Edexcel CHEMISTRY DOUBTS!!!!

[Malak]

I had the same doubt some time ago

Check my thread

That is not my question :S

I know by which mechanism they react but i can't explain why?
I have read the chemguide website too and I understand it but I can't write it/explain in some easy words 

[MKh]

That is not my question :S

I know by which mechanism they react but i can't explain why?
I have read the chemguide website too and I understand it but I can't write it/explain in some easy words 

Refer to the Edexcel A2 Chemistry Revision Guide. You might find what u r looking for over there.

[The Golden Girl =D]

Found it (:

EDIT :

There are TWO ways for you to approach this Question

Method ONE :

Primary Haloalkane , group around it is SMALL => hence easy to attack

tertiary Haloalkane , groups around it are Bulky/big => Hence difficult to attack

Method TWO :

primary NOT Stable   [there is an intermidate formed with a C⁺]

tertiary => Stable , surrounded by electron donating alkyl groups

[don't know what he meant by Alkyl groups though  ]

Please, Don't forget to Include me in Your Prayers =D

[Malak]

Thanks GG

Can anyone do this question for me :S

I don't understand why the volume is 12.5 :s

Q20 a(v)

Another doubt:

Why does diluting a weak acid increase its degree of dissociation? :S

[Malak]

Plus the above, can any also make trans-esterfication clear for me :S

Trans-esterification is fine for me now
but ofcourse I have another doubt

(attached) How is it D :S

Thanks

[Deadly_king]

Plus the above, can any also make trans-esterfication clear for me :S

Trans-esterification is fine for me now
but ofcourse I have another doubt

(attached) How is it D :S

Thanks

Firstly let's look for compounds which have optical isomers, i.e chiral centres!

This leaves us with only option B and D since A and C can in no way show chirality!

Now we look for E-Z isomers, only D is right for B has two methyl groups attached to the carbon containing the double bond! So it cannot show E-Z isomers.

Hope it helps

[KJ1995]

Thanks GG

Can anyone do this question for me :S

I don't understand why the volume is 12.5 :s

Q20 a(v)

Another doubt:

Why does diluting a weak acid increase its degree of dissociation? :S

When we add water the system trys to oppose the change (chattliers principle) by using up the water added which shifts equilibrium to the right, which means more [H+] and [CH3COO-], and less [CH3COOH].. putting those new values in the (K_a) equation will make the value of K_a bigger...and i also dont know why is it 12.5, can you please tell em when you find out?
Thank you

[~ Miss Relina ~]

Plus the above, can any also make trans-esterfication clear for me :S

Trans-esterification is fine for me now
but ofcourse I have another doubt

(attached) How is it D :S

Thanks

I'm sorry for interfering but are there any links which helped you out in Trans-esterification  point
Thanks in advance 

[Malak]

I'm sorry for interfering but are there any links which helped you out in Trans-esterification  point
Thanks in advance 

I used the text-book and the edexcel revision guide.
I dont think we need to know very much about it and the revision guide gave me enough detail needed.

[The Golden Girl =D]

Q.20 how come it's 12.5 ?!


Okay from the First part of the Question we know that in the Titration 0.25 moldm^-3 of NaOH and 0.125 moldm^-3 of Ethanoic Acid  !!


so What I did is I found the Ratio as follows =>       Acid : Base
                                                          
                                                                      0.125 : 0.25  divide by 0.125 for both

                                                      I  get a Ratio of => 1 : 2

And from the part (v) we get to know that  25 cm³ of the ACID is USED UP !!!  .. so to find the Volume at the End point (Where the salt/product is given indicating that the ACID is completely KILLED) is

                                                               1  : 2

                                                               25 :  X      

Cross multiply and You'll find that the Volume of the BASE is  =>                   12.5 cm³


I simply used my common sense and it worked


if it doesn't work Use the formula from the CIE board (not in the syllabus) ;

(C₁V₁) _acid =  (C₂V₂)_base.... and act smart

[Malak]

^ Oh, thank you

Another doubt (attached)

Q18 part c (ii)
Why do we use the moles and not the concentration given :s

[The Golden Girl =D]

I think most probably that is the case because when you put the Volumes , they cancel out.

I hope I helped (:

[~ Miss Relina ~]

what is the PH of 0.5 moldm^-3 ammonia solution given K_b= 1.8 x 10^-5

[Malak]

what is the PH of 0.5 moldm^-3 ammonia solution given K_b= 1.8 x 10^-5

You dont need to know that.
I know the old papers have this but if you check the specification. I doesn't mention Kb anywhere

[~ Miss Relina ~]

You dont need to know that.
I know the old papers have this but if you check the specification. I doesn't mention Kb anywhere

but I didn't see it  in old pastpapers actually it is in the Edxecel chemistry blue book