Edexcel A2 Chemistry 6ch04/05 JUNE 2015 Watch

lordoftheties
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#1021
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#1021
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Size:  98.9 KBAn explanation for the answer for part b) would be much appreciated.
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Wahid1
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(Original post by lordoftheties)
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Size:  98.9 KBAn explanation for the answer for part b) would be much appreciated.
According to these guys it square roots it *shrugs*


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bladex
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(Original post by lordoftheties)
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Size:  98.9 KBAn explanation for the answer for part b) would be much appreciated.
Is the answer D?
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HopefulDentist..
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(Original post by bladex)
You know the original concentration of aqeous ammonia is 4.0 mol dm-3 and you should have worked out that 3.84 mol dm-3 of ammonia is in the aqeous layer after they've been shaken which means 0.16 mol dm-3 must have been used to form the ammonia in trichloromethane. Sorry its quite hard to explain
ohh thanks so much!
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bladex
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(Original post by HopefulDentist..)
ohh thanks so much!
No problem! How are you feeling about kc calculations in general?
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HopefulDentist..
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(Original post by bladex)
No problem! How are you feeling about kc calculations in general?
tbh I'm okay with them most the time. I just didn't get the concept of the immiscible solvent thing- in my head there were 2 separate samples of ammonia dissolved in each of the solvents, I didn't get that there was one original ammonia solution that was then shaken in the hydrocarbon layer

you?
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simsid
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(Original post by lordoftheties)
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Size:  98.9 KBAn explanation for the answer for part b) would be much appreciated.
the answer is D.
this is because the new Kc is the original Kc value but to the power of a 1/2. this means that the original Kc value has to be square rooted to become the new value for Kc.
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rasil23
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(Original post by lordoftheties)
Name:  image.jpg
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Size:  98.9 KBAn explanation for the answer for part b) would be much appreciated.
You can test it mathematically which is what I do.
So for the original Kc, say I used 2 mol eql for HI, and 1 for H2 1 for I2
my kc value = 4

Now using the second equation and the same values, kc = 2
so answer is D
power to a half is just a fancy way of saying sq. root of the value, and since theres two of them it cancels out so its just 2/1 = 2
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bladex
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(Original post by HopefulDentist..)
tbh I'm okay with them most the time. I just didn't get the concept of the immiscible solvent thing- in my head there were 2 separate samples of ammonia dissolved in each of the solvents, I didn't get that there was one original ammonia solution that was then shaken in the hydrocarbon layer

you?
Ahh okay, and generally okay, i just get a bit confused when they fire all the information at you, in the exam ill have to stay calm and read the questions properly :/
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Nautic4l
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#1030
(Original post by Maham88)
here
Thank you
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HopefulDentist..
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(Original post by bladex)
Ahh okay, and generally okay, i just get a bit confused when they fire all the information at you, in the exam ill have to stay calm and read the questions properly :/
that's my main flaw, there hasn't been one exam where I haven't read atleast 2/3 questions wrong and lost silly marks
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Nautic4l
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#1032
(Original post by veniceswan)
I thought H2O is included in Kc calculation, are you talking about Ka
No H2O is sometimes not used when it's a 'solven't but I don't get when it acts as a solvent and when as a reactant.. ANYONE?

Also does anyone PLEASE have diagrams(labelled) of distillation and reflux apparatus?
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HopefulDentist..
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#1033
at endpoint is there an equal concentration of the salt and weak acid(or base)?
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Nautic4l
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#1034
(Original post by HopefulDentist..)
anyone know how to do 18cii on Jan 14 international paper? It's a one marker but I don't get it
Can you link me this paper please?
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HopefulDentist..
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(Original post by Nautic4l)
Can you link me this paper please?
It won't let me link it is there some way of attaching it?
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Maham88
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(Original post by HopefulDentist..)
It won't let me link it is there some way of attaching it?
you can find it here http://www.physicsandmathstutor.com/...4/#past_papers
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MeeraP07
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#1037
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#1037
Hey guys...
For Kc, if you have eqm moles of the reactants and products you divide them by the volume and then put those values in the Kc expression, right?
But if you have eqm concentrations, do you still need to divide by the volume?
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red122
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#1038
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#1038
Hi does anyone know why branched alkanes have lower entropy values ?


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Wahid1
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#1039
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#1039
Jan 2015 q24aiii anyone?


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HopefulDentist..
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#1040
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#1040
q22biv in june 2014 if someone could help that would be amazing!
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