Chemistry: Post your doubts here!

[Angelina_25]

Hey!

Could someone kindly show and explain me the mathematical way for doing the following questions:

Q6(ai) - Oct/Nov 2013, Paper 41 (Ans: 6)
Q6(ei) - Oct/Nov 2013, Paper 43 (Ans: 6)
Q9(di) - Oct/Nov 2010, Paper 43 (Ans: 4x4x4=64)

Q6 part ai) there are 6 possible combinations in which the tripeptides can be made ....which is (gly-ser-val ),(gly-val-ser) ,(val-ser-gly),(val-gly ser),(ser-val -gly ) and (ser-gly -val)
Q6 part eii) this one is also similar to the one i have showed above .....i bet u do this and get it right ....
Q9 part Di)there are four bases which is A C G and U in the mRNA strand ...therefore u have to do 4^3 (where 4 denotes the number of bases present and "3" stands fr the triplet)

 

[Namehere]

I've done this question so I know what precedes the question. But, this is primarily why I dislike when sections of the question are cropped out and asked.

Look at the diagrams above the part you've posted. It has a benzene, a cyclohexane, a straight chain alkane (I think butane?) and a branched alkane. It's telling you that benzene and the straight alkane carbon atoms are coplaner while the branched + cyclic carbons are not. THIS concept is to be used here. You need to identify which of A, B, C, D and E have rings OR branching. If they do, they're not co-planer.

I believe apart from B, all others are co-planer.

C is coplaner because of the C-O-C linkage. O is on a different plane but since we need to consider C atoms, it's co-planer. All the C's are in the same plane in this. If you don't get why C is co-planer, ask me and I'll draw it out.

I´ll go with Abby. Sorry for the incorrect above explanation. Could you anyways please tell me which year that paper is from?

 

[Namehere]

Q6 part ai) there are 6 possible combinations in which the tripeptides can be made ....which is (gly-ser-val ),(gly-val-ser) ,(val-ser-gly),(val-gly ser),(ser-val -gly ) and (ser-gly -val)
Q6 part eii) this one is also similar to the one i have showed above .....i bet u do this and get it right ....
Q9 part Di)there are four bases which is A C G and U in the mRNA strand ...therefore u have to do 4^3 (where 4 denotes the number of bases present and "3" stands fr the triplet)

Thank you! Although I´ve asked for the maths way (Q6´s)

 

[Namehere]

Q6 part ai) there are 6 possible combinations in which the tripeptides can be made ....which is (gly-ser-val ),(gly-val-ser) ,(val-ser-gly),(val-gly ser),(ser-val -gly ) and (ser-gly -val)
Q6 part eii) this one is also similar to the one i have showed above .....i bet u do this and get it right ....
Q9 part Di)there are four bases which is A C G and U in the mRNA strand ...therefore u have to do 4^3 (where 4 denotes the number of bases present and "3" stands fr the triplet)

Thank you!

Just to make sure, could you justify why it is 3! for the first question and 3P2 for the second question? I mean, why not 3P3 for the second question? (since it also gives 6).

 

[Angelina_25]

Thank you!

Just to make sure, could you justify why it is 3! for the first question and 3P2 for the second question? I mean, why not 3P3 for the second question? (since it also gives 6).

3! because it gives all the possible combinations without restrictions unlike permutations!!!!
And 3P2 because there are 3 aminoacids given and we asked to make dipeptides (only using 2 aminoacids )

 

[sitooon]

I´ll go with Abby. Sorry for the incorrect above explanation. Could you anyways please tell me which year that paper is from?

On 2009 , va'42 q5

 

[sumeru]

Thank you!

Just to make sure, could you justify why it is 3! for the first question and 3P2 for the second question? I mean, why not 3P3 for the second question? (since it also gives 6).

You are asked to make dipetide out of 3 amino acids......so in how many ways can you arrange 2 from 3..i.e 3p2.......

 

[Namehere]

I've done this question so I know what precedes the question. But, this is primarily why I dislike when sections of the question are cropped out and asked.

Look at the diagrams above the part you've posted. It has a benzene, a cyclohexane, a straight chain alkane (I think butane?) and a branched alkane. It's telling you that benzene and the straight alkane carbon atoms are coplaner while the branched + cyclic carbons are not. THIS concept is to be used here. You need to identify which of A, B, C, D and E have rings OR branching. If they do, they're not co-planer.

I believe apart from B, all others are co-planer.

C is coplaner because of the C-O-C linkage. O is on a different plane but since we need to consider C atoms, it's co-planer. All the C's are in the same plane in this. If you don't get why C is co-planer, ask me and I'll draw it out.

