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Chemistry: Post your doubts here!

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_22.pdf

Q2 B)i)
This is regarding Diagram.. could anyone plz draw it out For Mee!
Thnx in Advance :)!
1914889739.png


Each Na Connected to six Cl.Copied this from Endorsed Cambridge book.

*edit. Here is the image
3799868276.png
 
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At cathode always reduction takes place
At the anode always oxidation takes place
So the object to be plated is at the cathode & it will gain the electrons from copper
You can remember it this way if it helps:
CCR- Cathode cation reduction & RIG- reduction is gain
AAO- Anode anion oxidation & OIL- oxidation is loss
Thanks a lot, I got it now :)
If you don't mind, could you please help me with qn no. 3(c) of this paper? It's a similar one.. and I'm completely confused
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf
 
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Simple.
Since it is an Alcohol,
When you add an oxidising agent such as K2Cr2O7 In Primary Alcohol ( OH on the end of the carbon Chain) Attached so you get an Carboxylic acid Or If its an Secondary Alcohol( OH wala Carbon attached to other two Carbon) so you get a ketone .


Presence of an Acid and Alcohol gives you an Ester. Memorize the structure of a ester. Add the Carboxylic chain on Front and Alcohol chain on the end.

When in presence of Hydrogen Hallide.. OH on gerian-ol ,The OH will be replaces by Br


Dont think how hard the name is . They make you feel like that but Just think which Functional group it contains and if this particular functional reacts with this reagent what will be formed. List down all the reaction and reagents on a page this will help u alot and will be easy to memorized .
Thanks a lot
 
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View attachment 40878
Serious question people!! Please help. May Allah bless you all.

Been asked here the 8th time in 3 weeks and counting!

This stuff isn't in the syllabus or taught. This is precisely why it has shown you the 4 figures above the actual question. The examiner wants you to analyze those figures, see whats going on, and apply THAT to the question.

The question shows that all carbons in Benzene can be made to lie on the same plane, but not in cyclohexane. The question also shows that butane Carbons can be made to lie on the same plane, but methylpropane, also an alkane, cannot.
This tells us that if a compound has a ring, OR branching, it cannot be made to lie on the same plane.

All of them are co-planer apart from B.

C is coplaner because you don't have to consider the Oxygen atoms. The others are unbranched and two individual chains lying on the same plane.

Nice copy-pasta. I wonder why the writing style sounds eerily familiar. Oh rightt.. -.-

ON09 42 Q5a.
I've answered this before a day or two ago but I'll do it again anyway.
Look. This stuff isn't in the syllabus or taught. This is precisely why it has shown you the 4 figures above the actual question.... lying on the same plane.
 
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For the reaction of tollens reagent with an aldehyde the aldehyde gets converted into a carboxylic group. For the second one what you do is you remove the H2 from the 2,4 dinitrophenylhydrazine and from the other compound remove the O that is doubly bonded to the C. And then join both molecules together. The writing after the arrow is what should be written in the answer, the diagram before the arrow is for explanation.
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For the reaction of tollens reagent with an aldehyde the aldehyde gets converted into a carboxylic group. For the second one what you do is you remove the H2 from the 2,4 dinitrophenylhydrazine and from the other compound remove the O that is doubly bonded to the C. And then join both molecules together. The writing after the arrow is what should be written in the answer, the diagram before the arrow is for explanation.
View attachment 40925

Umm the stuff behind the arrow is also acceptable, and in fact it is recommended that it be used because it shows the placement of the Nitro groups (i.e 2,4)
 
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For the reaction of tollens reagent with an aldehyde the aldehyde gets converted into a carboxylic group. For the second one what you do is you remove the H2 from the 2,4 dinitrophenylhydrazine and from the other compound remove the O that is doubly bonded to the C. And then join both molecules together. The writing after the arrow is what should be written in the answer, the diagram before the arrow is for explanation.
View attachment 40925
thanks man
 
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Can anyone please help me with this question
calculate average bond energy of C-C bond in ethane given ΔHf(C2H6) = -85 KJ/mol , ΔHat(H)gas =218 KJ/mol and ΔHat(graphite) = 715 KJ/mol
 
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