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Chemistry: Post your doubts here!

iKhaled

iKhaled

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Oliveme said:
But why do we take the highest and lowest values?
Thank you.
Click to expand...
see in order for a reaction to occur the difference between the electrode potential of the two should be higher than 0.30v..if the value of the difference is below 0.3 a reaction may or may not occur but it will be impossible to notice so better to take the lowest one and the highest one. if we took the first 2 the top equation will go in the right direction and the second equation will go in the backward direction which means we will not have a nitrogen containing product at the end of the reaction. if we took the last 2 the difference in the electrode potential value is way less than 0.3v and also notice that the second equation will go in the backward direction and the third(last) equation will go in the forward direction which means both HNO3 from each reaction will cancel out just like how we cancel electrons so the best 2 to use is the top one and the bottom one because they have a value higher than 0.3 and also u r sure of having a nitrogen containing product at the end of the reaction.

i hope u do got it!
 
iKhaled

iKhaled

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iKhaled said:
see in order for a reaction to occur the difference between the electrode potential of the two should be higher than 0.30v..if the value of the difference is below 0.3 a reaction may or may not occur but it will be impossible to notice so better to take the lowest one and the highest one. if we took the first 2 the top equation will go in the right direction and the second equation will go in the backward direction which means we will not have a nitrogen containing product at the end of the reaction. if we took the last 2 the difference in the electrode potential value is way less than 0.3v and also notice that the second equation will go in the backward direction and the third(last) equation will go in the forward direction which means both HNO3 from each reaction will cancel out just like how we cancel electrons so the best 2 to use is the top one and the bottom one because they have a value higher than 0.3 and also u r sure of having a nitrogen containing product at the end of the reaction.

i hope u do got it!
Click to expand...
i am mistaken above in the part of the first 2 equations. if u used equation one and 2 a reaction will take place too and u will get a nitrogen containing product..but u can't take equation 2 and 3 since the nitrogen product will cancel each other and also the value of their electrode potential difference is waaayyy less than 0.3v. its recommended that u try to make the difference of the 2 electrode potential as high as u can thats why reaction one and 3 is the best to take but taking reaction one and 2 has no problem too.
 
Oliveme

Oliveme

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iKhaled said:
i am mistaken above in the part of the first 2 equations. if u used equation one and 2 a reaction will take place too and u will get a nitrogen containing product..but u can't take equation 2 and 3 since the nitrogen product will cancel each other and also the value of their electrode potential difference is waaayyy less than 0.3v. its recommended that u try to make the difference of the 2 electrode potential as high as u can thats why reaction one and 3 is the best to take but taking reaction one and 2 has no problem too.
Click to expand...
Thank you so, so much. That was probably the best explanation I could have gotten. Thanks once again. (y)
 
Tasneem Infinity

Tasneem Infinity

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I am new to this site..can anyone tell me where do we go to login? That login tray just pops in any time and i have to wait for its arrival...!.

Nd where to we go to post questions?

n can i get a complete list of resources for this coming up cambridge igcse examination in oct/nov?

m sorry 2 whoevr if m posting dis in d wrong place...
 
Soldier313

Soldier313

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Tasneem Infinity said:
I am new to this site..can anyone tell me where do we go to login? That login tray just pops in any time and i have to wait for its arrival...!.

Nd where to we go to post questions?

n can i get a complete list of resources for this coming up cambridge igcse examination in oct/nov?

m sorry 2 whoevr if m posting dis in d wrong place...
Click to expand...
it's quite simple, when you just type http://www.xtremepapers.com/community

Screen shot 2013-05-15 at 1.03.46 PM.png

this is a screenshot of the page, then you just tap on the log in or sign up, and you can put in your username and password

you get pst papers, and other resources by tapping on the home button, that's towards the left of the screenshot,
or you can just type in the search box on the right, for whatever you require.

For more help, you just go to the help forums: https://www.xtremepapers.com/community/forums/help-ideas-suggestions.4/

:)

PS: sorry to the mods for going off topic, just helping out a newbie :)
 
Soldier313

Soldier313

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Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot :)
 
biba

biba

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KurayamiKimmi said:
Asalamo Alaikum everyone!
Can anyone please help me with this A2 question:-
Q 3 part d(ii) of this paper?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_42.pdf
Click to expand...
okay in d ( i ) we found the molecular formula as SiCl3... now in d ii the first Mr is 133 that is SiCl3 cuz (28+35+35+35=133)... now we have 247... let's see 247-133=114, we knw the formula of 133, find the formula of 114 which is SiCl2O (28+35+35+16=114) ,add both formulas of 133 and 114 i.e SiCl3 + SiCl2O = Si2Cl5O(Cl 3Si-O-SiCl 2 )
do the same for 263.. 263 - 247 = 16,meaning only O is added hence add Si2Cl5O + O= Si2Cl5O2 (.Cl 3Si-O-SiCl 2-O)
..
 
biba

biba

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Soldier313 said:
Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot :)
Click to expand...
I can draw the structures but I dunno how to explain them to you..
I made some notes on NMR u can check it out , may b it helps :
 

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sagar65265

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Soldier313 said:
Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot :)
Click to expand...

For the 2007 specimen, 9 (b) & (d):

The first clue is the empirical formula calculated for the earlier part question; it is C8H10O, and since the number at the top of each peak, that describe the number of hydrogen atoms per group, add up to 10, it looks like the molecular formula is also C8H10O. This is very useful information!

The first two peaks from the right hand side of the graph are typical of a CH3CH2 group;
i) the first group has 3 hydrogen atoms and it is a triplet, so it is attached to a carbon atom with 2 hydrogen atoms attached to it.
ii) the seconds peak is a carbon atom with 2 hydrogen atoms attached to it and since it is a quadruplet, it is attached to a carbon atom with 3 hydrogen atoms on it; the two of them describe each other!
The last peak describes 4 equivalent hydrogen atoms and is at about 6.8 ppm; this indicates either a phenol group or a benzene group (this can be confirmed by checking the empirical formula; subtracting the 2 carbons and 5 hydrogens from the CH3CH2 group gets us a remainder of C6H5O - this looks an awful lot like a phenol group with one extra substituted side chain!)

From this information, it looks like the molecule is an ethyl group attached to a phenyl group; since the 4 phenyl hydrogens are equivalent, the -OH group and the -CH2CH3 seem to be attached to opposite ends of the molecule - and that should be the answer!

d) If the isomer shows no change after adding D2O, this means that there is no -OH group in this isomer, else the third peak on this NMR spectrum would have disappeared. Since there is no change, there is no -OH OR -COOH group or any other ionising group, and there are only two groups that follow this pattern of containing only 1 oxygen - a carbonyl group or an ethoxy group.

Since a carbon double bonded to an oxygen can never occur on a benzene ring and this occurring on the side chain would remove some hydrogens, voiding the "isomer" statement, the compound is an ethoxy compound, mostly ethoxybenzene (or whatever the actual name is, sorry i'm not being more specific!). The observable changes would have to be the loss of the third peak from the right, the peak caused by a methoxy group (at about 3.3 - 4.0 ppm) and the change in splitting of the CH3 peak, I guess.

I'll try out the others ASAP, sorry this is all i've been able to do so far!

Hope this helped!
Good Luck for all your exams!
 
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