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%
This discussion is closed.
LeaX
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(Original post by jojo1995)
To anyone on here... are any of you gap year students?
I think I'll be having a Gap year before going to uni (well it looks that way) but i was wondering, if any of you are gap yearstudents ... what are you/ did you do on your gap year?

thanks
I'm taking a gap year but doing more a levels during it so it'll be studying part time and then hopefully getting a job and learning to drive.
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jojo1995
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(Original post by LeaX)
I'm taking a gap year but doing more a levels during it so it'll be studying part time and then hopefully getting a job and learning to drive.
Thank you I'm sure you'll have a great time !
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jojo1995
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Could anyone help me with this pls ? What is the colour mn 2+ turns in naoh and in nh3 and also for the deprotonation state there is a colourchange due to mn2+ being oxidised , what is it pls?

It's just the rev guide Cgp and the facer say different things
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F1's Finest
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(Original post by jojo1995)
Could anyone help me with this pls ? What is the colour mn 2+ turns in naoh and in nh3 and also for the deprotonation state there is a colourchange due to mn2+ being oxidised , what is it pls?

It's just the rev guide Cgp and the facer say different things
It depends whether if the NaOH is a couple of drops, or in excess.

Also, depends whether the NH3 is a couple of drops, or in excess.

Mn2+ gives a white/dark ppt for a couple of drops of NaOH, for NaOH in excess, the precipitate won't dissolve further.

Mn2+ gives a white/dark ppt with a couple of drops of nh3 but in excess, again, the ppt does not dissolve.
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jojo1995
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(Original post by James A)
It depends whether if the NaOH is a couple of drops, or in excess.

Also, depends whether the NH3 is a couple of drops, or in excess.

Mn2+ gives a white/dark ppt for a couple of drops of NaOH, for NaOH in excess, the precipitate won't dissolve further.

Mn2+ gives a white/dark ppt with a couple of drops of nh3 but in excess, again, the ppt does not dissolve.
Thank you dark precipitate ? Dark what pls ?
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F1's Finest
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(Original post by jojo1995)
Thank you dark precipitate ? Dark what pls ?
A brown(ish) colour, as the precipitate has darkened as a result of it being oxidised rapidly by oxygen in air.
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jojo1995
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(Original post by James A)
A brown(ish) colour, as the precipitate has darkened as a result of it being oxidised rapidly by oxygen in air.
Thank you so much !!
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itsjordan
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Does anyone know when propanenitrile is heated under reflux with HCl/H2SO4 to form propanoic acid whether the by-product is NH3 or NH4Cl?

One book says propanoic acid and NH3 is produced, the other says propanoic acid and NH4Cl...
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Reversal
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Does anyone want to help me build a list of reactions that we need to learn off-by-heart? They're always the things that get me stuck, so I'm just wanting to start revising them now (I'm just going to go through the edexcel book looking at the reactions it looks like are important, can someone tell me which ones you need to know?)

Transition Metals:


Copper:

Reaction of Copper with hot Sulfuric acid:

Cu (s) + 2H2SO4 (l) >>> CuSO4 (g) SO2 (g) + 2H2O (l)

Reaction of Copper with oxygen in the presence of CO2 (rusting)

Copper oxidised very slowly, forming CuCO3.Cu(OH)2

Reaction of alkaline solution of copper (II) sulfate with an aldehyde: (Fehling's Test)

2Cu2+ (aq) + 2OH- (aq) + 2e- >>> Cu2O (s) +H2O
(Copper(II) ions are complexed with 2,3,-dihydroxybutanedioates to stop Copper(II) hydroxide being formed)

Copper (II) sulfate solution with ammonia solution, precipitate of hydrated Copper (II) hydroxide formed:

[Cu(H2O)6]2+ (aq) + 2NH3 (aq) >>> [Cu(H2O)4(OH)2] (s) +2NH4+ (l)

Adding more ammonia solution to above reaction:
Ligand Exchange, ammonia molecules replace four water molecules in complex. Forms deep, inky-blue ions.


[Cu(H2O)4(OH)2] (s) + 4NH3(aq) >>> [Cu(NH3)4(H2O)2]2+ (aq) + 2OH- (aq) + 2H2O (l)


Does this help anyone? Feel free to quote and add
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NutterFrutter
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(Original post by itsjordan)
Does anyone know when propanenitrile is heated under reflux with HCl/H2SO4 to form propanoic acid whether the by-product is NH3 or NH4Cl?

One book says propanoic acid and NH3 is produced, the other says propanoic acid and NH4Cl...
You'd have a mixture of both (initially NH3 produced which then can react with HCl to produce NH4Cl).
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F1's Finest
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(Original post by Reversal)
Does anyone want to help me build a list of reactions that we need to learn off-by-heart? They're always the things that get me stuck, so I'm just wanting to start revising them now (I'm just going to go through the edexcel book looking at the reactions it looks like are important, can someone tell me which ones you need to know?)

