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Chemistry: Post your doubts here!

scouserlfc

scouserlfc

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yousef said:
For converting Methyl benzene To Chloromethyl benzene .. in the Free radical substitution ...we need
Cl2 and hf as reagent and condition .. as stated in marking scheme ..
My question is what is ( hf ) ?
Click to expand...

Which MS has this answer ?? Plz let us all know :p

hf reminds me of that formula of Energy required for electrons to jump up from one energy level to one higher than that,where h is Plancks constant and f is frequency :p :p ;)
 
yousef

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scouserlfc

scouserlfc

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danial 234 said:
and mcq no 35
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf
thnx in advance
Click to expand...

The answer is C .
1 is not one of the answers as CO does not react with O2 in atmosphere,all of these reactions which convert CO to CO2 occur in some heated environment like on a catalytic converter so 1 is not the answer.
The other Nitrogen and Sulfur both oxidize in air. Nitrogen easily goes from NO +1/2O2 ---> NO2 while SO2 you may think doesnt form SO3 without the conditions that are present in Contact process but in reality this happens. NO2 + SO2 ---> SO3 + NO . Now this NO again converts back to NO2 by NO +1/2O2 ---> NO2 . You might say that SO2 is not reacting with O2 but it is with NO2 but NO2 is a catalyst here and so in reality u can say it reacts with the O2 provided by NO2 . Theres another way u cud answer this without even caring abt whether option 3 is correct is by finding that 1 is wrong so 2 and 3 is the answer which corresponds to C.

danial 234 said:
i need help wid dis mcq no 31
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
Click to expand...

For this question again you need to prove option 1 wrong and then you have your answer. You have 1 mole of a monomer which has 6.02*10^23 particles now lets assume we polymerise these monomers and form a polymer will the polymer also be 1 mole meaning will it also have those 6.02*10^23 particles. the answer quite clearly is no because polymer is large and if u start with 1 mole of monomer you will never form 1 mole of polymers but a fraction of this. In reality you will form 1/(6.02*10^23) moles of polymer. this is option 3 and so it is correct. Once again dont wait to find if option 2 is correct since you know that 1 is wrong tick C because there is not anyway to have 3 in answer without having option 1 .

Please feel free to ask any problem regarding these explanations .:)
 
danial 234

danial 234

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scouserlfc said:
The answer is C .
1 is not one of the answers as CO does not react with O2 in atmosphere,all of these reactions which convert CO to CO2 occur in some heated environment like on a catalytic converter so 1 is not the answer.
The other Nitrogen and Sulfur both oxidize in air. Nitrogen easily goes from NO +1/2O2 ---> NO2 while SO2 you may think doesnt form SO3 without the conditions that are present in Contact process but in reality this happens. NO2 + SO2 ---> SO3 + NO . Now this NO again converts back to NO2 by NO +1/2O2 ---> NO2 . You might say that SO2 is not reacting with O2 but it is with NO2 but NO2 is a catalyst here and so in reality u can say it reacts with the O2 provided by NO2 . Theres another way u cud answer this without even caring abt whether option 3 is correct is by finding that 1 is wrong so 2 and 3 is the answer which corresponds to C.



For this question again you need to prove option 1 wrong and then you have your answer. You have 1 mole of a monomer which has 6.02*10^23 particles now lets assume we polymerise these monomers and form a polymer will the polymer also be 1 mole meaning will it also have those 6.02*10^23 particles. the answer quite clearly is no because polymer is large and if u start with 1 mole of monomer you will never form 1 mole of polymers but a fraction of this. In reality you will form 1/(6.02*10^23) moles of polymer. this is option 3 and so it is correct. Once again dont wait to find if option 2 is correct since you know that 1 is wrong tick C because there is not anyway to have 3 in answer without having option 1 .

Please feel free to ask any problem regarding these explanations .:)
Click to expand...
thanx a lot
 
NaNinG

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dunno if im interupting ny1s Question....first tym
Oct/Nov 2011 ppr 5, question #2 the graph part
im stuck at the labelling axes n plotting part of the question,as in which combination to take on both x n y axis....some1 pls help :/
 
daredevil

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NaNinG said:
dunno if im interupting ny1s Question....first tym
Oct/Nov 2011 ppr 5, question #2 the graph part
im stuck at the labelling axes n plotting part of the question,as in which combination to take on both x n y axis....some1 pls help :/
Click to expand...
link? :)
 
daredevil

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NaNinG said:
http://www.sheir.org/a-level-chemistry-51-nov2011.pdf
yeah so heres the link....Q2b,the graph part,
editing to my prev question would say that log(rate of reaction) is taken n the values r neg ....waot abt the reciprocal of AT,should i take log of that as well,
see if u can help :)
Click to expand...
no no u just have to take log for the rate of reaction!
1) first find the rate of reaction for each experiment.... [one of ur coloumns' heading is "rate of reaction (1/time) s^-1"
2) then find the log(rate of reaction) ... this will be the heading of ur next coloumn.
3) and the third coloumn will be name "1/absolute temperature K^-1 "
Then plot the 2nd and 3rd coloumns on the graph .....
Temperatuere is ur independant variable so plot 1/absolute temperature on the x axis and the log(rate of reaction) on the y axis .... which means ur graph will be of the 4th quadrant as shown in the attatchment....
P.S. log DOES NOT have units
 

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NaNinG

NaNinG

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daredevil said:
no no u just have to take log for the rate of reaction!
1) first find the rate of reaction for each experiment.... [one of ur coloumns' heading is "rate of reaction (1/time) s^-1"
2) then find the log(rate of reaction) ... this will be the heading of ur next coloumn.
3) and the third coloumn will be name "1/absolute temperature K^-1 "
Then plot the 2nd and 3rd coloumns on the graph .....
Temperatuere is ur independant variable so plot 1/absolute temperature on the x axis and the log(rate of reaction) on the y axis .... which means ur graph will be of the 4th quadrant as shown in the attatchment....
P.S. log DOES NOT have units
Click to expand...
thanx 4 the quick response dear....i'll plot n see if i got further doubts :)
 
NaNinG

NaNinG

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daredevil said:
okay sure. np :)
Click to expand...
well,im kinda gettin the intercept at x....m i doin it right,n then im completely messing up wid finding the slope....(i know,not so hard of a question,been long since i last attempted a graph question,kinda forgetting :p)...pls see through :)
 
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