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Chemistry: Post your doubts here!

étudiante

étudiante

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Jaf said:
Can ethene be polymerized in a single reaction or no?
I remember seeing a question that tests the knowledge of this twice (maybe more) and now I can't remember the answer.
Click to expand...
wat do u mean by single reaction? :confused:
 
Scafalon40

Scafalon40

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Jaf said:
Can ethene be polymerized in a single reaction or no?
I remember seeing a question that tests the knowledge of this twice (maybe more) and now I can't remember the answer.
Click to expand...
Yes, it can:
Conditions

Temperature:about 200°C
Pressure:about 2000 atmospheres
Initiator:a small amount of oxygen as an impurity
Wanna answer my question above?
 
étudiante

étudiante

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Jaf

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Scafalon40 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q34
Ans b
Why is statement 3 false?
Ca ions will react with carbonic acid to give CaCO3, no?
Click to expand...
Carbonic acid? :confused: Where's the carbon in the question?
I don't have a solid, convincing answer for this either. The best I can come up with is that even if calcium ions react with the acids, they still remain in the solution as calcium ions (unlike the other two reactions above) and thus the equilibrium remains unaffected.
If I hadn't seen the answer to the question and it had come in an exam, I would have ticked B.
 
Scafalon40

Scafalon40

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Jaf said:
Carbonic acid? :confused: Where's the carbon in the question?
I don't have a solid, convincing answer for this either. The best I can come up with is that even if calcium ions react with the acids, they still remain in the solution as calcium ions (unlike the other two reactions above) and thus the equilibrium remains unaffected.
If I hadn't seen the answer to the question and it had come in an exam, I would have ticked B.
Click to expand...
That is exactly why I mentioned carbonic acid.
Ca ions will remain in solution as ions, because in most cases Ca salts are soluble.
But CaCO3 is not soluble, is it?
 
leadingguy

leadingguy

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usualy in paper 5 planning and analysis question 2 there is a part asking to draw construction lines
on the graph.:confused: what are these construction lines?? I think these are the lines to take two separated co-ordinates
from the graph and then we have to calculate gradient??? is this so???
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s08_qp_5.pdf
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s08_ms_5.pdf

in this paper qstn 2 part, f same is being asked can any one solve it with explanation ???
 
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Soulgamer

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CaptainDanger said:
Q-39? 1 is wrong so C is the answer but how is 2 correct?
9701_s07_qp_1.pdf
9701_s07_ms_1.pdf

Q-21??
9701_s10_ms_11.pdf

Q-37? How is 3 correct?
9701_w04_ms_1.pdf

Q-39?
9701_w08_ms_1.pdf
Click to expand...

Q39. The answer is C. Option 2 is correct as when concentrated H2SO4 is added, dehydration or elimination occurs. OH is lost from one carbon atom and H is lost from the neighboring carbon atom. Alkenes are formed. In this case, look at the OH. H can be removed from the upper carbon and from the lower carbon hence two alkenes are formed. Option 3 is valid as secondary alcohol will oxidize to a ketone.

Q21. This question is about free radical substitution. As we know in propagation step one Cl free radical takes Hydrogen form the alkane forming HCL and a alkyl free radical(the alkane chain devoid of one hydrogen in its simplier words).
Look at the three CH3's that surround the central Carbon. If even one Hydrogen is removed from them. They form the same alkyl free radical. Now look at CH3-CH2-C. Here hydrogen can be removed from either CH3 or CH2, like this

*CH2-CH2-C (* means free radical)
CH3-*CH-C
Hence 3 free radicals can form i.e *X

Q37. Remember the termination step. Two free radicals join to form a molecule. Break this molecule CH3CHCl-CHClCH3 into this CH3*CHCl and *CHClCH3. These are two free radicals formed in the propagation step. Whenever these questions come. Break the central bond and check if two free radicals are formed.


Q39. For this question you have to make all the possible free radicals formed and then break off the molecules given and check the if the free radicals match with your's. For 3 break of the molecule like this:

CH3*CH2 *CH(CH3)CH2CH3
By just looking at the last product you see the number of carbon has increased by 1 which is not possible.

btw in this question they are talking about this free radical *C3H7 formed during the propagation step.
Now look at this. This free radical can either look like this: *CH2-CH2-CH3. Lets call this 'A'

or it can look like this: CH3-*CH-CH3. Lets call this 'B'

Now in termination we combine two free radicals so what combination can we form with these two?
Lets see, we can get A+A
B+A
and B+B
Note that just combine the free radicals like in B+A
CH3
|
*CH + *CH2-CH2-CH3 like this
|
CH3

Hope I helped :)
 
USMAN Sheikh

USMAN Sheikh

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Soulgamer said:
Q39. For this question you have to make all the possible free radicals formed and then break off the molecules given and check the if the free radicals match with your's. For 3 break of the molecule like this:

CH3*CH2 *CH(CH3)CH2CH3
By just looking at the last product you see the number of carbon has increased by 1 which is not possible.

btw in this question they are talking about this free radical *C3H7 formed during the propagation step.
Now look at this. This free radical can either look like this: *CH2-CH2-CH3. Lets call this 'A'

or it can look like this: CH3-*CH-CH3. Lets call this 'B'

Now in termination we combine two free radicals so what combination can we form with these two?
Lets see, we can get A+A
B+A
and B+B
Note that just combine the free radicals like in B+A
CH3
|
*CH + *CH2-CH2-CH3 like this
|
CH3

Hope I helped :)
Click to expand...

IN this i got all ur point bt why did we eleminate ans 3 as it also has 6 carbons (ur statement ) why ???? kindly clarify it waiting for ur reply ????
 
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Soulgamer

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you cant break them off equally as they won't be a perfect free radical. Its kinda hard to explain here. practice making termination products,
 
DragonCub

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JD REBORN said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Q34
Click to expand...
Mercury level rises in the right-hand limb, this means that the pressure in R increases.
In 1, according to Le Chartelier's Principle, a rise in temperature would favour the forward reaction, thus the equilibrium position shifts closer to the products. In the reaction, every 1 mole of reactants forms 2 moles of products. As more reactants are converted to products, the total mole number (reactant plus product) in R gets larger, so the pressure becomes higher. 1 is correct.
In 2, since the mole number of reactants and products in the reaction is the same, where the equilibrium shifts to shall have no effect on the pressure since the total mole number stays unchanged. 2 is incorrect.
In 3, there is definitely no change since the two bulbs are filled with the same gas. 3 is incorrect.
Answer is D.
 
Scafalon40

Scafalon40

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Saad Sarfraz said:
Man when caco3 cant be formed as the hydroxide ions are neutralised by the acid furthermore if caco3 is formed itll make the solution alkaline and thus preventing tooth decay so C cant be the option as they asked why tooth enamel is dissolved more readily when saliva is acidic :)
Click to expand...
Carbonic acid has the formula H2CO3
The carbonate ion is what reacts with the Ca ion, it has nothing to do with the proton, as you say...
 
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