Chemistry: Post your doubts here!

[AbbbbY]

View attachment 39749
AbbbbY asma tareen

Kc = [Pesticide in Hexene]/[Pesticide in water]

Assuming X dissolves in hexene,

8 = x/0.005-x
Solving for x you get 0.00444g


(shorter method used bec 2 marks only + same volumes)

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

q8 q22 q39
q20 do we need to draw isomers by ourselves or is there a formula?
q24 why is it not B?
q25 what type of reaction is D?
q30 can someone please show the chiral carbons?

20: None that I know of. There's 2^n for cis-trans tho.

24: Just draw out the ester. CH3C(Br)(CH2Br)COOC(CH3)2(CO2H)
B cannot be because it's showing the two Carboxylic terminals linking up. That can't happen.

25: It's 2,4 DNPH reacting with a carbonyl, so it's addition elimination

30: http://i.imgur.com/CzMA2hs.jpg

 

[NinjaInPyjamas]

20: None that I know of. There's 2^n for cis-trans tho.

24: Just draw out the ester. CH3C(Br)(CH2Br)COOC(CH3)2(CO2H)
B cannot be because it's showing the two Carboxylic terminals linking up. That can't happen.

25: It's 2,4 DNPH reacting with a carbonyl, so it's addition elimination

30: http://i.imgur.com/CzMA2hs.jpg

Thank you! Also can you help in q8 q22 q39 please

 

[AbbbbY]

Thank you! Also can you help in q8 q22 q39 please

Oh sorry. Missed those.

8: D
Easiest question to confuse kids. It's basically a Hess' Law question borrowing concepts from Born-Haber cycles. Can be solved in a heart beat if you're going the AS+A2 way. Can be solved in two heart beats if you're not. Your hess' cycle will look something like this.

At A, initially atomization is happening. The value for that is given. If you remove two electrons from that, Ca2+ gaseous ion will be formed. That's the first 2 i.e's added (data booklet). At B, you're hydrating the gaseous ion. That's enthalphy of hydration to give you Ca2+(g)

Capiche?

22: C
CH3CHO + HCN -> CH3CHOHCN -> CH3CHOHCOOH
C2H5CHO -> C2H5COOH

Difference: 90-74 = 16

39: A

1: 1 mol -> 0.62 mol = 62%
2: 1 mol -> 0.62 mol = 62%
3: 1 mol -> 0.613 mol = 61.3%. Close enough.

 

[Snow Angel]

Hi, are there any tips to get full marks in chemistry as level practical paper 3, if there r plz share

 

[AbbbbY]

What about the color of the flames? Only magnesium is yellow and sulfur is blue.. Do the rest burn with a white flame?

Mg isn't yellow! THat's a criminal mistake! Have you guys seriously never thrown magnesium salts on the burners during your practicals? I guess my batch was the only one full of devils.

Magnesium burns with a very bright while flame. Sodium burns with a yellow flame.

Oh also. Try Barium some time. Burns with an interesting green flame

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_22.pdf

Can someone please explain in detail q1 (c) and (d) please?

1 c:

i- O2
ii- CO2
iii- 40-30 = 10
iv- 50-30 = 20


Explanations:

You gotta realize whats going on.
HC is mixed with Oxygen and burnt.
CO2 is produced but some unreacted O2 is also present. So, the gas mixture is CO2+O2
This CO2+O2 mixture is then reacted with KOH. The KOH reacts with the CO2 leaving behind O2 gas.

i So, the gas left behind at the end is O2.
ii The gas produced is obviously CO2.
iii 40cm3 of gas was left behind of which 30 remained after KOH so 10 was CO2
iv 30cm3 of O2 was left behind. So, 50-30 was used.

d:

Hydrocarbon + O2 -> CO2 + H2O
10 : 20 : 10 : irrelevant (complete reaction)
so, I can say that 1 mol of the HC reacts completely with 2 mols of O2 to give 1 mol of CO2

CxHy -> xCO2 so x = 1 since CO2 = 1

Using your equation in the first part, x+y/4 O2. We know that's 2 so x+y/4 = 2. Solve for y and you get 4.
Thus, CH4.

 

[sitooon]

Which Can be coplaner and which CAnt ? ?

 

[Namehere]

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)

Thank you very much in advance!

 

[hope4thebest]

Simple! It's pretty much an AS question.

E is a di-ol. Just add 2 OH's across the double bond.

ii)
Reaction I: Cold + dilute KMnO4
Reaction II: K2Cr2O7 + H+ + warm


__

Incase you don't get whats going on, the cold KMnO4 will add a di-ol across the double bond. Cr2O72- + heat oxidises primary alcohols to aldehydes (if reflux then carboxylic), secondary to ketones, and tertiary to nothing. The upper ol is primary and the lower is tertiary so stays as is.

Thank You! I am usually bad in picking skeletal formula. TY

 

[hope4thebest]

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)

Thank you very much in advance!

difficult questions

 

[Namehere]

Which Can be coplaner and which CAnt ? ?

Carbon atoms can be coplanar: A (all sp2 hybridized), C (all sp3 hybridized)
Carbon atoms can´t be coplanar: B (benzene ring is sp2, cyclic alkane is sp3), D (up until the alcohol all is sp3 hybridized, the carboxlyic acid is sp2 hybridized), E (the group on the right is sp3, whereas the left group is all sp2)

Hope it is right and that it helps

 

[AbbbbY]

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)

Thank you very much in advance!

6(ai): 3! = 6.
6(ei) 3P2 = 6
9(di):
Your initial selection, i.e U, G, A or C has 4 possible selections.
Once that's selected, you have 4 further possible selections.
And when that's selected you have 4 final selections.

4x4x4 = 64
Remember the basic 'box' method of solving Permutations and Combinations questions? That's whats used here. The concept.

 

[AbbbbY]

Which Can be coplaner and which CAnt ? ?

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.

 

[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