Why is Oxygen on a different plane? Sorry for asking so many (stupid) questions.

 

[AbbbbY]

Why is Oxygen on a different plane? Sorry for asking so many (stupid) questions.

Ummm.. See the wedges

It has something to do with the lonepair + hybridization but I can't quite explain it.

See this.

The red part is the Oxygen.

 

[sitooon]

1 b(vi)

 

[AbbbbY]

1 b(vi)

Could you ttell me your answer to v? Would save me the trouble of going through all the steps.

Edit: Nevermind. Done them all anyway.

Original Solution: Fe3+ = 1.5 in 1000 so 0.15 in 100
Solution after dissolving copper several times: Fe2+ = Answer to part v/10 =0.6/10 = 0.06

Remaining Fe3+ = 0.15 - 0.06 = 0.09

Fe3+ : Cu
2 : 1
0.09 => 0.045mol

0.045 mol = 0.045*63.5 = 2.86g

 

[sitooon]

Could you ttell me your answer to v? Would save me the trouble of going through all the steps.

Edit: Nevermind. Done them all anyway.

Original Solution: Fe3+ = 1.5 in 1000 so 0.15 in 100
Solution after dissolving copper several times: Fe2+ = Answer to part v/10 =0.6/10 = 0.06

Remaining Fe3+ = 0.15 - 0.06 = 0.09

Fe3+ : Cu
2 : 1
0.09 => 0.045mol

0.045 mol = 0.045*63.5 = 2.86g

[Fe2+] = 1.5 × 10–3 × 1000/2.5 = 0.6 (mol dm–3) ecf from (iv)
Your ans is right
I got stuck myself at this stage :
"
Original Solution: Fe3+ = 1.5 in 1000 so 0.15 in 100
Solution after dissolving copper several times: Fe2+ = Answer to part v/10 =0.6/10 = 0.06

Remaining Fe3+ = 0.15 - 0.06 = 0.09 "
If you can elaborate this point if possible as how you did it .

 

[AbbbbY]

[Fe2+] = 1.5 × 10–3 × 1000/2.5 = 0.6 (mol dm–3) ecf from (iv)
Your ans is right
I got stuck myself at this stage :
"
Original Solution: Fe3+ = 1.5 in 1000 so 0.15 in 100
Solution after dissolving copper several times: Fe2+ = Answer to part v/10 =0.6/10 = 0.06

Remaining Fe3+ = 0.15 - 0.06 = 0.09 "
If you can elaborate this point if possible as how you did it .

Original conc is 1.5 dm3

this means:

1.5 mols in 1000 cm3
so
0.15 mols in 100 cm3

Conc of Fe2+
0.6 mols in 1000 cm3
so
0.06 mols in 100cm3

 

[Haya Ahmed]

Hey ! .. Can someone help me here !
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf

Question 34,35,36

Thanks ^_^

 

[ahmed abdulla]

Paper 5 , http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_53.pdf

Question 2b(iii) ( moles )
Question 3 , how to Complete the table > no formulas are given

& Whats is the max. Volume a gas syringe can hold which we are supposed to label here ?

AbbbbY

 

[itallion stallion]

Can someone plz help me with this chemistry question. 9701 october November 2011 paper 41 q6 bii.
Thanks!

 

[goodluckayesha]

Can someone plz help me with this chemistry question. 9701 october November 2011 paper 41 q6 bii.
Thanks!

this is beta-pleated
you have to draw either an alpha-helix or beta-pleated and show the hydrogen bonding between them

 

[AbbbbY]

Hey ! .. Can someone help me here !
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf

Question 34,35,36

Thanks ^_^

34:

I'd go with B. 3 is definitely wrong because calcium IONS wont react with H+, but solid will. 1 and 2 are correct. 2 is most definitely correct, and so I'm left with B.

35:
1- It does.
2- It does.
3- It doesn't because it's basic itself.

so, A

36:
B?

1- It is. NaClO3.
2- It does. +5 in NaClO3 and -1*5 in 5NaCl.
3- It doesn't. It's going from 6*+1 to +1 and 5*+1 so definitely not [R].

 

[AbbbbY]

this is beta-pleated

Umm since they're beta pleated sheets, shouldn't they be running in opposite directions. I.e, CO under NH

 

[goodluckayesha]

Umm since they're beta pleated sheets, shouldn't they be running in opposite directions. I.e, CO under NH

they r not in diff directions.... in beta pleated hydrogen bonds form btw diff polypeptide chains and the r diff polypeptide chains
this diagram is lso given in my TB