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Chemistry: Post your doubts here!

Abu mota

Abu mota

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h4rriet said:
6. Use pV=nRT.
8. Volume is proportional to temperature at constant pressure. You can tell from pV=nRT.
27. Unreactive towards mild oxidising agents = tertiary alcohol. We can get only D from a tertiary alcohol.
29. B and C are not acidic; A is not chiral. That leaves D.
30. The acid used to make this ester is hexanoic acid. Only C shows a hexanoate.
38. Step 1 will give us the product so we won't need steps 2 and 3.

21. Any C=C bond breaks and is replaced with a C=O bond.
29. Draw propanone first, then break the double bond and add an H to the O and a -CN bond. The remove the CN and put CO2H in its place.
30. (30+30)-18(water).
39. Compounds 1 and 2 are formed by the reaction of C3H7 radicals. Compound 3 is formed by the reaction of a C4H9 radical with a C2H5 radical. These last can't be made if only propanone is used.
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I'm sorry, but I still don't understand q.6, q.29 why is c not acidic(there is co2h bonded to it).
 
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strangerss

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strangerss said:
I think the answer is B , am I right?
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Okay , in A ethanol with sulphuric acid is a hydrolysis reaction , so ethanol loses a water molecule and nothing happens to sulfuric because I think it acts as a catalyst , thus nothing gains or losses oxygen. In C it's also an example of the hydrolysis of an ester , therefore the ester splits into it's acid and alcohol and again nothing gains or loses anything , the hydroxyl group returns to the alcohol from the water and the carboxylic acid gains it's hydrogen back . in D there is no reaction because a ketone doesn't react with fehlings or tollents reagent. However in B tollents reagent is made of silver nitrate dissolved in ammonia right? , so the silver ions will oxidise the aldehyde into a carboxylic acid(oxidation) while they themselves are reduced to silver atoms after gain electrons(reduction) that's why a silver mirror appears on the inside of the tube . I hope I made sense :)
 
Aymen Ezazi

Aymen Ezazi

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strangerss said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf here you go :D
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Q3-- they did not ask to see the data booklet but it will be a lot easier if u just go through it , A is not correct bcoz it is Mg and its second ionisation energy is not high thn sodium and aluminium, C is not correct because its is silicon and its 2nd ionisation energy is not higher thn Al and P, its not D bcoz its S and evn its 2nd ionisation energy is not higher thn either of their neighbours, so its B
 
cute97

cute97

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syed1995 said:
5 b .. whenever a state change occurs from liquid to gas.. the bonds are broken.. and hydrogen bonding is only present where H is joined with N, O or F.

6 A .. because of Hydrogen Bonding.. The IMF are high.. while to consider a gas to be an ideal gas the inter molecular forces need to be zero.

11 A .. because the Acid will react with OH- to decrease the concentration of OH- ions which will shift equilibrium to right hand side and more product forms...

33 D .. 2 Is wrong because Triple bond is in products and non-reversible reaction so that won't affect how the reactants react.. and N=N bond is present in N2H4 not triple bond.

40 B 1 is correct.. 2 is correct as 2 OH groups are present (OH) and -OH in COOH .. H bonded with O is hydrogen bonding, 3 is wrong since it will form a ketone from secondary alcohol.


I'd suggest that you revise the chapter on bonding from the book or ask your teacher to give you revision on the topic.. since your concepts on Hydrogen bonding seem to be a little weak :)
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THAAAANKKK YOOOUUU :D I appreciate that and thanks for your advice ;)
 
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strangerss

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Aymen Ezazi said:
Q3-- they did not ask to see the data booklet but it will be a lot easier if u just go through it , A is not correct bcoz it is Mg and its second ionisation energy is not high thn sodium and aluminium, C is not correct because its is silicon and its 2nd ionisation energy is not higher thn Al and P, its not D bcoz its S and evn its 2nd ionisation energy is not higher thn either of their neighbours, so its B
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then why didn't they tell us to refer to the data booklet? the answer B is phosphorous? But how it's p orbital has 1 electron so it lost two electrons but they're asking for the electron configuration of it's second ionisation energy?
 
