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Chemistry: Post your doubts here!

KZW

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17
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In the question, they give you the initial number of moles of each substance present in the equilibrium. All of them should change by the same amount since the ratios are all 1:1, so lets say the change in the number of moles is x.

Therefore, the equilibrium concentrations are
(0.70–x) for CO2 and H2, and (0.30+x) for CO and H2O.

You know Kc is 1.44, and Kc = [products]/[reactants] yes? So pop those values in, and solve as a quadratic.
1.44= (0.30+x)^2 /(0.70-x)^2 etc, and solve for x.

It should give you x= 0.25, and so you substitute it back into the (0.70-x) for CO2 and H2, and (0.30+x) for CO and H2O.


i) The 30cm^3 is the excess oxygen that wasn't reacted with A.
ii) CO2 is the only other gas that is produced, so it should be quite obvious.
iii) Total volume of gas left after combustion = 40cm^3. If 30cm^3 is oxygen, then 10cm^3 should be CO2.
iv) Original volume of oxygen=50cm^3, and there is 30cm^3 of oxygen left. Therefore volume of oxygen reacted is 20cm^3 :)

Yea i did type it wrong -_- its Q4 (b) (v) What is the relationship between the repeat unit of polymer D and the repeat unit of PVA.

Basically they are referring to the number of each element in the monomers. They both have the same amount of each element, so you could say they're isomers, or they have the same molecular formula.
 
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In the question, they give you the initial number of moles of each substance present in the equilibrium. All of them should change by the same amount since the ratios are all 1:1, so lets say the change in the number of moles is x.

Therefore, the equilibrium concentrations are
(0.70–x) for CO2 and H2, and (0.30+x) for CO and H2O.

You know Kc is 1.44, and Kc = [products]/[reactants] yes? So pop those values in, and solve as a quadratic.
1.44= (0.30+x)^2 /(0.70-x)^2 etc, and solve for x.

It should give you x= 0.25, and so you substitute it back into the (0.70-x) for CO2 and H2, and (0.30+x) for CO and H2O.



i) The 30cm^3 is the excess oxygen that wasn't reacted with A.
ii) CO2 is the only other gas that is produced, so it should be quite obvious.
iii) Total volume of gas left after combustion = 40cm^3. If 30cm^3 is oxygen, then 10cm^3 should be CO2.
iv) Original volume of oxygen=50cm^3, and there is 30cm^3 of oxygen left. Therefore volume of oxygen reacted is 20cm^3 :)



Basically they are referring to the number of each element in the monomers. They both have the same amount of each element, so you could say they're isomers, or they have the same molecular formula.

Ok. but what i dont get is why they are not the same empirical formula? Because if 2 compounds have the same molecular fomula, doesnt that mean they have the same empirical formula as well >< confused.
 
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Chemistry help !, I couldn't find information related to that dicarboxylic acid reacts with little c.H2SO4 to give a cyclic compound, in neither AS level nor A2 Level. It is such a totally weird question.

Question 5 (d) (i)

Could you guys plz help me?
 

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100

Messages
18
Reaction score
4
Points
13
In the question, they give you the initial number of moles of each substance present in the equilibrium. All of them should change by the same amount since the ratios are all 1:1, so lets say the change in the number of moles is x.

Therefore, the equilibrium concentrations are
(0.70–x) for CO2 and H2, and (0.30+x) for CO and H2O.

You know Kc is 1.44, and Kc = [products]/[reactants] yes? So pop those values in, and solve as a quadratic.
1.44= (0.30+x)^2 /(0.70-x)^2 etc, and solve for x.

It should give you x= 0.25, and so you substitute it back into the (0.70-x) for CO2 and H2, and (0.30+x) for CO and H2O.



i) The 30cm^3 is the excess oxygen that wasn't reacted with A.
ii) CO2 is the only other gas that is produced, so it should be quite obvious.
iii) Total volume of gas left after combustion = 40cm^3. If 30cm^3 is oxygen, then 10cm^3 should be CO2.
iv) Original volume of oxygen=50cm^3, and there is 30cm^3 of oxygen left. Therefore volume of oxygen reacted is 20cm^3 :)



Basically they are referring to the number of each element in the monomers. They both have the same amount of each element, so you could say they're isomers, or they have the same molecular formula.




thanks KZW but ur not suposed to do quadratic its given in the sylabus! but i found out that u have to cancel the the power by adding a square root on both side :)
but anyway thanks for ur help!!!!
 

100

Messages
18
Reaction score
4
Points
13
In the question, they give you the initial number of moles of each substance present in the equilibrium. All of them should change by the same amount since the ratios are all 1:1, so lets say the change in the number of moles is x.

Therefore, the equilibrium concentrations are
(0.70–x) for CO2 and H2, and (0.30+x) for CO and H2O.

You know Kc is 1.44, and Kc = [products]/[reactants] yes? So pop those values in, and solve as a quadratic.
1.44= (0.30+x)^2 /(0.70-x)^2 etc, and solve for x.

It should give you x= 0.25, and so you substitute it back into the (0.70-x) for CO2 and H2, and (0.30+x) for CO and H2O.



i) The 30cm^3 is the excess oxygen that wasn't reacted with A.
ii) CO2 is the only other gas that is produced, so it should be quite obvious.
iii) Total volume of gas left after combustion = 40cm^3. If 30cm^3 is oxygen, then 10cm^3 should be CO2.
iv) Original volume of oxygen=50cm^3, and there is 30cm^3 of oxygen left. Therefore volume of oxygen reacted is 20cm^3 :)



Basically they are referring to the number of each element in the monomers. They both have the same amount of each element, so you could say they're isomers, or they have the same molecular formula.




thanks KZW but ur not suposed to do quadratic its given in the sylabus! but i found out that u have to cancel the the power by adding a square root on both side :)
but anyway thanks for ur help!!!!
 
