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Chemistry: Post your doubts here!

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I think C. (Na can react with both -CO2H and OH, there is 3 in C so -3 but NaOH only with -CO2H so -1) I'm not sure though as I didn't revise from a long time.

I have a question : http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf

Number 9, the answer is D but I don't know how to get it.

i didn't quite get it :/
I get the part where Na reacts with both -CO2H and -OH but after that i don't get it ..
and
yes it is C.
 
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I did draw an ICE, but I don't understand how they got 4(b-x) at the bottom. I just got (b-x)^2

If x moles of HI dissociates, then H2 and I2 gain x/2 moles (mole ratio 2:1). Amount of HI left = b-x. Now, Kp=p(I2)xp(H2)/p(HI)^2. [(x/2)*(x/2)]/(b-x)^2 will simplify to D.
 
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i didn't quite get it :/
I get the part where Na reacts with both -CO2H and -OH but after that i don't get it ..
and
yes it is C.

Na reacts with both groups to form ONa- which is charged (-1), there is three so 3 ONa- will form

NaOH is the same thing but with only -CO2H group.
 
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Can somebody explain those?
cccccc.png
q23 - What do they mean by mole fraction is 0.5? How does that affect the identity of X?
cccccc.png
q33 - Why is it 1? Why isn't it 2?
cccccc.png
q35 -How does nitrogen undergo a redox reaction? It only goes from +3 to 0.
cccccc.png
q31 - Where do ionic bonds form? All elements present are non metals?
 
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Its C aluminum carbide
Al4C3+12H2O------->4AL(OH)3+3CH4

CH4+2O2----->2H2O+CO2

Moles of CO2=72/24000=0.003
moles of Al4C3= 0.003/3=0.001 (as ratio of Al4C3:CH4 is 1:3)
now mass of Al4C3= 0.001 into (27+27+27+27+12+12+12)=0.144g
____________________________________________________________________________
dis was done by jiyad ahsan (y)


Righttt i understood the process but how do we know its Al4C3? :/
 
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Righttt i understood the process but how do we know its Al4C3? :/

I did it a bit differently and got the correct answer as well.

72/24000 = 0.003 moles of carbon dioxide.
mass of carbon = 0.003 * 12 = 0.0036g.
0.144 - 0.0036 = 0.108 g of X (Aluminium) left.
0.108/27 = 0.004 of Aluminium
Ratio of Al to C
0.004 : 0.003
4 : 3
therefore it is Al4C3
 
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