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Chemistry: Post your doubts here!

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http://papers.xtremepapers.com/CIE/Cambridge International%
UntitledCHEM.jpg
20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w09_qp_11.pdf
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Some1 did dis b4... im tryin to understand..
 
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answer is B
amount of sulphite use = conc x vol
= 0.1 x (25/1000)
=2.5x10^-3mol
2 electrons are lost
amount of electrons lost = 2 x 2.5 x 10^-3 ==== 5 x 10^-3 mol
amount of electorns gained by metallic salt = amount of electrons lost by sulphite
= 5 x 10^-3 mol
amount of metallic salt used = 0.1 x (50/1000) ==== 5 x10^-3
amount of electrons gained PER MOLE of salt = (5 x 10^-3)/(5x10^-3) = 1 unit
hence oxidation state of metallic salt decreases by 1 unit meaning the oxidation satate becomes +2 from +3!!! this is for the question 9 , some other person posted it :D

Thankss but i never saw such a working :( i think my teacher missed explaining or wat (n)
 
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see da above post
i got it look
remove a aldehyde group and the ring you are left with C13 H26 now acording to alkene general formulae there should have been 26 hydrogen as each double bond removes two hydrogen two double bond will remove the hydrogen . hence three double bond in cyclo hexene and 2 extra makes 5 hence A syed1995 daredevil
 
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quote="Alice123, post: 607701, member: 44056"]see da above post[/quote] can any of you help in the question i posted above? Thanks :D
 
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Use of the Data Booklet is relevant to this question.
Ethyl ethanoate can be obtained from ethanoic acid and ethanol by the following reaction.
CH3CH2OH + CH3CO2H CH3CO2CH2CH3 + H2O
Ethanol (30 g) and ethanoic acid (30 g) are heated under reflux together, and 22 g of ethyl
ethanoate are obtained.
What is the yield of the ester?

Please help!! thankyou
 
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Hay there,

Thanks for helping out, but its still not solved :(
Option B has H and H on both sides.. (not H and Cl, if that would been, then it would be a sure answer, but its not like that)
Option C has Cl and Cl on both sides..


And I'll be grateful if you explain the question number 10 of the same paper. Why its answer is C, whats wrong in B ??\

Regards
no no on the left side it has O and on the right it has H in B part. in C it has O on one side and Cl on the other side. that is what i meant by OPPOSITE poles. thus dipoles are formed.

in B of q10 the activation energy is really high so the reactoin will take a long time to proceed bcz it will need more time to attain such a high amount of energy tto activate the reaction.
 
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i got it look
remove a aldehyde group and the ring you are left with C13 H26 now acording to alkene general formulae there should have been 26 hydrogen as each double bond removes two hydrogen two double bond will remove the hydrogen . hence three double bond in cyclo hexene and 2 extra makes 5 hence A syed1995 daredevil
okay woaaah!! :O umm... i think if i took the time of drawing out the whole friggin structure i cud maybe get the hang of it - or maybe not :p
i remember our sir made us do it in class and after a little while and stumps from sir i got it and now i forgot it again :p so thanks agaiinn! :p
 
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Q1's answer is C. from the data u can infer that 50 cm3 of O2 was used to burn 10cm3 of hydrocarbon (CxHy)
make up an equation for the unknown hydrocarbon for ur aid and u'll c that 1 mol of CxHy requires 5 moles of O2 to produce 3 mol of CO2. 6 O go into the CO2 and the remaining 4 (from 5 O2) go to the H2O which means that 4 H2O molecules were formed. so we have 8 H and 3 C thus the ansewr.

Q2's answer is bcz 2 mol of NaN3 make 3 mol of N2 which means 1 mol makes 1.5 mol of N2 directly. then the Na produced also makes N2.
wth 2 mol of NaN3 we get 2 mol of Na so with 1 mol we get 1 mol of Na.
10 mol of Na produce 1 mol of N2 so 1 mol produces 0.1 mol
so total N2 produced is 1.5 + 0.6 = 1.6mol

i don't get q10 so if u get an answer for that tag me ^_^
 
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Just a tip for Organic Chemistry questions: My teacher always said "Draw the structure in a way you understand..." so the key is to draw the structure then solve, even if skeletal formula is given, write it in structural or displayed form then work out the answer...
 
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