Messages
887
Reaction score
466
Points
73
Chemistry help !, I couldn't find information related to that dicarboxylic acid reacts with little c.H2SO4 to give a cyclic compound, in neither AS level nor A2 Level. It is such a totally weird question.

Question 5 (d) (i)

Could you guys plz help me?

lets see look buddy there are two possible reactions to this
1.conc sulfuric acid will form an alkene but the number of C atoms will remain the same
2. conc sulfuric acid is also used for esterification esterification will increase the number of C atoms wont they.
well the examiner says there is a cyclic compound meaning there will be minimum of 6 C atoms
he didnt provide us with an acid or alcohol ?? well he doesnt had to as X is both alcohol and acid
 
Messages
887
Reaction score
466
Points
73
Chemistry help !, I couldn't find information related to that dicarboxylic acid reacts with little c.H2SO4 to give a cyclic compound, in neither AS level nor A2 Level. It is such a totally weird question.

Question 5 (d) (i)

Could you guys plz help me?

lets see look buddy there are two possible reactions to this
1.conc sulfuric acid will form an alkene but the number of C atoms will remain the same
2. conc sulfuric acid is also used for esterification esterification will increase the number of C atoms wont they.
well the examiner says there is a cyclic compound meaning there will be minimum of 6 C atoms
he didnt provide us with an acid or alcohol ?? well he doesnt had to as X is both alcohol and acid
 
Messages
14
Reaction score
0
Points
1
lets see look buddy there are two possible reactions to this
1.conc sulfuric acid will form an alkene but the number of C atoms will remain the same
2. conc sulfuric acid is also used for esterification esterification will increase the number of C atoms wont they.
well the examiner says there is a cyclic compound meaning there will be minimum of 6 C atoms
he didnt provide us with an acid or alcohol ?? well he doesnt had to as X is both alcohol and acid


Just wondering, then why does it form a cyclic compound rather than straight line ester?
 
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Points
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Just wondering, then why does it form a cyclic compound rather than straight line ester?
actually for a straight line ester u need one alcohol plus one carboxyllic acid. here as we take two X each has alcohol carboxyllic acid . alcohol and carboxyllic acid react forming an ester group two ester group isnt that hard see
 
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Can anyone please tell me where i can find the marking scheme for chemistry paper 1 5070 ? :)
Thanks.
 
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140
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conversion of indigo (C16H10N2O2) to -> C16H28N2O2
total of 18 H gained, so must be 9H2 in the reactants
ie. full equation something like this C16H10N2O2 + 9H2 -> C16H28N2O2

next we calculate mol of indigo
n(C16H10N2O2)= 2.5 / (12x16 +10 + 2x14 + 2x16) = 2.5/262 = 9.54x10^-3 mol
and we know n(H2) = 9n(C16H10N2O2) from above equation so n(H2) = 9 x 9.54x10^-3 = 0.0859 mol

under room temperature and pressure, 1mol of gas occupies 24dm^3 volume so
V(H2) = 24 x 0.0859 = 2.06 dm^3
 
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25
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explain the following:
a) why are most metals strong ,but ionic solids are brittle?
b) why is an alloy of copper and tin stronger than either copper or tin alone?
 
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561
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Points
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conversion of indigo (C16H10N2O2) to -> C16H28N2O2
total of 18 H gained, so must be 9H2 in the reactants
ie. full equation something like this C16H10N2O2 + 9H2 -> C16H28N2O2

next we calculate mol of indigo
n(C16H10N2O2)= 2.5 / (12x16 +10 + 2x14 + 2x16) = 2.5/262 = 9.54x10^-3 mol
and we know n(H2) = 9n(C16H10N2O2) from above equation so n(H2) = 9 x 9.54x10^-3 = 0.0859 mol

under room temperature and pressure, 1mol of gas occupies 24dm^3 volume so
V(H2) = 24 x 0.0859 = 2.06 dm^3

thanks...:)
 
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Is there anyone appearing As chemistry 9701/22 this oct-nov 2013?? The exam in my country got postponed for some reason and I guess they'll take it after two days. So, if any of you have given this paper or any variant 21 or 23, can u post here the questions u remember? a slight help would be remembered forever. I AM A VERY WEAK STUDENT, plz someone help me:(
 
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140
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5bi
production of O2
given cathode equation from question: 2H2O(l) + 2e– -> H2(g) + 2OH–(aq), find E value in data booklet = -0.83V
now we find anode equation, we know the reaction was neutral at the start so reactant must be ph7 (H2O) so must be O2 + 4H+ + 4e– ⇌ 2H2O = +1.23V
E = (1.23 – (–0.83)) = 2.06V

5bii
cathode reaction is still the same 2H2O(l) + 2e– H2(g) + 2OH–(aq) -0.83V
anode for production of Cl2 Cl 2 + 2e– -> 2Cl – +1.36V
Eo = (1.36 – (–0.83)) = 2.19V

5ci
Cl2:O2 ratio as [Cl-] increases
we look at the the equation from production of O2 and Cl2
in production of O2, Cl- concentration is not related so there will be no change
in production of cl2, if you increase Cl- concentration equilibrium of Cl 2 + 2e– -> 2Cl – shifts back to the reactants (left) so E value decrease (less positive)

5cii
same as part i, because [Cl-] increase, equilibrium shifts to reactants , E value decrease so more Cl2 is formed
O2 does not change
so Cl2(g) : O2(g) increases
 
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