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strangerss

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strangerss

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cute97 said:
more doubts plz :$
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q5 for the pi bond isnt it overlapping of two p orbitals, Q11 how to do such ques, Q21, Q39 what are the H2SO4 reactions ??
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9, Q19, Q28
thanks in advance :D
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question 9 , I don't know:eek: any one help??.. question 19 it's A right? draw the displayed formula of the compounds and try to make a chiral center from the elements given to you, if you can't then it won't form an optical isomer , like in iv there's also a double bond and the only carbon with a single bond must have either two hydrogen's or two bromine get ?..in 28 it's C right? if you draw the displayed formula of the compound u will see that the double bond is still there , that's because nabh4 will only reduce the aldehyde group to it's primary alcohol , the other options are insensible because for example in A addition of hydrogen in that method will not reduce the aldehyde , it will only make the double bond single. hope you understood:)
 
Aymen Ezazi

Aymen Ezazi

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strangerss said:
then why didn't they tell us to refer to the data booklet? the answer B is phosphorous? But how it's p orbital has 1 electron so it lost two electrons but they're asking for the electron configuration of it's second ionisation energy?
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yes it lost two electrons , but note that the second one is from an orbital closer to the nucleus , so has high 2nd ionisation energy, in C the 2nd ionisation energy will be little higher thn A bcoz its in p sub shell but 2 different orbitals, and A will have least 2nd ionisation energy bcoz its in s sub shell and both electrons in same orbital so will have greater repulsion.. i hope this made it clear :)
 
Aymen Ezazi

Aymen Ezazi

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strangerss said:
alsalam allaykom , http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf question 2 , I don't know what to understand from the graph, any help please?
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the third cross in this have the least ionisation energy that means it is from group one , and that would be sodium , and before sodium comes a noble gas , neon which will have the highest ionisation energy, and before neon is fluorine so that will be it :)
 
ahmed abdulla

ahmed abdulla

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Aymen Ezazi said:
the third cross in this have the least ionisation energy that means it is from group one , and that would be sodium , and before sodium comes a noble gas , neon which will have the highest ionisation energy, and before neon is fluorine so that will be it :)
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can u help with this :)
 

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Aymen Ezazi

Aymen Ezazi

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ahmed abdulla said:
can u help with this :)
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Q40 is easy, just look for an intermidiate, because there is always a curve downwards like this in the middle when ever they ask anything related to an intermediate, and in only 1 there is an intermediate formed when the Br is broken, THN the intrmediate forms thn OH group attaches.. in 2 and 3 no intermediate forms , there is only direct substituton..

the nxt question its not A bcoz alcohol reacts with H2SO4 to give an alkene, which is DEHYDRATION and not redox.. not C bcoz an ester is HYDROLYSED when heated with acid to give alcohol and carboxylic acid again no redox.. and its not D bcoz its a ketone and ketone DOES NOT REACT at all with Fehlings reagent and Tollen's reagent.. it is B bcoz the aldehyde gets OXIDISED to carboxylic acid and the Ag+ ions presnt in the solution gets REDUCED to Ag metal which forms a silver mirror in the test tube... i hope u got it :)
 
ahmed abdulla

ahmed abdulla

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Aymen Ezazi said:
Q40 is easy, just look for an intermidiate, because there is always a curve downwards like this in the middle when ever they ask anything related to an intermediate, and in only 1 there is an intermediate formed when the Br is broken, THN the intrmediate forms thn OH group attaches.. in 2 and 3 no intermediate forms , there is only direct substituton..

the nxt question its not A bcoz alcohol reacts with H2SO4 to give an alkene, which is DEHYDRATION and not redox.. not C bcoz an ester is HYDROLYSED when heated with acid to give alcohol and carboxylic acid again no redox.. and its not D bcoz its a ketone and ketone DOES NOT REACT at all with Fehlings reagent and Tollen's reagent.. it is B bcoz the aldehyde gets OXIDISED to carboxylic acid and the Ag+ ions presnt in the solution gets REDUCED to Ag metal which forms a silver mirror in the test tube... i hope u got it :)
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thanks ... if u look at ur textbook u will find that even 2 will form intermediate ie. SN2 ... so thats probably NOT the reason
 
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