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

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Thanks i did the question now. Although i still don't quite understand how finding the mol of hcl and naoh and subtracting the two gives me the mol left in equilibrium of ethanoic acid.

The thing is that HCl here acts as the catalyst in this reaction but because HCl is a strong acid than ethanoic acid, it also reacts with NaOH. Now, with NaOH and HCl reacting, we know they have a 1:1 ratio(For HCl and NaOH as well as for CH3COOH) but you see, the base has around 0.045 moles and HCl is 0.005 moles which have reacted so this means that the rest of the acid remaining here is CH3COOH.

(edited 7 years ago)

Reply 101

Funky_Giraffe

Could someone please explain how you could deduce the second half equation without knowing it? Really appreciate your help :smile:

Reply 102

samb1234

Original post by Funky_Giraffe

Could someone please explain how you could deduce the second half equation without knowing it? Really appreciate your help :smile:

From as, what 2 possible products are there if i oxidise a primary alcohol?

Reply 103

Funky_Giraffe

Original post by samb1234

From as, what 2 possible products are there if i oxidise a primary alcohol?

aldehyde or carboxylic

Reply 104

ST_123

Original post by sabahshahed294

Ah ok that makes sense now. Thanks a lot!

Reply 105

mizzyportia

how is everyone revising for Chemistry sorry if its a stupid question :blush:

Reply 106

sabahshahed294

Original post by mizzyportia

how is everyone revising for Chemistry sorry if its a stupid question :blush:

Going through notes because I keep on forgetting stuff from it tbh lol :P

Reply 107

sabahshahed294

Original post by ST_123

Ah ok that makes sense now. Thanks a lot!

You're welcome :smile:

Reply 108

samb1234

Original post by mizzyportia

how is everyone revising for Chemistry sorry if its a stupid question :blush:

Made a lot of detailed notes and then will do a lot of past papers probably

Reply 109

Funky_Giraffe

Hey, does someone mind explaining exactly why C is the correct answer here? I put A :/

Reply 110

samb1234

Original post by Funky_Giraffe

Hey, does someone mind explaining exactly why C is the correct answer here? I put A :/

What is the bronsted lowry definition of a base, and how is it relevant here

Reply 111

Funky_Giraffe

Original post by samb1234

What is the bronsted lowry definition of a base, and how is it relevant here

A base is a proton acceptor. Is it something to do with the inductive effect of the alkyl group in CH₃NH₂ which makes it more likely to accept a proton?

Reply 112

samb1234

Original post by Funky_Giraffe

A base is a proton acceptor. Is it something to do with the inductive effect of the alkyl group in CH₃NH₂ which makes it more likely to accept a proton?

Essentially yes. We know from tertiary halognenoalkanes from as that part of the reason they react via sn1 is that methyl groups are slightly electron donating, reducing the magnitude of the +ve charge on the Carbon. The same thing happens here - the methyl group slightly increases the negativity of the nitrogen compared to just NH3. The C6H5NH2 is similar to phenol - the lone pair of the nitrogen is partially delocalised into the ring system making it less -ve and therefore it doesn't pick up a proton as easily

Reply 113

Funky_Giraffe

Original post by samb1234

Thanks!!

Reply 114

ayvaak

I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.

Reply 115

ST_123

https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/January%202013%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf

For 18ai) im confused as to how to find the order of Br-. Hoping someone could explain it. Thanks

Reply 116

Funky_Giraffe

Original post by ayvaak

I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.

The systematic, logical way of going through this would to recognise that transition metals have an electronic structure of the form [Ar] 3d^x 4s^y . Working from Ti to Cu on the periodic table, you've got

Ti - [Ar] 3d¹ 4s²
V - [Ar] 3d² 4s²
Cr - [Ar] 3d³ 4s²
Mn - [Ar] 3d⁵ 4s¹

...and so on

Now, the thing you should note is that the difference in energy between the 4s- and 3d- sub-shells is minimal. Therefore, you should not expect a big jump between the last electron being removed from the 4s and the first from the 3d (electrons are removed from the 4s before the 3d)

So, you should be thinking all of this when you see the question, and bearing it all in mind, study the options. If this were a transition metal you would only get a jump between the 3rd and 4th ionisation energies onwards, because that's when the first Transition Metal, Titanium, loses it's first electron that's in neither the 3d nor 4s sub-shell (see the electronic structure of Ti above), so it's the first opportunity for a big jump in ionisation energy. So disregard options A and B immediately because their big 'jump' is before the 3rd and 4th ionisation energies. Notice how option D doesn't seem to have any jumps between the ionisation energies. So by default the answer is C.

(edited 7 years ago)

Reply 117

ayvaak

How many Unit 4 and Unit 5 past papers have people managed to track down? (Im trying to find out if Im missing any) There seems to be quite a few with the addition of international and january papers :smile:

Reply 118

samb1234