Chemistry P4| A2 only

[periyasamy]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_41.pdf
Question 3 part d)iii) plzzz explain me how to find the concentration of the acid and the salt ...
Thanks in advance!

I guess this is the way.
Ok for the conc of acid,based on eqn u know that 1 mole of propanoic acid react with 1 mol of naoh
thus u find the mole of naoh by mv/1000
u will get 0.05 mol.However only o.o3 of it could react with the acid based on eqn.
so the mole remaining in buffer (unreacted acid) is 0.05-0.03=0.02mol
,so 0.02/100 *1000=0.2
N for the second part remove 0.03 mol from 0.05 mol

 

[Angelina_25]

I guess this is the way.
Ok for the conc of acid,based on eqn u know that 1 mole of propanoic acid react with 1 mol of naoh
thus u find the mole of naoh by mv/1000
u will get 0.05 mol.However only o.o3 of it could react with the acid based on eqn.
so the mole remaining in buffer (unreacted acid) is 0.05-0.03=0.02mol
,so 0.02/100 *1000=0.2
N for the second part remove 0.03 mol from 0.05 mol

Thanku!

 

[sweetjinnah]

Hei guys.Can anyone help me with this terribly hard question from may june 13 paper.Thank u.

for ligand exchange the liagand changes on the products side of reaction.
for redox oxidation no. changes..
the reaction which has a salt nd water at the prroducts side its acid-base.
nd u only have to place one tick against each reaction.. i hope u got it try doing urself

 

[periyasamy]

for ligand exchange the liagand changes on the products side of reaction.
for redox oxidation no. changes..
the reaction which has a salt nd water at the prroducts side its acid-base.
nd u only have to place one tick against each reaction.. i hope u got it try doing urself

Thanks a lot.

 

[zackle09]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
can someone pleaseeeeeee explain question 5b(iv)!!

 

[Farez]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
can someone pleaseeeeeee explain question 5b(iv)!!

There's only 1 tertiary H, and there are 9 primary H's. The probability of the Cl replacing the tertiary H, i.e. probability of J forming, is 21 (because 1 * 21), and the probability of Cl replacing one of the primary H's, i.e. probability of K forming, is 9 (because 9 * 1).

Therefore, the ratio of J/K is 21:9.

 

[zackle09]

There's only 1 tertiary H, and there are 9 primary H's. The probability of the Cl replacing the tertiary H, i.e. probability of J forming, is 21 (because 1 * 21), and the probability of Cl replacing one of the primary H's, i.e. probability of K forming, is 9 (because 9 * 1).

Therefore, the ratio of J/K is 21:9.

THANKYOUU!