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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

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samb1234
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Original post by PlayerBB
I thought we didn't include solid molecules in the equilibrium constant because their concentrations were too low compared to other reactants that any changes would be insignificant and that's why they're constant, however, it seems like i got something wrong, so can someone help? Why the answer is A and not C ?


the 'concentration' of a solid is just it's density
Original post by samb1234
the 'concentration' of a solid is just it's density


Ah, so its density won't be changing and that's why it is constant ?

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samb1234
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Original post by PlayerBB
Ah, so its density won't be changing and that's why it is constant ?

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yes. Density of a solid is just a constant number.
Original post by samb1234
yes. Density of a solid is just a constant number.


Ah thanks! :smile:

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Original post by genevievelaw
I found this (it suggests why OH produces a broad peak), look on page 4


Thank you, pretty much sums up the answer to my question. :h:

Original post by sabahshahed294
From what I know, it's something to do with on how the bonds react when they are exposed to infrared. When a bond is exposed to infrared, what happens is that they undergo stretching or bending. This is to do with the frequency absorbed and every particular bond absorbs particular frequencies which is why they produce distinct peaks.


Original post by samb1234
Within your large sample of alcohol you'll have hydrogen bonding to different extents in different molecules, which affects the amount of energy needed to get the bond to stretch so because of this difference for a large sample you are going to get a wide number of frequencies at which some of the bonds stretch


I understand how IR spectrometry works, but for our spec we just need to know that different bonds give different peaks due to different amounts of absorption, right? A friend of mine was talking about protons and I was 90% sure he was confusing it with NMR lol.
Original post by Ayman!
Thank you, pretty much sums up the answer to my question. :h:





I understand how IR spectrometry works, but for our spec we just need to know that different bonds give different peaks due to different amounts of absorption, right? A friend of mine was talking about protons and I was 90% sure he was confusing it with NMR lol.


Sounds like proton NMR to me XD
Reply 606
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samb1234
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Original post by Ayman!
Thank you, pretty much sums up the answer to my question. :h:





I understand how IR spectrometry works, but for our spec we just need to know that different bonds give different peaks due to different amounts of absorption, right? A friend of mine was talking about protons and I was 90% sure he was confusing it with NMR lol.


yeah you don't need to know the details just how to read them
Can't wait till this is all over not gonna lie lol.
Original post by samb1234
yeah you don't need to know the details just how to read them


how drab

Original post by Don Pedro K.
Can't wait till this is all over not gonna lie lol.


U4 is fine

U5 is death.
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samb1234
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Original post by Ayman!
how drab



U4 is fine

U5 is death.


Lol ngl I'm hoping to get as high in this is possible to minimise what I need in unit 5. And yeah it's one of those things in chem that you just sort of accept happen/work rather than understand unfortunately.
U5 is depressing...I just hope U4 is descent because January 16 was too easy! :s-smilie:
Original post by samb1234
Lol ngl I'm hoping to get as high in this is possible to minimise what I need in unit 5. And yeah it's one of those things in chem that you just sort of accept happen/work rather than understand unfortunately.


Same here haha, two kids from the year above us had 120/60 in U4/U6 respectively and 96 in U5 to get them the A*. Doesn't seem like a bad plan :lol:
Reply 612
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ayvaak
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Original post by samb1234
yes. Density of a solid is just a constant number.



Does density stay the same because turning a solid into a gas is not feasible because of high energy required to reverse phase change?
Original post by Dinaa
Hi, I think the answer is D because a larger radius means lattice enthalpy is less negative. Larger radius means it is easier to bond with due to weaker attraction.


Original post by samb1234
It's D because the smaller the ionic radius, the greater the charge density so the stronger the attraction to the lone pairs on oxygens so hence more energy is released


Oh yes makes sense! Thanks :smile:
What exactly do we need to know about chromatography? (HPLC and Gas-liquid)

I'm so confused I don't think we were ever taught this area.. Worrying lol.
Original post by dental17
What exactly do we need to know about chromatography? (HPLC and Gas-liquid)

I'm so confused I don't think we were ever taught this area.. Worrying lol.


Off the top of my head:

--- How they work -

What the mobile and stationary phase are for each (as in what state the phases are and what substance the phases are made of).

What causes the difference in retention time (the amount of attraction to one of the phases because of ...)

Perhaps working out the retention time.

--- Whichever one is to be used -

Using GC or HPLC.

If something is missing then I'm sure someone else or I will say so.

But make sure you definitely are familiar with the mobile phase and stationary phase stuff since questions often appear where they ask you what they are and the differences etc.
Original post by samb1234
yes. Density of a solid is just a constant number.


Is the density assumed constant, since there is some small mass change in a reaction like a decomposition?
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_H_V
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Help with 2016 IAL paper question 10, (answer is B) I cant understand why

The pH of three solutions with concentration 1.0 mol dm–3 was measured.
Solution 1 NH3
Solution 2 CH3COONa
Solution 3 NH4Cl

Which of the following shows the three solutions in order of increasing pH?
A 1, 2, 3
B 3, 2, 1
C 3, 1, 2
D 2, 3, 1
Screen Shot 2016-06-12 at 20.13.03.png

anyone help with part b) please? I've really confused myself. why is the answer D and not C?
thanks
Original post by Neha121101
Screen Shot 2016-06-12 at 20.13.03.png

anyone help with part b) please? I've really confused myself. why is the answer D and not C?
thanks


The rate equation involves only propanone and hydrogen ions. The concentration of iodine therefore has no effect on the rate. Hence, when the concentration of propanone is doubled, the rate will also double; not quadruple. The doubling of iodine concentration will have no effect on the rate since the order of reaction with respect to iodine is 0 (since it is not involved in the rate equation). Therefore, the answer is D :smile:

Also, you wouldn't be able to tell whether C was correct or not due to the fact that we are not told whether the forward reaction is exothermic or endothermic, and thus cannot make inferences about what a change to temperature will do to the rate constant :smile:!

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