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Biology; Chemistry; Physics: Post your doubts here!

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For question 6 see that the expt 2 and 3 the end point of volume evolved is same thus the number of moles ought to be the same only the rates of reaction are diffrent.
The 2009 paper.
 
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I dont know the middle one but the last one is Hydrogen because it's lighter and would diffuse into the porus pot as the it's Mr is 2. The lightest would diffuse the fastest.
 
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Q6 Paper 09
a) In experiment 1 the number of moles of acid was less than in experiment 2:
1 mole occupies 1 dm space. So moles are directly proportional to volume. More volume = More moles but exp 1 has more volume so it has more moles than exp 2

(b)In experiment 3 the calcium carbonate was more finely powdered than in experiment 1
The amount was larger thus the volume was larger. We have no evidence that fine power was used.

(c)In experiments 2 and 3 the number of moles of acid was the same.
Same Volume = Same Moles

(d) In experiment 3 the concentration of the acid was higher than in experiment 2.
Wrong as it's rate of reaction is lower than exp 2.
 
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Q6 Paper 09
a) In experiment 1 the number of moles of acid was less than in experiment 2:
1 mole occupies 1 dm space. So moles are directly proportional to volume. More volume = More moles but exp 1 has more volume so it has more moles than exp 2

(b)In experiment 3 the calcium carbonate was more finely powdered than in experiment 1
The amount was larger thus the volume was larger. We have no evidence that fine power was used.

(c)In experiments 2 and 3 the number of moles of acid was the same.
Same Volume = Same Moles

(d) In experiment 3 the concentration of the acid was higher than in experiment 2.
Wrong as it's rate of reaction is lower than exp 2.
Thank you for explaining but read the statement of the question which says equal volume of both HCl and CaCO3 was used then how can the vloume of hydrogen in experiment 1 be greater?
 
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I dont know the middle one but the last one is Hydrogen because it's lighter and would diffuse into the porus pot as the it's Mr is 2. The lightest would diffuse the fastest.
I get that it would diffuse fastest but what's aabout the bubbles in the water and that whole apparatus? :/
 
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I get that it would diffuse fastest but what's aabout the bubbles in the water and that whole apparatus? :/
Hydrogen produces bubbles doesnt it? Plus it's a porous pot where the air is present. Therefore light gases tend to move into the porous pot, would be carried by the long tube thingy due to diffusion and produce bubbles in water. It's more of a tukka question if you notice. :p I guess.
 
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http://papers.xtremepapers.com/CIE/Cambridge International O Level/Chemistry (5070)/5070_s07_qp_4.pdf
Question 8 part L. Why are we taking the Mr of CxHy as 14? When are we supposed to do that?
What we do is that we take the mr of all the elements known and skip the CxHy. The mr of the rest of the compound is 90. We've already cacluated the total mr so we subtract 90 from 118. It becomes 28. So you can just calculate it yourself that in 28, there can be two carbon atoms that make 24 and 4 hydrogen atoms hence 28.
Edited. I made a mistake in calculation. Mahnoorfatima
 
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Why do we use electric heater during distillation of Methylbenzene and benzene? is it because they are flammable or that to provide uniform heating?
 
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Anyone please explain all. :/


Q4) 5.00 g of an organic compound G contains 2.73 g of carbon, 0.45 g of hydrogen, and 1.82 g
oxygen.
[Ar: H,1; C,12; O,16]
Its empirical formula is
(a) CHO
(b) CH4O
(c) C2H4O
(d) C2H2O

Its easy. We have to find the emprical formula.

2.73 carbon, divide by mr of carbon which is 12, we get 0.23.
0.45 divided by mr of hydrogen which is 1 so we get 0.45.
1.92 oxygen, similarly divided by its mr of 16, we get 0.11.

Now we have
C H O
0.23 0.45 0.11

Now divide it by the smallest number of the three which here is 0.11

o.23/o.11 = 2
0.45/o.11= 4
o.11/o.11 = 1

Therefore emperical formula is C2H4O.
 
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Thank you for explaining but read the statement of the question which says equal volume of both HCl and CaCO3 was used then how can the vloume of hydrogen in experiment 1 be greater?
They don't mention concentration of the three equal volumes. CaCO3 is in excess as well.
Experiment 1: Low Conc as volume of gas is least.
Exp 2: Same as 1 as volume of gas is same but rate of reaction is more.
Exp 3: Very High as same volume of HCL gives very high volume of gas.
 
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Q4) 5.00 g of an organic compound G contains 2.73 g of carbon, 0.45 g of hydrogen, and 1.82 g
oxygen.
[Ar: H,1; C,12; O,16]
Its empirical formula is
(a) CHO
(b) CH4O
(c) C2H4O
(d) C2H2O

Its easy. We have to find the emprical formula.

2.73 carbon, divide by mr of carbon which is 12, we get 0.23.
0.45 divided by mr of hydrogen which is 1 so we get 0.45.
1.92 oxygen, similarly divided by its mr of 16, we get 0.11.

Now we have
C H O
0.23 0.45 0.11

Now divide it by the smallest number of the three which here is 0.11

o.23/o.11 = 2
0.45/o.11= 4
o.11/o.11 = 1

Therefore emperical formula is C2H4O.
Thank you so much for explaining but this is not what I asked for. :p
 
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They don't mention concentration of the three equal volumes. CaCO3 is in excess as well.
Experiment 1: Low Conc as volume of gas is least.
Exp 2: Same as 1 as volume of gas is same but rate of reaction is more.
Exp 3: Very High as same volume of HCL gives very high volume of gas.
they say in excess to equal masses. Though CaCO3 was in excess but the amount was same in all three experiments.
 
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