start tbh. Check the ms.
But the top position could be any group, could rotate ring until different top position, im sure there is a method for choosing the start number?
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phaseshift
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#1143
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(Original post by Knoyle quiah)
But the top position could be any group, could rotate ring until different top position, im sure there is a method for choosing the start number?
My understanding was that, in this case, where there is an -OH group attached to the benzene ring, that becomes position 1.
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SKK94
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(Original post by LeaX)
Hey, could someone please just check my thought process is correct.

When a redox reaction produces a solid, the equilibrium moves to the right (so the constant increases) so the value for the cell's emf will increase? Which explains why, despite a negative Ecell, it could still occur?
That's right, though, I'm not sure whether it is the equilibrium constant or the reaction coefficient that is affected by the equilibrium shift. (I thought the equilibrium constant is only affected by temperature?)

And I think it's suitable to say the cell's emf increases and becomes positive?
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SKK94
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(Original post by James A)
My gut feeling is A. Could be wrong though.....
Haha, I though it was A too.
Right one is B

(Original post by GeorgeL3)
I think the answer is B (for the parent ion peak at least) because the RAM of bromine is 80, this however is an average resulting from the mass of the two most common stable isotopes of Bromine; one being Br-79, and the other being Br-81.
Br-80 is much less common and only has a half life of about 18 minutes so it won't appear in the mass spectrum.
Therefore there will be one peak for the bromoethane with Br-79 in it (29 + 79 = 108) and one peak for the bromoethane with Br-81 in it (29 + 81 = 110).
Woah! Thanks!

In normal Mr calculations, we are supposed to use 80/79.9 for bromine, unless specified otherwise right?
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LeaX
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(Original post by SKK94)
That's right, though, I'm not sure whether it is the equilibrium constant or the reaction coefficient that is affected by the equilibrium shift. (I thought the equilibrium constant is only affected by temperature?)

And I think it's suitable to say the cell's emf increases and becomes positive?
Thank you.
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GeorgeL3
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(Original post by SKK94)
Haha, I though it was A too.
Right one is B



Woah! Thanks!

In normal Mr calculations, we are supposed to use 80/79.9 for bromine, unless specified otherwise right?
Yeah if it was like a mole calculation or something then you just use 80. The 79 and 81 are just for the mass spec and RAM stuff.
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GeorgeL3
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(Original post by Knoyle quiah)
But the top position could be any group, could rotate ring until different top position, im sure there is a method for choosing the start number?
I think they actually use a different naming system that we don't learn about using:
'ortha' for adjacent substituents (eg 1,2)
'meta' for being next door but one (eg 1,3)
'para' for opposite substituents (eg 1,4)

Luckily we don't have to learn this though and hopefully they won't mind what number we use.
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Knoyle quiah
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can someone explain to me clearly the method for solvent extraction? dont understand it fully from reading from books
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Knoyle quiah
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I know the answer is D because it has no hydrogens to form HCL, but how is C possible? can someone show the equations of the other answers please
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senz72
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(Original post by posthumus)
x
Hi mate, I was wondering how is it all going and what are your plans for chemistry. What are you aiming for? What do you need to get the grade? If you did unit 1, how did it go?
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itsjordan
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#1152
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Can't find a unit 4 thread but does anyone know how to work out question 21(a)(ii) in edexcel unit 4 January 2012 paper? It's about orders of reaction and initial rate... The mark scheme doesn't explain how to get the answer.


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GeorgeL3
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(Original post by Knoyle quiah)
I know the answer is D because it has no hydrogens to form HCL, but how is C possible? can someone show the equations of the other answers please
The reaction for C is just the same as normal except it has a methyl group in place of a Hydrogen. It would be like this:
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GeorgeL3
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(Original post by itsjordan)
Can't find a unit 4 thread but does anyone know how to work out question 21(a)(ii) in edexcel unit 4 January 2012 paper? It's about orders of reaction and initial rate... The mark scheme doesn't explain how to get the answer.


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I haven't done unit 4 since the Jan exam but I'll give it a go.
You work out in a)i) that k = 1.03 x 10-4
Then since you are given the rate equation:
rate = k[CH3CH2Br][OH-]
And you know that the concentration of both reactants is 0.02, you can just sub everything in to find out the initial rate:
rate = (1.03 x 10-4) x (0.02) x (0.02) = 4.1 x 10-8 moldm−3s−1
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Knoyle quiah
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(Original post by GeorgeL3)
The reaction for C is just the same as normal except it has a methyl group in place of a Hydrogen. It would be like this:
Ahh thanks , didnt know there could be a methyl group instead of a H, but makes sense cheers
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itsjordan
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(Original post by GeorgeL3)
I haven't done unit 4 since the Jan exam but I'll give it a go.
You work out in a)i) that k = 1.03 x 10-4
Then since you are given the rate equation:
rate = k[CH3CH2Br][OH-]
And you know that the concentration of both reactants is 0.02, you can just sub everything in to find out the initial rate:
rate = (1.03 x 10-4) x (0.02) x (0.02) = 4.1 x 10-8 moldm−3s−1
Ohhhhh I get it, thank you!


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JoshL123
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#1157
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Hi everyone! Just wondering if you had tips in regard to timing for this paper. Its definitely my biggest problem!
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JRP95
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(Original post by JoshL123)
Hi everyone! Just wondering if you had tips in regard to timing for this paper. Its definitely my biggest problem!
Yeah dont worry about the time and you'll get the 40 odd ums you need for an A, sorted
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Knoyle quiah
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Can someone explain how to work out the answer for this? answer is D but how, i thought sodium carbonate gets rid of acid impurities, and i havnt come across sodium sulphate as a drying agent
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phaseshift
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(Original post by Knoyle quiah)
Can someone explain how to work out the answer for this? answer is D but how, i thought sodium carbonate gets rid of acid impurities, and i havnt come across sodium sulphate as a drying agent
It's anhydrous sodium sulphate as it says in the question stem.
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