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Chemistry: Post your doubts here!

abcde

abcde

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smzimran said:
Iam not sure about this but i think that is because the 1 and 2 carbon atoms are not bonded to 4 atoms, the C-H bond is broken and H+ lost before an external bond is formed there is a pie electron cloud in the benzene ring....
Hope iam corect :)
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I was treating the benzene ring as a cycloalkane...The carbon-carbon double bonds in it escaped my mind. Thanks! :)
 
DANGERBP

DANGERBP

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abcde said:
W.S!
Kp is simply the equilibrium constant in terms of partial pressures. Since gases can occupy any container, there volume is indefinite. The Kp provides a better measure of the equilibrium constant in reactions involving gases as conc. = moles/volume, which is difficult to determine.
To find partial pressure, use: p = n/N x P, where
n: no. of moles of the particular gas (whole partial pressure is being found)
N: total no. of moles (of all gases)
P: total pressure (of all gases).

Once you have the partial pressures, Kp is found in exactly the same way as Kc. Just use the partial pressures where you'd use concentrations.
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mm is the calculation included in a level ???
 
smzimran

smzimran

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DANGERBP said:
shit man !!! i have no idea what tht is !!! our stupid teacher havent even gave us any info about this !! :mad: thought it was for a2 ! ughhhhh
im screwed :cry:
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Its not that difficult as it seems to be, last year when i was in AS, i also felt worried on seeing these "weird" calculations but once you get the trick, you'll call it a piece of cake...:)
 
L

legion

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what is the appearance of ammonium chloride? and the colour of aqueous of aqueous ammonia?
 
tom ed

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smzimran said:
In that method n is the number of double bonds + the number of chiral centers however as there are no chiral centres in consideration here, the double bonds must be counted.
I use this formula:
Total Number of isomers = 2^(a+b)
where
a is the number of double bonds
b is the number of chiral centres
Click to expand...
....ok ths sounds authentic =P......but still are u sure abt ths one....cuz i cant find it in any of the text books i have.....sorry .... no offence...:p
 
smzimran

smzimran

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tom ed said:
....ok ths sounds authentic =P......but still are u sure abt ths one....cuz i cant find it in any of the text books i have.....sorry .... no offence...:p
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I always solved these type of questions when i was in AS using this method and never got wrong, its absolutely correct. Dnt worry:)
 
tom ed

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thank u ..:).anyways....its good to see u helping out others even when u r out of ths trouble....if u dont mind...may i ask u ...how did u manage the practical part in AS.....
 
DANGERBP

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smzimran said:
Its not that difficult as it seems to be, last year when i was in AS, i also felt worried on seeing these "weird" calculations but once you get the trick, you'll call it a piece of cake...:)
Click to expand...
ya but man we havent took any calculation regarding equilibrium !! im afraid he havent gave us even more stuff =/ ide kill him if so ! -_-
 
abcde

abcde

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smzimran said:
In that method n is the number of double bonds + the number of chiral centers however as there are no chiral centres in consideration here, the double bonds must be counted.
I use this formula:
Total Number of isomers = 2^(a+b)
where
a is the number of double bonds
b is the number of chiral centres
Click to expand...
This formula will only yield the total number of stereo-isomers. Yes? Structural isomers will still have to be taken into account.
 
smzimran

smzimran

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abcde said:
This formula will only yield the total number of stereo-isomers. Yes? Structural isomers will still have to be taken into account.
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Yes the formula is for total number of STEREO isomers only
 
smzimran

smzimran

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DANGERBP said:
thx dude ! i hope ! =p
btw how did you do in ur a levels ??
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Ur welcome..
I got 3 'a', i have written it down in the information section of my profile.
BTW, the a levels isnt complete yet, still doing A2....:)
 
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