thankssee attachment for the explanation
hope you can understand
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thankssee attachment for the explanation
hope you can understand
thanksLet CH3CO2H = x cm3
Let CH3CO2Na = (100-x) cm3
pH = pKa + log(salt/acid)
5.5 = -lg(1.79E-5) + log(salt/acid)
5.5 = 4.75 + log(salt/acid)
log(salt/acid) = 0.75
Salt/acid = 0.1778
(100-x)/x = 0.1778
100-x = 0.1778x
100 = 1.1778x
x = 100/1.1778 = 85
100-x = 15
2 products. One will be organic and other is CO2.http://maxpapers.com/wp-content/uploads/2012/11/9701_s13_qp_23.pdf
question 5 part a iv
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_ms_22.pdf2 products. One will be organic and other is CO2.
Organic product will be HO2CCO2H (Break the alkene, you get CHO it further oxidises to COOH and OH is oxidised to COOH as well)
I think it is A.Help with this question please View attachment 57985
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_ms_22.pdf
question 4 part a
product of
C if I write CH3(CH2)2CH2OH instead of CH3CH2CH2CH2OH
and product of
E if I write CH3CH=CHCOOH instead of CH3CH=CHCO2H
Fine?
http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w09_ms_22.pdf
question5 part b
for equation if I CH3CH2CH2CH2OH instead of C4H9OH
Fine?
The answer is AI think it is A.
What is your doubt?
About whether it is A or B?
It can't be D since I don't think that an ester can dissolve in water completely.The answer is A
But yes I'm confused between A or B... even D seems likely
Can u point out some reasons why the Ans is A and not B or D
Ooh okay..! ThanksIt can't be D since I don't think that an ester can dissolve in water completely.
Then you have A and B.
It can't be B since alcohols are not acidic enough to react with aqueous solutions of strong alkalis.( However,they can react with the recatice metals.)
WelcomeOoh okay..! Thanks
can u plz tell why the moles for A is halved !! ??? OCT/06 q2 part iv
thanks !Dibasic acid means it releases 2 moles of H+ per mole of acid.
So ratio of dibasic acid : NaOH is 1: 2
Dibasic acid means it releases 2 moles of H+ per mole of acid.
So ratio of dibasic acid : NaOH is 1: 2
Hey is this question for AS or A2 ?guys can you work out this question...........................
An excess of sparingly soluble solid, M(OH)3 was added to 200cm3 of 0,1 mol/dm3 KOH. The mixture was shaken, allowed to reach equilibrium at rtp and the filtered.
25.0cm3 of the filtrate was titrated with 0,2mol/dm3 of HCl of which 18,0cm3 were required to reach end point.....
(i) Calculate the number of moles of hydroxide ions in the filtrate...........
(ii)Find the concentration of hydroxide ions in the filtrate..........which are from M(OH)3
(iii)Hence deduce the concentration of M3+ ions
(iv)State what is meant by solubility of a substance and hence give the solubility of M(OH)3 in 0.1 mol/dm3 KOH
(v) Use your answer to calculate Ksp value of M(OH)3........................
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