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

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Original post by GeorgeL3
x


They are brilliant thanks ! ... all I'll be using this evening :biggrin: Love the way you have followed the spec too, which I always find difficult to do :tongue: :smile:
Original post by iwantopas19
asas.png

this is really simple :/ but i dont remember how to do it nw
cn someone plz enlighten me


I think for this question we have to first analyse their trend for their overall ionisation energies. Sounds ridiculous but yeah this is the tricky part.
If you try measuring the difference of the ionisation energies between the 1st and 2nd, and also between the 7th and 8th ionisation energy you will notice there is a large gap between them.

Since transition elements have an incompletely-filled 3d orbital, their ionisation energies take place in the 3d orbital. Their first ionisation energies are low, and they have a huge gap between 1st and 2nd, we assume they are from the 4s1 and 3d10 orbital.
The difference in ionisation energy remains small/almost constant till the 7th and 8th I.E. Here they occur between 3d5 and 3d4 orbital therefore a huge amount of enery is needed to remove electrons from 3d5 as they are half-filled (very stable). choose the one with the largest difference.

So the answer should be C?





Original post by LeaX
thank you so much :smile: so does the anhydrous calcium chloride remove the ether too?


It removes iodine from ether so it forms solid calcium iodide. ether will be seperated by it. You're welcome :smile:
Original post by bubblegummer
I think for this question we have to first analyse their trend for their overall ionisation energies. Sounds ridiculous but yeah this is the tricky part.
If you try measuring the difference of the ionisation energies between the 1st and 2nd, and also between the 7th and 8th ionisation energy you will notice there is a large gap between them.

Since transition elements have an incompletely-filled 3d orbital, their ionisation energies take place in the 3d orbital. Their first ionisation energies are low, and they have a huge gap between 1st and 2nd, we assume they are from the 4s1 and 3d10 orbital.
The difference in ionisation energy remains small/almost constant till the 7th and 8th I.E. Here they occur between 3d5 and 3d4 orbital therefore a huge amount of enery is needed to remove electrons from 3d5 as they are half-filled (very stable). choose the one with the largest difference.

So the answer should be C?









GREAT! thanks! :smile:
Reply 1624
Avatar for jojo1616
jojo1616
Original post by iwantopas19
GREAT! thanks! :smile:


You could just use the information in the data booklet for.this ques

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Original post by iwantopas19
asas.png

this is really simple :/ but i dont remember how to do it nw
cn someone plz enlighten me


Original post by bubblegummer
I think for this question we have to first analyse their trend for their overall ionisation energies. Sounds ridiculous but yeah this is the tricky part.
If you try measuring the difference of the ionisation energies between the 1st and 2nd, and also between the 7th and 8th ionisation energy you will notice there is a large gap between them.

Since transition elements have an incompletely-filled 3d orbital, their ionisation energies take place in the 3d orbital. Their first ionisation energies are low, and they have a huge gap between 1st and 2nd, we assume they are from the 4s1 and 3d10 orbital.
The difference in ionisation energy remains small/almost constant till the 7th and 8th I.E. Here they occur between 3d5 and 3d4 orbital therefore a huge amount of enery is needed to remove electrons from 3d5 as they are half-filled (very stable). choose the one with the largest difference.

So the answer should be C?




My class including the teacher got stuck on this for ages !!!

Then afterwards I realised that the ionisation energies of elements including the transition metals that we study are in there ! :facepalm:
Guys, do we need to know all the reactions and mechanisms for methylbenzene? Also, for sulphonation reaction is the electrophile formed from H2SO4 and SO3?

Original post by anony.mouse
My class including the teacher got stuck on this for ages !!!


Then afterwards I realised that the ionisation energies of elements including the transition metals that we study are in there ! :facepalm:


Guess your teacher went nuts looking for the answer :P We have too many stuff to remember from unit 1 to 5 and sometimes i have no idea which one to apply :frown:
(edited 10 years ago)
Original post by EHT1
Can anyone please explain to me their method of balancing equations using oxidation numbers? Thanks


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Firstly I balance everything apart from oxygen and hydrogen
Then I balance the oxygens by adding water
Balance the hydrogens by adding H+
Balance the charge by adding electrons to one side!

I hope this helps a little bit ...
Original post by posthumus
Don't think it's necessary at all thought it's good to know what is getting oxidized and reduced...




