Qualification > Sciences
ALL CIE CHEMISTRY DOUBTS HERE !!
Deadly_king:
--- Quote from: Physics09 on May 04, 2012, 07:00:52 pm ---Exam in three days! Urgent help please!
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w07_qp_1.pdf
Q: 30 and 21
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21. C-Pb bond is a very weak bond...........So tetra methyl lead acts a source of methyl radicals ;)
30. Upon breaking the ester bonds, we can have only two possible acids formed namely CH3CH2CH2CO2H or CH3CO2H.
Hence answer is B ;)
--- Quote from: Physics09 on May 04, 2012, 07:00:52 pm ---http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s08_qp_1.pdf
Q:23 and 27
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23. Answer is D due to high molecular weight which will prevent the hydrocarbon from rising upon water ;)
27. The formula C10H14O indicates that it should be an alcohol and the fact that it is unreactive to mild oxidising agents, it cannot be a primary one ;)
So you just have to eliminate all the answers which upon hydration would give a primary alcohol ;) This leaves us with B, C and D! But only D gives a tertiary alcohol which we all know is not affected by mild oxidising agents ;)
--- Quote from: Physics09 on May 04, 2012, 07:00:52 pm ---http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_1.pdf
Q: 28
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28. The reaction is one forming an ester ;) So to form an ester we usually need and alcohol and a carboxylic acid ;)
If you break the ester bond, you'll get as products CH3CO2H and CH3CH2CH(CG3)CH2OH, this is because the second carbon atom of 2methyl-butyl methanoate contains a double bond with O which indicated that it should be the acidic carbon atom ;) The rests just fits in ;)
Check the answers, if they are good, then my explanation are correct ;) IF not, then am afraid I can't help :/
AM afraid I don't currently have enough time to check the marking schemes or my notes........I just replied according to my knowledge :/
Hope it helps :)
NidZ- Hero:
can Someone please Explain me in details the answerrs of the Following ?
- Q3
- Q5
- Q8
- Q11
- Q26
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FIGCSE%2FChemistry%2FCIE%2F2002+Nov/
paper 1
Ghost Of Highbury:
--- Quote from: NidZ- Hero on May 05, 2012, 03:16:48 pm ---can Someone please Explain me in details the answerrs of the Following ?
- Q3
- Q5
- Q8
- Q11
- Q26
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FIGCSE%2FChemistry%2FCIE%2F2002+Nov/
paper 1
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First I.E + Second I.E is same for Co and Al (757 + 1640 for Co) and (577+1820) = 2397. Use data booklet for IEs
A will have a dipole because Chlorine atoms and Fluorine atoms can be places in different carbons, and due to different electronegativities,
a net dipole remains. Same reason for C and D. The answer, C2Cl4 doesn't have a net dipole because Cl atoms have same Electronegativity and thus
net dipole cancels and is zero.
A 2 g sample of hydrogen at temperature T and of volume V exerts a pressure p. p = nrT/V = 2RT/V (moles of H2 = 2/1 = 2)
check which option gives same pressure. For e.g A gives p = nRT/V = RT/V (n = 2/2 = 1). B gives p = 4RT/V
C gives correct answer because n = 1 + 1 = 2.
2HI <-> H2 + I2
Initially b 0 0
Equilibrium b-x b/2 b/2
(b^2/4)/(1-b)^2
26. First break the double bonds, add the -OH groups and make the diols. NOw, for oxidation to ketone, 2 hydrogen atoms should be lost, one from the -OH group, one from the same carbon atom. And remember u need a diketone, so 2 ketone groups.
It isnt possible to get a diketone in A and D because you'll only get one ketone group in each. From B and C, B gives u an aldehyde after losing th 2 h-atoms, C is the answer.
Twinkle Charms:
Flask X contains 1dm3 of helium at 2kPa pressure and flask Y contains 2dm3 of Neon at 1kPa pressure.
If flasks are connected at constant temperature, what is the final pressure??
Arthur Bon Zavi:
--- Quote from: Twinkle Charms on May 06, 2012, 07:16:29 am ---Flask X contains 1dm3 of helium at 2kPa pressure and flask Y contains 2dm3 of Neon at 1kPa pressure.
If flasks are connected at constant temperature, what is the final pressure??
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Total volume = 1dm3 + 2dm3 = 3dm3
PV = nRT
As temperature is constant, PV is constant, so P1V1 = P2V2
Flask X : (2)(1) = P2(3)
P2 = 2/3 kPa
Flask Y : (1)(2) = P2(3)
P2 = 2/3 kPa
Total pressure = sum of partial pressures
Total pressure = 2/3 + 2/3 = 4/3 kPa
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