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Chemistry: Post your doubts here!

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Please someone help me out in this!
I am clueless
Q9
A: initial: 2 moles of P
equilibrium: x moles of R, therefore there must be 2x moles of Q (as the ratio of Q to R is 2:1). P will be 2-x moles (we subtract moles of R from moles of P since they are the same in equilibrium)
Add them up: x + 2x + 2 - x = 2x + 2 so A is incorrect

B: initial: 2 moles of P
equilibrium: x moles of R, therefore there must be 2x moles of Q (as the ratio of Q to R is still 2:1). P will be 2-2x moles (we subtract Q from P this time because P is the same as Q in equilibrium in this equation)
Add them up: x + 2x + 2 - 2x = x + 2 so B is the correct answer
I solved this question by considering each option one at a time.
Option B)
.......2P <=> 2Q + R
I.......2...........0.......0
C....-2x........+2x....+x
E....2-2x........2x......x

2-2x + 2x + x
= 2 + x
 
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Please someone help me out in this!
I am clueless
Q14
first write down the equation
3Ba(NO3)2+10Al---->3N2+5Al2O3+3BaO

Mr of Ba(NO3)2 = 137 + 14*2 + 16* 6 = 261
looking at the mole ratio, 3 moles Ba(NO3)2 gives 3 moles N2.
moles of Ba(NO3)2= 3*10^-3
moles of N2= 3*10^-3
Moles=volume/24
Volume of N2= 0.072 dm^3
the options given are in cm^3
so 0.072*1000=72 cm^3
 
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Please someone help me out in this!
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Q.14
Equation
X(NO₃)₂ -------> XO + 2NO₂ + ½O₂

5/(X + 124) = (5 - 3.29)/(X + 16)
5(X + 16) = 1.71(X + 124)
5X + 80 = 1.71X + 212.04
5X - 1.71X = 212.04 - 80
3,29X = 132.04
[ X ≈ 40 ]

Calcium has da Mr of 40
Answer: B
 
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Q.14
Equation
X(NO₃)₂ -------> XO + 2NO₂ + ½O₂

5/(X + 124) = (5 - 3.29)/(X + 16)
5(X + 16) = 1.71(X + 124)
5X + 80 = 1.71X + 212.04
5X - 1.71X = 212.04 - 80
3,29X = 132.04
[ X ≈ 40 ]

Calcium has da Mr of 40
Answer: B
omg :cry: this is so much to do :confused: this should be a 2 mark question in paper 2 instead or may be they should have given the Mr's at least to make it a little faster
thanks i got it now :)
 
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Y
well i am not sure about 9 amd 11
25) it should be B na bcoz there are 3n carbon and 6n Hydrogen so if we balance the equation we get 9 moles of oxygen on the product side so for balancing the moles of oxygen in reactact obviously it should 4.5 because it already O2 so simple calculation for verfication --> 4.5 *2 ==> 9 so the overall equation is balanced :)
I hope you got it!
Yeah got it
Thanks :)
 
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http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Can anyone explain question number 13 ans B, 28 ans D and 35 ans B.

Questions are;
In 13 why can't A be correct.
In 28 why is D correct, I think D is correct as tertiary alcohols don't oxidise even by strong oxidising agents right?
In 35 I don't get a point.

And in 19 is NO a reducing agent or not?
Q13 i have problems with equations too >.< but here i wouldn't have gone for A
lets look at other reaction of NaOH something you might be familiar with
like this one NaOH + НСl = NaCl + Н2О
why not NaClOH2?
Sulphur dioxide is acidic its just like HCl, it is weaker though
now you are left with B C or D
when acid+ base water is release
so not D and C isn't balanced
but i guess we need to learn this kind of equations
here is a link http://www.allreactions.com/index.php/group-1a/natrium/sodium-hydroxide just have a quick look
Q28 yes you are right
Q35
Carbon monoxide is neutral, and does not react like the other two acidic gases.
 
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Q
hey in the previous page i posted some same doubts n they got solved check them out :)
Q1 i can solve tht :)
The ans is A bcz we need to find the concentration so write the eq C = N multiply V
1.0*10^-2 multiplied by 25 = 0.25
0.25/50 (the water sample) = 5*10^-3
Ans in step 2/2 =2.5*10-3
Basically wt we did was c=n*v/water sample=ans/2
Hope u got it :D
 
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Q

hey in the previous page i posted some same doubts n they got solved check them out :)
Q1 i can solve tht :)
The ans is A bcz we need to find the concentration so write the eq C = N multiply V
1.0*10^-2 multiplied by 25 = 0.25
0.25/50 (the water sample) = 5*10^-3
Ans in step 2/2 =2.5*10-3
Basically wt we did was c=n*v/water sample=ans/2
Hope u got it :D
:) yup i got all of them Hadi Murtaza solved them :D
thanks anyways
 
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