More ligands so I suspect great splitting... therefore the colour would darken

I think the colour should be lighter. Greater splitting means more energy absorbed, so light with shorter wavelength (darker colour) is absorbed, then the complementary colour should be lighter maybe?
Reply 1629
Avatar for kabOOmm
kabOOmm
in jan 2013 paper can somebody please explain question 14(d)(iv) If copper(I) iodide is treated with nitric acid, rather than sulfuric acid, a blue solution is still formed but no pink solid. Use the standard electrode potentials on page 15 of your data booklet to explain this. Quote any data that you use.

if no pink solid forms, then shouldn't the Ecell be -ve. i dont get itt!!!
Original post by kabOOmm
in jan 2013 paper can somebody please explain question 14(d)(iv) If copper(I) iodide is treated with nitric acid, rather than sulfuric acid, a blue solution is still formed but no pink solid. Use the standard electrode potentials on page 15 of your data booklet to explain this. Quote any data that you use.

if no pink solid forms, then shouldn't the Ecell be -ve. i dont get itt!!!

CU+ turns into cu2+, E is +0.15V, this is on the left hand side, so the right hand cell should be more positive than 0.15V. Next part I am not sure but I guess its because the Ecell value for NO3- + 3H+ turns into HNO2 + H2O is more positive than the E cell value of CU+ to Cu, so the nitrate ions react instead of copper getting formed.

Anyone else have any suggestions?
what are the conditions needed for makin paracetamol ?
Original post by iwantopas19
what are the conditions needed for makin paracetamol ?


Depends on what your starting reagents are.
Original post by iwantopas19
what are the conditions needed for makin paracetamol ?


paracetamol is an N substituted amide , so you react an amine with an acyl chloride
reactions with acyl chlorides are at room temperature
OK how much experimental procedure may have to describe? I can think of these ones

recrystalisation
melting point determination
steam distillation
iodine thiosulfate titrations
dichromate titrations
producing Chromium (II)
Original post by iwantopas19
what are the conditions needed for makin paracetamol ?


The CGP revision guide says ethanoyl chloride is added to a concentrated aqueous solution of the amine, a violent reaction occurs, which produces a solid, white mixture of the products. So I think it's under standard conditions?

Oh and in the paracetamol case, the solid products should be brown but they are usually stained brown with unreacted phenylamine.
Original post by DrewYouTwo
OK how much experimental procedure may have to describe? I can think of these ones

recrystalisation
melting point determination
steam distillation
iodine thiosulfate titrations
dichromate titrations
producing Chromium (II)


Solvent extraction! I got tested on this in unit 6b last month and I lost all 3 marks....
Reply 1637
Avatar for kabOOmm
kabOOmm
Original post by clydeshen411
CU+ turns into cu2+, E is +0.15V, this is on the left hand side, so the right hand cell should be more positive than 0.15V. Next part I am not sure but I guess its because the Ecell value for NO3- + 3H+ turns into HNO2 + H2O is more positive than the E cell value of CU+ to Cu, so the nitrate ions react instead of copper getting formed.

Anyone else have any suggestions?



but they said in the mark scheme
Copper (formed (by disproportionation)) is oxidized(by nitric acid) must be stated in words
stand alone mark


does this mean that copper(pink solid) did form but got oxidized? why couldn't this happen with the sulphuric acid too?
Original post by clydeshen411
The CGP revision guide says ethanoyl chloride is added to a concentrated aqueous solution of the amine, a violent reaction occurs, which produces a solid, white mixture of the products. So I think it's under standard conditions?

Oh and in the paracetamol case, the solid products should be brown but they are usually stained brown with unreacted phenylamine.


Original post by DrewYouTwo
OK how much experimental procedure may have to describe? I can think of these ones

recrystalisation
melting point determination
steam distillation
iodine thiosulfate titrations
dichromate titrations
producing Chromium (II)


Original post by xxEmma95xx
paracetamol is an N substituted amide , so you react an amine with an acyl chloride
reactions with acyl chlorides are at room temperature


Original post by James A
Depends on what your starting reagents are.



different answers... :biggrin:

thank you everyone :smile:
Does anyone know how to explain why there's 196g of Chromium (III) sulfate in Jan 2013 paper Question 2C? Shouldn't it 392g at 1mol/dm3?

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