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Chemistry: Post your doubts here!

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samir2012

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Need URGENT help with chemistry P4 applications......i already have the booklet but i found it rather incomplete. Anything else out there to give me a COMPLETE summary of the application chapters?? plz reply and help me quickly :)
 
iKhaled

iKhaled

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Silent Hunter said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_42.pdf

its question 4 (c) ? how to do it ....... by the way any tips for this paper..... its just so confusing at times :\ :\


iKhaled

or any one?
Click to expand...
hey..

first of all quote the values of the half cells from the data booklet u have Fe^3+/Fe^2+ = 0.77.... O2+H^+/H20 = 1.23 we have put this value because this is when it is oxidized under acidified conditions and for the oxidation in the alkaline conditions u have Fe(OH)3/Fe(OH)2 + OH- = -0.56 and O2+H2O/OH- = 0.40 (this value is when it is oxidized under alkaline conditions so lets calculate the E cell of both reactions

E(acid conditions) = 1.23 = 0.77 = 0.46
E(alkaline conditions) = 0.4-(-0.56) = 0.96

note that the higher the value of the E cell the more likely the reaction will occur..values below 0.3 v means a reaction may not happen or it is very slow and any value less than zero means the reaction will not happen at all..so as u can see when its in alkaline conditions it is more likely to happen..so yeah thats all

hope u got it!
 
Student12

Student12

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Can anyone explain the effect of concentration when doing the prefrentional discharge? HELPPP!
 
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Taha.Sajid

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sagar65265

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iKhaled said:
yes this is very weird :O i want the explanation of this too hmm...
Click to expand...

Think of the reaction in this way; replace all the Silicon with Carbon atoms and Chlorine atoms with Hydrogen atoms (only to visualize the reaction, not to do it in reality!) so that you get a structure that is equivalent to ethane, with 1 "C-C" bond and 6 "C-H" bonds. They are basically asking you for the enthalpy change when the "ethane" molecule splits to give two "methane" molecules, right?

The simplest way to do this is to remove the "C-C" bond and add a "Hydrogen" atom in between to each "methyl" radical to give two "methane" molecules, so if you do just that, you'll find that the only bonds needed to be broken are a "C-C" bond and a "H-H" bond.

In reality, this translates to breaking a Silicon-Silicon bond and a Chlorine-Chlorine bond, and forming two Silicon-Chlorine bonds to give the final structure!

Hope this helped!
Good Luck for all your exams!
 
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raamish

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Can anybody show me the diagram i must draw of tertiary structure of protein asked in qs 8a j08 chemistry ppr(A2). Can somebody post the picture of the diagram they drew. And in qs 8b) can somebdy explain me why we didnt use all the tripeptides given. bcuz i used all. N terminal end and C terminal end peptides i drew correct. Please help thanku :)
 
knowitall10

knowitall10

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raamish said:
Can anybody show me the diagram i must draw of tertiary structure of protein asked in qs 8a j08 chemistry ppr(A2). Can somebody post the picture of the diagram they drew. And in qs 8b) can somebdy explain me why we didnt use all the tripeptides given. bcuz i used all. N terminal end and C terminal end peptides i drew correct. Please help thanku :)
Click to expand...
Ok..we didn't use all the polypeptides given because we have to use each only once. I hope ur understanding what I'm trying to say. Try it agian and you'll get the right answer inshAllah:) As for the tertiary structure: we have to draw the types of bonding. So i think a scribble would do as long as you label any random area with hydrogen, disulfide, ionic, or hydrophobic interactions some how somewhere. I'm not uploading a photo of a drawing:
A) cz I still have to sit my butt down and pledge that I'm starting past papers once and for all-which i can't really do with the jokes thread open:p:p
B) because my drawing will display my actual IQ level...and my baisti will take place:D

Despite my uselesssness, I hope i helped you..if not, please forgive me....
:):)
 
knowitall10

knowitall10

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Another important concept not mentioned in our books:

For Diazotization and coupling reactions:
Step 1 is the reaction between phenylamine and nitrous acid (nitric (III) acid), HNO2, to give a diazonium salt. This is what we know.
But nitrous acid is unstable and has to be made in a test tube, then the phenyl amine is added:
Nitrous Acid is made using sodium nitrite (sodium nitrate(III)) and dilute HCl:
NaNO2 + HCl ------> HNO2 + NaCl and then, the production of benezene diazonium chloride takes place (which we all know about)
 
knowitall10

knowitall10

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Ring Activating groups (Substitute in the 2,4 and 6 positions): -alkyl, -OH, -NH3
Ring De-activating groups (substitute at the 3 and 5 positions): -COOH, -NO2, -carbonyl, -Cl2

If there're any more anyone wants to add to the list, or if I'm wrong, please do mention it:)
Your help is appreciated!!!
 
knowitall10

knowitall10

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Silent Hunter said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf

its Q 3 (e) (iv) ? How to do it ? :\ :\ ...... (oh man!chem and maths on same day with phy the day after is so annoying .....) :( :( :cry:

knowitall10
iKhaled

or anyone having chem p4
?

JazakAllah
Click to expand...
Assalamu 'Alaykum:)
"And when calamity befell them, they said 'To Allah do we belong and to Him shall we return"-Surah Al-Baqarah:):)
Faith, Sister, faith!!!
What's the answer? Have you tried the formula Pka=-log Ka?
 
Silent Hunter

Silent Hunter

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knowitall10 said:
Assalamu 'Alaykum:)
"And when calamity befell them, they said 'To Allah do we belong and to Him shall we return"-Surah Al-Baqarah:):)
Faith, Sister, faith!!!
What's the answer? Have you tried the formula Pka=-log Ka?
Click to expand...

JazakAllah ..... yeah i tried .... but coudlnt get it .... why cant we use the values of concentrations as it is? how to it step by step ? can you explain abit?

and by the way am a boy. :p
 
knowitall10

knowitall10

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Silent Hunter said:
JazakAllah ..... yeah i tried .... but coudlnt get it .... why cant we use the values of concentrations as it is? how to it step by step ? can you explain abit?

and by the way am a boy. :p
Click to expand...
Oh! LOL! I'm sorry :p
Can you please tell me the answer? I'll try working backwards from there:)
 
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