Chemistry: Post your doubts here!

[littlecloud11]

Asslamu Alikum wa rahmatullahiView attachment 21507 wa barakatoho..

Can someone explain why the answer for 3 (a) (ii) is K (potassium)

Well, the general trend for the ionic radii is that it decreases as we move from left to right across the periodic table (owing to an increasing nuclear charge of the ions as we move towards the right). The greatest number of electronic shells in the ion is obviously present in elements of the third period compared to the other two and abiding by the aforementioned trend potassium would be the largest cation.

 

[littlecloud11]

SALAAAAAAM PEOPLE
does anyone have question papers from june 2000 till 2002 + the marking schemes? i want paper 1, 2 and 3 plz if not possible only paper 1 if not possible only the marking schemes ..thnx

Have you tried Scribd? They might have some of the papers.

 

[yubakkk]

the answer is doubled because the equation says
"for decomposition of every 2 moles of NaHCO3,
1 mole of H2O and CO2
are produced.
so when you take Mr of one mole of H2O and CO2, you need to consider that there should have been 2 moles of NaHCO3

i think mole of nahco3 / mole of gas =2/2 so 2 2 cancel hence we have not to do *2 please show me with written steps.

 

[PhyZac]

Well, the general trend for the ionic radii is that it decreases as we move from left to right across the periodic table (owing to an increasing nuclear charge of the ions as we move towards the right). The greatest number of electronic shells in the ion is obviously present in elements of the third period compared to the other two and abiding by the aforementioned trend potassium would be the largest cation.

Jazaki Allahu khaira....thank you so much. May Allah reward you for your help. Ameen.

 

[PhyZac]

SALAAAAAAM PEOPLE
does anyone have question papers from june 2000 till 2002 + the marking schemes? i want paper 1, 2 and 3 plz if not possible only paper 1 if not possible only the marking schemes ..thnx

http://www.freeexampapers.com/

 

[Muhammad Bin Anis]

i think mole of nahco3 / mole of gas =2/2 so 2 2 cancel hence we have not to do *2 please show me with written steps.

whats your answer?
isnt it 2.71?

 

[yubakkk]

whats your answer?
isnt it 2.71?

ya answer is 2.71 but how?

 

[aseelz]

http://www.freeexampapers.com/

I cant find marking schemes for paper 1 !! I have for p2 and p3 any further help? Thnx

 

[aseelz]

I c

Have you tried Scribd? They might have some of the papers.

I couldnt find..if u did plz send me the link i would be very grateful

 

[Translucent231]

11g of ethylethanoate were mixed with 18cm3 of 1mol dm-3 hydrochloric acid in a flask and allowed to stand at constant temperature until equilibrium had been reached The contents of the flask were titrated with 1M sodium hydroxide solution and 106cm3 of the alkali were required. Assuming that 18cm3 of 1M hydrochloric acid contains 18g water, calculate the equilibrium constant Kc.

heres the equation : CH3COOC2H5 (l) + H2O(l) ⇋ CH3COOH(l) + C2H5OH(l)​

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​

 

[PhyZac]

I cant find marking schemes for paper 1 !! I have for p2 and p3 any further help? Thnx

Download the examiner report. It has the mark scheme of Paper1. "er" is the symbol for examiner report.

 

[Muhammad Bin Anis]

ya answer is 2.71 but how?

if you take the Mr of one mole of CO2 and H2O , then you must take two moles of NaHCO3

(2 x 84)/(44+18)

 

[scouserlfc]

i wanted 2 know da explanation for V and W please?

V is CH3CH2CHOHCH2OH because cold KMnO4 acts on it remember the reaction of alkenes where the double bond breaks to give and two OH attach ! thats what happens

W is CH3CH2COCHO this because now the oxidizing agent causes V having at one end primary alcohol and in the centre the secondary alcohol to oxidize and you know the products are aldehyde and ketone respectively

 

[salvatore]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_21.pdf

Pleaseee help me with question no. 3 (c) and 4 (c). I do not understand how to solve it.. an explanation will be much appreciated.
Waiting for a reply.
Thanks!

 

[yubakkk]

plz answer

 

[yubakkk]

 

[yubakkk]

i need help in above question.
why percentage are doubled?

 

[soumayya]

i need help in above question.
why percentage are doubled?

% error = 0.1/9 * 100
(0.05 for weighing more than 9.00 + 0.05 for weighing less than required)

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_21.pdf

Pleaseee help me with question no. 3 (c) and 4 (c). I do not understand how to solve it.. an explanation will be much appreciated.
Waiting for a reply.
Thanks!

I'm reposting this.. anyone?

 

[littlecloud11]

I'm reposting this.. anyone?

3c) You'll have to draw a double hump diagram for this. It's essentially the same as the simpler energy diagram except we break the single hump into 2 to represent the intermediate steps. There should be two peaks and the first one should be higher than the second. If you recall free radical substitution mechanism you know the intermediate steps. The first hump represents the formation of methyl free radical and HCl and the second represents the chloromethane formation. The reason the second hump is lower is because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one. The activation energy is labeled for the first hump.

4c) Ok, the final product is a ketone. Ketones come from the oxidation of secondary alcohols, so the intermediate must be a secondary alcohol. As you start with an Alkene you have to introduce an -OH group into the molecule so you use steam with conc. H3PO4. The result is CH3CH2CH(OH)CH3.
(For solving synthetic routes it helps if you go backwards, try thinking of the source of the product)

Hope this helps!