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AS Chemistry P2 Prep.

Saad (سعد)

Saad (سعد)

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MysteRyGiRl said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_22.pdf
umm frst question b ii) ? how did we get da answer
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The mole ratio of HCl and NaOH is the same, in the reaction:

NaOH + HCl --> NaCl + H2O.

So the number of moles of NaOH reacting with the HCl is the same as the no. of moles of HCl we used; 0.005. (In the question, we are told that we used 0.005 mol of Hydrogen Chloride. As the NaOH is in excess, all of the HCl will react with it - 0.005 mol of HCl and 0.005 mol of NaOH react to produce 0.005 mol of NaCl and water each).
 
Saad (سعد)

Saad (سعد)

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TheMan123 said:
The examiner isn't there to punish you, they are there to test your knowledge, they will accept both because they are both correct
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They will accept both, but they will be more inclined to award an A* to the one they preferred and was correct, than the one that they accepted because it was just correct.
 
MysteRyGiRl

MysteRyGiRl

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Saad (سعد) said:
The mole ratio of HCl and NaOH is the same, in the reaction:

NaOH + HCl --> NaCl + H2O.

So the number of moles of NaOH reacting with the HCl is the same as the no. of moles of HCl we used; 0.005. (In the question, we are told that we used 0.005 mol of Hydrogen Chloride. As the NaOH is in excess, all of the HCl will react with it - 0.005 mol of HCl and 0.005 mol of NaOH react to produce 0.005 mol of NaCl and water each).
Click to expand...

thnx and da c) part as well plzz :)
 
Saad (سعد)

Saad (سعد)

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MysteRyGiRl said:
thnx and da c) part as well plzz :)
Click to expand...

This one's a bit difficult... I dislike Equilibria. >_<

From (b)(iv) we know that 0.04 of NaOH reacted with the ethanoic acid; NaOH is ROH in the table. As the mole ratio of CH3COOH and NaOH is the same, 0.04 mol of NaOH react with 0.04 mol of CH3COOH.

This leaves behind 0.10 - 0.04 = 0.06 mol of CH3COOH and NaOH at equilibrium. So write 0.06 under CH3COOH and NaOH in the table.

That 0.04 mol which dissociated from CH3COOH and NaOH will form 0.04 mol of the products; 0 + 0.04 = 0.04. So each of the CH3COONa and H2O has 0.04 mol; write 0.04 mol in the table under their respective headings.

And you're done with (c)(i). For (c)(ii), simply use the way of finding Kc, and you have the concentrations, so do a simple calculation in the Calculator and you're done.

Hope that helps insha'Allah.
 
mominzahid

mominzahid

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Saad (سعد) said:
Not if its heat under reflux. HOOCCOOH is what will be formed.
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no it wasnt mentioned heat under reflux the compound was given and i had to give the structural formula of the product formed on reaction with hot kmno4/H+
is my answer right?
 
MysteRyGiRl

MysteRyGiRl

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Saad (سعد) said:
This one's a bit difficult... I dislike Equilibria. >_<

From (b)(iv) we know that 0.04 of NaOH reacted with the ethanoic acid; NaOH is ROH in the table. As the mole ratio of CH3COOH and NaOH is the same, 0.04 mol of NaOH react with 0.04 mol of CH3COOH.

This leaves behind 0.10 - 0.04 = 0.06 mol of CH3COOH and NaOH at equilibrium. So write 0.06 under CH3COOH and NaOH in the table.

That 0.04 mol which dissociated from CH3COOH and NaOH will form 0.04 mol of the products; 0 + 0.04 = 0.04. So each of the CH3COONa and H2O has 0.04 mol; write 0.04 mol in the table under their respective headings.

And you're done with (c)(i). For (c)(ii), simply use the way of finding Kc, and you have the concentrations, so do a simple calculation in the Calculator and you're done.

Hope that helps insha'Allah.
Click to expand...
eee jazakaALLAH khair :D lol m blind :/
 
Saad (سعد)

Saad (سعد)

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mominzahid said:
no it wasnt mentioned heat under reflux the compound was given and i had to give the structural formula of the product formed on reaction with hot kmno4/H+
is my answer right?
Click to expand...

If its not under reflux, still, unless they say 'the product was distilled' or 'removed' before complete reaction or something like that, then its an aldehyde. If not, its a carboxylic acid.
 
Saad (سعد)

Saad (سعد)

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RGBM211 said:
which gas deviates more from the ideal behaviour He,N2,CO2,CH4

Anyone ?
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He is inert, so no. The rest are all non-polar; but CO2 should be the answer, as it has the largest mass, and therefore the largest strength of van der Waals' forces among all of the remaining gases.

He = No. Its the closest to ideality.
CH4 = More ideal than N2.
N2 = More ideal than CO2.
CO2 = Your answer. The most deviant via the Ideal Gas Theory.
 
T

TheMan123

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Saad (سعد) said:
They will accept both, but they will be more inclined to award an A* to the one they preferred and was correct, than the one that they accepted because it was just correct.
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I think A*s are awarded based on being in the top few percent of highest marks, if two people have the same marks, and one put COOH, and the other put CO2H, they would both get the same grade. An A* just means you got a really high mark
 
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