Transition Metals:


Copper:

Reaction of Copper with hot Sulfuric acid:

Cu (s) + 2H2SO4 (l) >>> CuSO4 (g) SO2 (g) + 2H2O (l)

Reaction of Copper with oxygen in the presence of CO2 (rusting)

Copper oxidised very slowly, forming CuCO3.Cu(OH)2

Reaction of alkaline solution of copper (II) sulfate with an aldehyde: (Fehling's Test)

2Cu2+ (aq) + 2OH- (aq) + 2e- >>> Cu2O (s) +H2O
(Copper(II) ions are complexed with 2,3,-dihydroxybutanedioates to stop Copper(II) hydroxide being formed)

Copper (II) sulfate solution with ammonia solution, precipitate of hydrated Copper (II) hydroxide formed:

[Cu(H2O)6]2+ (aq) + 2NH3 (aq) >>> [Cu(H2O)4(OH)2] (s) +2NH4+ (l)

Adding more ammonia solution to above reaction:
Ligand Exchange, ammonia molecules replace four water molecules in complex. Forms deep, inky-blue ions.


[Cu(H2O)4(OH)2] (s) + 4NH3(aq) >>> [Cu(NH3)4(H2O)2]2+ (aq) + 2OH- (aq) + 2H2O (l)


Does this help anyone? Feel free to quote and add
For example, the Fehlings test, all you gotta remember is Cu2+ ions reacting, to form Cu2O, then you know how to balance the equation out in the exam, without having to memorise the other parts of the formula.

Also,

Copper oxidised very slowly, forming CuCO3.Cu(OH)2

They wouldn't ask for you to recall this formula.

You will, however be given electrode potentials and asked whether the rusting of copper will take place or not.

For the last equation, it's not exclusive to copper only, it's exclusive to the other hydrated hydroxides, Cobalt to Silver.

Namely amine complexes as you were saying ^^
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itsjordan
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(Original post by NutterFrutter)
You'd have a mixture of both (initially NH3 produced which then can react with HCl to produce NH4Cl).
Oh alright, I thought if you used HCl as catalyst you would get NH4Cl, and if you used H2SO4 you would get NH3.
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NutterFrutter
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(Original post by itsjordan)
Oh alright, I thought if you used HCl as catalyst you would get NH4Cl, and if you used H2SO4 you would get NH3.
I can't remember the reaction very well, I even thought you had to use a weak acid for this reaction? :lol:

With the HCl, you'd get NH3 and then NH4Cl.
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Reversal
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(Original post by James A)
For example, the Fehlings test, all you gotta remember is Cu2+ ions reacting, to form Cu2O, then you know how to balance the equation out in the exam, without having to memorise the other parts of the formula.

Also,

Copper oxidised very slowly, forming CuCO3.Cu(OH)2

They wouldn't ask for you to recall this formula.

You will, however be given electrode potentials and asked whether the rusting of copper will take place or not.

For the last equation, it's not exclusive to copper only, it's exclusive to the other hydrated hydroxides, Cobalt to Silver.

Namely amine complexes as you were saying ^^
So do we just need to remember the various reactions of transition metals with aqueous sodium hydroxide and aqueous ammonia and the organic pathways, and everything else can be worked out using the ionic equations/electrode potentials in the data booklet?
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F1's Finest
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(Original post by Reversal)
So do we just need to remember the various reactions of transition metals with aqueous sodium hydroxide and aqueous ammonia and the organic pathways, and everything else can be worked out using the ionic equations/electrode potentials in the data booklet?
http://www.edexcel.com/migrationdocu...4%20250510.pdf
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Reversal
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I knew someone was going to post this My bad!
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F1's Finest
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(Original post by Reversal)
I knew someone was going to post this My bad!
It's not to insult you in any way, but it makes life easier for both of us, also it's good to read through the spec, it shows you exactly what you need to know.

A good idea is to print off the pages for unit 5, then just tick the bullet points as you cover them :borat:
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Reversal
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(Original post by James A)
It's not to insult you in any way, but it makes life easier for both of us, also it's good to read through the spec, it shows you exactly what you need to know.

A good idea is to print off the pages for unit 5, then just tick the bullet points as you cover them :borat:
Thank you I am suddenly much clearer on what we have to know! I think I'll do that, cheers for the advice

This edexcel book is so rubbish, does anyone else use it?
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F1's Finest
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(Original post by satnamsinghrai)
Does anyone know why a condenser is important in reflux?
It prevents the volatile reaction mixture from escaping. As the reaction is taking inside the pear shaped flask, volatile gas will rise above, however the condenser cools the vapour, allowing it to react fully with the reaction mixture, hence a higher yield of product is obtained.
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mrhedgehog
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(Original post by James A)
It prevents the volatile reaction mixture from escaping. As the reaction is taking inside the pear shaped flask, volatile gas will rise above, however the condenser cools the vapour, allowing it to react fully with the reaction mixture, hence a higher yield of product is obtained.
Thank you very much
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