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3:20 AM
How the formula Intensity, I' = I $e^(\usepackage{ upgreek }
\upmu
mathmode)-$
I don't know latex simply tried.
How the formula in the first line of the solution derived?
 
3:46 AM
Its the beer-lambert law, From what I know this is not currently a part of the JEE syllabus, and its derivation is not something that is easy at this level...
 
 
1 hour later…
5:04 AM
@satan29 Thanks.
the derivation of it in the first part of it video satisfactory.
 
@CHEMOJEE what's the question? Is it that we have to find intensity when it passes from one medium to other ?
 
@RoverThis is the question
 
@CHEMOJEE ok
 
 
4 hours later…
9:05 AM
Guys , for this Q. Is the answer d
I think it should be 3 methylhex 3 ene
 
@SrijanM.T 3- methyl hex-3-ene
@SrijanM.T Yes
 
9:25 AM
Then . @Rover All the options are wrong right ?
 
9:45 AM
@JohnRennie Hello sir
 
@SrijanM.T hi :-)
 
Can I ask a Q of Chem I.e structure related.
Carbon type and functional group
 
I doubt I can help with organic chemistry as I don't know much about it.
 
Ohk.
[![enter image description here][1]][1]


[1]: https://i.stack.imgur.com/JP9qo.jpg

Is this the diagram for butan - 2 - al?
If you can it’s good
 
You can certainly ask and see if anyone else can answer.
 
9:46 AM
Also ok , if not
@JohnRennie ok sure
 
"al" means an aldehyde doesn't it?
Isn't that a ketone?
Butanone, also known as methyl ethyl ketone (MEK), is an organic compound with the formula CH3C(O)CH2CH3. This colourless liquid ketone has a sharp, sweet odor reminiscent of acetone. It is produced industrially on a large scale, but occurs in nature only in trace amounts. It is partially soluble in water, and is commonly used as an industrial solvent. It is an isomer of another solvent, tetrahydrofuran. == Production == Butanone may be produced by oxidation of 2-butanol. The dehydrogenation of 2-butanol using a catalyst is catalyzed by copper, zinc, or bronze: CH3CH(OH)CH2CH3 → CH3C(O)CH2CH3...
 
Yes, double bond O
Ohk. I got it @JohnRennie
 
:-)
An aldehyde has the C=O at the end of the chain and a ketone has it in the middle of the chain.
The reason we distinguish between the two is that the double bond at the end is much more reactive.
 
10:25 AM
Ohk sir. Thank you very much .
@JohnRennie Ohk.
 
11:04 AM
@SrijanM.T yes
 
Thnx
 
 
6 hours later…
5:37 PM
How to solve this problem?
 

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