Chemistry: Post your doubts here!

[knowitall10]

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

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
B) because my drawing will display my actual IQ level...and my baisti will take place

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

 

[knowitall10]

@EVERYONE WHO'S IN A2!!!!!!!!!!

 

[knowitall10]

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]

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!!!

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/... 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 .....)

knowitall10
iKhaled

or anyone having chem p4
?

JazakAllah

 

[knowitall10]

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 .....)

knowitall10
iKhaled

or anyone having chem p4
?

JazakAllah

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]

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?

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.

 

[knowitall10]

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.

Oh! LOL! I'm sorry
Can you please tell me the answer? I'll try working backwards from there

 

[veteran]

Calculate the pH of the buffer formed when 10.0 cm3 of 0.100 mol dm–3 NaOH is added to 10.0 cm3 of 0.250 mol dm–3 CH3 CO2H, whose pKa
= 4.76.

 

[knowitall10]

Calculate the pH of the buffer formed when 10.0 cm3 of 0.100 mol dm–3 NaOH is added to 10.0 cm3 of 0.250 mol dm–3 CH3 CO2H, whose pKa
= 4.76.

That's the same question the brother posted right before you Can you please tell me the answer? I'll try to work backwards.

 

[veteran]

That's the same question the brother posted right before you Can you please tell me the answer? I'll try to work backwards.

The examiner used this pH = 4.76 + log (0.05/0.075) and answers is 4.58 or 4.6

 

[knowitall10]

The examiner used this pH = 4.76 + log (0.05/0.075) and answers is 4.58 or 4.6

Yes. He used the formula:
pH= pKa +log ( [salt]/[acid] )

 

[veteran]

Yes. He used the formula:
pH= pKa +log ( [salt]/[acid] )

Yeah but what i dont get is how he came up with those values, can you give be a step by step solution? i would really appreciate ir.

 

[Silent Hunter]

Yes. He used the formula:
pH= pKa +log ( [salt]/[acid] )

yeah! how does the values of the green part came?

 

[knowitall10]

Yeah but what i dont get is how he came up with those values, can you give be a step by step solution? i would really appreciate ir.

Ok...help me out with this alright? Let's work it together:
What i get is that it will form the salt sodium ethanoate....what next?

 

[knowitall10]

veteran and Silent Hunter If i don't do what the mark scheme says, i'm getting a pretty much close answer of 4.36.

 

[raihan1904]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
Q2 part (b) explanation plzz
plzz tag me in the explanation of this question!!
thanks
best ov luck evry1

 

[veteran]

Ok...help me out with this alright? Let's work it together:
What i get is that it will form the salt sodium ethanoate....what next?

i think i got it. I calculated the equilibrium concentration between the acid and the salt being produced. I assumed that the acid will produce only half of the CH3COO- ions so moles =10/1000 x 0.250 and divide the answer by 2 to get the moles of the ethanoate ion. Then do the same for NaOH moles 10/1000 x 0.1.. and divide the answer by two to get the value of sodium ions. That is the same as the concentration of sodium ethanoate. Then get concetration of the salt. Subtract the moles of salt from that of the moles of ethanoate ions and then use that formula. I hope that makes some sense.

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
Q2 part (b) explanation plzz
plzz tag me in the explanation of this question!!
thanks
best ov luck evry1

Here-
iKhaled explained this earlier.

C40H82 -------> C16H34 + C24H48

a good way of explaining this question is to see how many bonds u have before the reaction and how many bonds u have after the reaction for example...

before the reaction we have 39 C-C bonds and 82 C-H bonds...after the reaction we have 15+22 = 37 C-C bonds and one C=C bond and 48+34 = 82 C-H bonds

so we have broken 39-37 = 2 C-C bonds and formed 1 C=C bond
bond broken = 2 x 350 = +700
bond formed = -610 = -610
enthalpy = + 90 kjmol^-1 !!

thats it..the mark scheme got the answer as +180 because he has used a different equation in cracking instead of forming C24H48 as our alkene he formed 2 moles of C12H24 and in that way we will have 2 C=C bonds instead of one and 4 C-C bonds instead of 2 because one mole of C12H24 contains 1 C=C bond and 10 C-C bonds so the 2 moles will contain 20 c-c bond and 2 C=C bonds 20+15= 35 so 39-35 = 4 C-C bonds so we had broken 4 C-C bonds and formed 2 C=C bond..in that way u will get 180 kj per mole instead of 90..both are correct though depending on the equation u made

i hope u got it!!

 

[knowitall10]

i think i got it. I calculated the equilibrium concentration between the acid and the salt being produced. I assumed that the acid will produce only half of the CH3COO- ions so moles =10/1000 x 0.250 and divide the answer by 2 to get the moles of the ethanoate ion. Then do the same for NaOH moles 10/1000 x 0.1.. and divide the answer by two to get the value of sodium ions. That is the same as the concentration of sodium ethanoate. Then get concetration of the salt. Subtract the moles of salt from that of the moles of ethanoate ions and then use that formula. I hope that makes some sense.

Yeah...it makes sense.... but why do we divide by 2? But That's one decent explanation!