Chemistry: Post your doubts here!

[Oliveme]

Thank you so much for the help.

Also,
Can you please tell the effect of electron withdrawing and donating groups attached to benzene ring?

I mean, for example if Cl is going to add in benzene in presence of AlCl3
I get confuse on which position to add the Cl (ortho, meta or para), if electron donating is attached and on which to add if electron withdrawing is attached.

I need a little briefing on it.

Your help is appreciated.
Best Regards

Electron-donating groups are -OH (phenol), -NH2 and -CH3 (methyl) - They activate positions 2,4, and 6 on the benzene ring. (2 and 6 are basically the same positions - depends on the way you look at it.)
Electron-withdrawing groups are -NO2 and -COOH groups. They activate positions 3 and 5 on the benzene ring.
Halogens are exceptions. They're meanies. They are electron-withdrawing but activate positions 2, 4 and 6.

 

[gary221]

Electron-donating groups are -OH (phenol), -NH2 and -CH3 (methyl) - They activate positions 2,4, and 6 on the benzene ring. (2 and 6 are basically the same positions - depends on the way you look at it.)
Electron-withdrawing groups are -NO2 and -COOH groups. They activate positions 3 and 5 on the benzene ring.
Halogens are exceptions. They're meanies. They are electron-withdrawing but activate positions 2, 4 and 6.

loved ur explanation..

 

[Oliveme]

loved ur explanation..

Thank you!

 

[Oliveme]

Can someone please help with p34 Oct/Nov 2012 question( g) I didn't know how to find the mean :S

From the list of given numbers, you eliminate 62 because that's way out of given range. Notice that all numbers, except 62, are around about 54.
So mean = (56+54+56+53)/4 = 54.75 and round it off to 54.8 - Always look out for a value that is too out of the range or say is the furthest.

 

[marz_katy]

can any1 xplain me colorimetry??

 

[Jiyad Ahsan]

can any1 xplain me colorimetry??

you mean calorimetry? right?
well its a technique used to measure enthalpy change of reactions.. by experimental calculations.. i'll explain how using a neutralisation reation
lets say you have 50cm^3 of acid and the same amount of an alkali (also assuming that 1 cm^3 weighs 1 gram)
and specific heat capacity is 4.2 joules per gram per degree celsius (assuming same as water)
measure the initial temperature (take it as 22 degrees celsius)
then measure the final temperature when the reaction is complete (take that as 28 degrees celsius)
apply the formula q = m c (change in) T , where m is mass, c is the heat capacity, T is the temperature
so, q = 100 x 4.2 x 6
q=2520 Joules, and thats your enthalpy change of reaction.. hope it helped

 

[xhizors]

you mean calorimetry? right?
well its a technique used to measure enthalpy change of reactions.. by experimental calculations.. i'll explain how using a neutralisation reation
lets say you have 50cm^3 of acid and the same amount of an alkali (also assuming that 1 cm^3 weighs 1 gram)
and specific heat capacity is 4.2 joules per gram per degree celsius (assuming same as water)
measure the initial temperature (take it as 22 degrees celsius)
then measure the final temperature when the reaction is complete (take that as 28 degrees celsius)
apply the formula q = m c (change in) T , where m is mass, c is the heat capacity, T is the temperature
so, q = 100 x 4.2 x 6
q=2520 Joules, and thats your enthalpy change of reaction.. hope it helped

sorry but colorimetry is different
its used to identify the intensity of light being absorbed and emitted when measuring the Rate of Reaction when increasing concentrations!!
I can help further if you like too!

 

[Hirdayesh.B.Shrestha]

How many significant figure do we use in our answers? Is it three or two? Or is it what the values in the question is in?

 

[strangerss]

From the list of given numbers, you eliminate 62 because that's way out of given range. Notice that all numbers, except 62, are around about 54.
So mean = (56+54+56+53)/4 = 54.75 and round it off to 54.8 - Always look out for a value that is too out of the range or say is the furthest.

Oh that explains , thanks a lot for the help jazak alALLAH 5er

 

[marz_katy]

yes plz can u help me to knw more about dis plzzz=)

 

[saadgujjar]

any guess about chemistry paper 34

 

[Jinkglex]

How many significant figure do we use in our answers? Is it three or two? Or is it what the values in the question is in?

usually 3, unless specified otherwise

 

[VinnCh]

Hello, I need help here. For 9701_w09_qp_42 , question 2d (ii), I don't quite get how the question is done.

 

[biba]

Hello, I need help here. For 9701_w09_qp_42 , question 2d (ii), I don't quite get how the question is done.

um in c (i) we calculated [Ag+] = 7.1x10^-7
now they are asking us to calculate [NH3] that wil give us this value of [Ag+].. and they have given ys the concentration on silver nitrate... here we have to assume that due to ligand substitution silver nitrate got converted into the complex [Ag(NH3)2] so the concentration of this complex is also 0.1 mol/dm3...
ur expression for kc is kc = [Ag(NH3)2] / [Ag+][NH3]^2
u know kc,[Ag+] and [Ag( NH3)2].... find [NH3]

 

[messi10]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf

Its a Ka question:
Q no 3 (e) part iv

Can you please help me out.

I know it has used the formula, pH=pKa + log (salt/acid)
but how the values?

 

[VinnCh]

um in c (i) we calculated [Ag+] = 7.1x10^-7
now they are asking us to calculate [NH3] that wil give us this value of [Ag+].. and they have given ys the concentration on silver nitrate... here we have to assume that due to ligand substitution silver nitrate got converted into the complex [Ag(NH3)2] so the concentration of this complex is also 0.1 mol/dm3...
ur expression for kc is kc = [Ag(NH3)2] / [Ag+][NH3]^2
u know kc,[Ag+] and [Ag( NH3)2].... find [NH3]

Ah.. So it it based on an assumption. I understand now. Thanks alot! =)

When R=H, RNH2= NH3 and [Ag(RNH2)2]+ = [Ag(NH3)2]+
So your Kc expression is -
Kc= [Ag(NH3)2]+/[Ag+] [NH3]^2
The question asks you to find the conc of NH3 when the conc. of silver ions is the same as you answer to ci.
So [Ag+] = ci = 7.1* 10^-7
[Ag(NH3)2] =.1 (given in dii)
therefore
1.7*10^7 = .1/ (7.1*10^-7) * [NH3]^2
[NH3]= .091 mol/dm^3

Hello, thanks for replying too, but I was actually wondering why do they use [Ag(NH3)2] as 0.1 , but biba explained it to me already, so thanks anyways! =)

 

[Nab900]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_43.pdf

Can Someone please explain me Question 4b (i) and 4(c)

 

[biba]

Ah.. So it it based on an assumption. I understand now. Thanks alot! =)



Hello, thanks for replying too, but I was actually wondering why do they use [Ag(NH3)2] as 0.1 , but biba explained it to me already, so thanks anyways! =)

welcum !

 

[iKhaled]

F

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_43.pdf

Can Someone please explain me Question 4b (i) and 4(c)

For 4bi..u never did that titration in ur life ?

 

[Nab900]

F
For 4bi..u never did that titration in ur life ?

i did but wht i actually want to know is that why doesn't the colour change from green to red-brown??? y is it changing to colourless or pale yellow???