Chemistry: Post your doubts here!

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Anyone explain these two questions to me please? Thanks first! :Yahoo!:
FYI, the answer given for Q10 is D whereas for Q11 is D.
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milkboyz said:
Anyone explain these two questions to me please? Thanks first! :Yahoo!:
FYI, the answer given for Q10 is D whereas for Q11 is D.
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Making the solution more alkaline means increasing the cocentration of hydroxide ions. To use up the extra hydroxide ions the equilibrium would shift towards the backwards reaction and produce more V2+ ions, so D will be in favour of it.

I still can't see Q11.
 
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WTH.....examiner report and ms are contradictory ....examiner report is ryt???or ms???
 

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hassam said:
WTH.....examiner report and ms are contradictory ....examiner report is ryt???or ms???

The examiner report is correct. You have to find the mass of CaSO3 lost. So you'll find the moles of CaSO4 and since the mole ratio of CaSO3 to CaSO4 is 1:1, you can use the same moles to calculate mass of CaSO3.
 
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chemistry paper 3 , confused about the mark scheme language ! :(
please help if you can .

I am using ABCs for ions here.

A + ..... (arrow) 5 B + ....

i. No. of moles of A = n = v * c ( v and c given ) = 0.000240
ii. No. of moles of B = 5 * 0.000240 = 0.00120

AND THE MARK SCHEME READS :
ans to (i) × 5
and
ans to (ii) × 10
with no additional steps



Confusion : "ans to (ii)" given IN the answer to (ii) ?? What the red line mean ??

<The qp and the ms attached.>
 

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started AS..help needed :(
Q- Use ur knowledge of chemistry to suggest a method of removing the iron oxide coating from the sand grains that might be applied on an industrial scale.Justify ur answer. (3)

So sand grains are silicon(IV)oxide...ummm ...So are we goin to titrate them ? :unknown:
 
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Chemistry Equilibrium question pwease help !

Can anyone pretty please help me with this question ? ^.^
coz seriously , its driving me crazy >-> :%) :%)

A mixture contains 0.5 mol of ethanoic acid ,0.5 mol of ethanol,0.1 mol of ethyl ethanoate and 0.1 mol of water was set up and allowed to come to an equilibrium at 298K.
CH3CO2H + C2H5OH = CH3CO2C2H5
Calculate the amount of ,in moles, of each of the substance present at the equilibrium, if Kc is 4.0.
Note-(the reaction is reversible)
 
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Re: Chemistry Equilibrium question pwease help !

KurayamiKimmi said:
Can anyone pretty please help me with this question ? ^.^
coz seriously , its driving me crazy >-> :%) :%)

A mixture contains 0.5 mol of ethanoic acid ,0.5 mol of ethanol,0.1 mol of ethyl ethanoate and 0.1 mol of water was set up and allowed to come to an equilibrium at 298K.
CH3CO2H + C2H5OH = CH3CO2C2H5
Calculate the amount of ,in moles, of each of the substance present at the equilibrium, if Kc is 4.0.
Note-(the reaction is reversible)

which year queestion is this????
 
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I.A= Initial Amount
A.R= Amount Reacted
A.F= Amount Formed
E.A= Equilibrium amount

Btw, your equation is wrong, you have missed out H2O and that mus have been the cause of problems.
 

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Your concept for this molecular energy distribution is not correct. With a catalyst, the number of molecules possessing energy that is equal to or more than the activation energy also increase. In this graph the Y axis represent the number of molecules, not the energy, and line X is the activation energy beyond which there are more molecules than Y. So the answer is C.
 
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16 and 17 are simple concepts that you need to learn, just check out the respective chapters to clear them up.

Q18. For this question, you just need to calculate the mole ratios and convert them to mass as following.
Write the respective equations: NaCl->Na+ + Cl-
2Cl- -> Cl2
2H+->H2
Na + OH- -> NaOH

Calculate moles of NaCl= 58.5/58.5=1kmoles
Moles of Cl2 formed by 1:2 ratio=0.5kmoles according to the mole ratio, so 0.5*71(mass of Cl2)=35.5kg
Moles of H2 formed by 1:2 ratio=0.5, 0.5*2=1Kg
Moles of NaOH due to 1:1 ratio=1kmole, 1*40=40Kg.
So the answer is A.

Hope this helped. :)
 
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