Chemistry: Post your doubts here!

[gary221]

For 38, D is the only ans bcoz
Halogenoalkanes in which all the hydrogen atoms in the alkane have been replaced by halogen atoms are difficult to burn, thus used as fire retardants...

 

[hm12]

Fellows need help with these paper 1 questions​

others have explained some of them but im not entirely satisfied.​

​

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf​

http://www.xtremepapers.com/papers/...and AS Level/Chemistry (9701)/9701_w08_er.pdf​

Q3 how do we work out the answer of this question?(which is D)​

Q31 Why is the answer not A?​

Q38 Why is the answer not B?(someone in a previous post said that flourine has to be present in order for a hydrocarbon to be a flame retardant. Can someone confirm that and tell me why?)​

​

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf​

Q9 Why is the answer not A?​

Q31 Why cant the answer be B? The second statement seems correct.​

for the first paper i.e 08 Q3 electronic configuration of Li is 1 S^2 and 2 S^1 and for Cr it is 1S^2, 2S^2, 2P^6,3S^2,3P^6,3D^5 and 4 S^1 so means both Li and Cr have one electron is their S orbital so answer is D for Q31 its not A because permanent dipole occurs in molecules or compounds which have a difference of electronegativity and for Q38 read the following link http://www.chemguide.co.uk/CIE/section103/learningcd.html

 

[gary221]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf​

Q9 Why is the answer not A?​

Q31 Why cant the answer be B? The second statement seems correct.​

For 9, the ans is nt A, bcoz the total mass has to be taken ie mass of HCl + mass of NaOH which = 100 cm3
then delta H =(mass *specific heat capacity * temp change)/no. of moles
so the ans is C

 

[hmlahori]

for the first paper i.e 08 Q3 electronic configuration of Li is 1 S^2 and 2 S^1 and for Cr it is 1S^2, 2S^2, 2P^6,3S^2,3P^6,3D^5 and 4 S^1 so means both Li and Cr have one electron is their S orbital so answer is D for Q31 its not A because permanent dipole occurs in molecules or compounds which have a difference of electronegativity and for Q38 read the following link
http://www.chemguide.co.uk/CIE/section103/learningcd.html

For 3, v hv 2 see which pair of atoms hv 1 electron in thr s orbital...
In A, the electronic configuration of Ca is [Ar], 4S2...ie 2 electrons in s orbital thus A is wrong
In B, the e.c. of Cu is [Ar], 3d10, 4S1...bt Be is 1S2, 2S2.....thus B is wrong.
In C, the e.c. of H is 1S1...bt He is 1S2...thus C is wrong.
In D, Li = 1S2, 2S1...n Cr = [Ar], 3d5, 4S1....so D is correct

Isnt the electronic configuration of Cr = 1s2 2s2 2p6 3s2 3p6 4s2 3d4?
Doesnt the 3d orbital contain 10 electrons and come after the 4s orbital as it has a higher energy?

 

[hm12]

Isnt the electronic configuration of Cr = 1s2 2s2 2p6 3s2 3p6 4s2 3d4?
Doesnt the 3d orbital contain 10 electrons and come after the 4s orbital as it has a higher energy?

yes but Cr and Cu are the two exceptions they have only one electron in the s orbital if you study A2 you would know it better in the topic transition metals

 

[Jaf]

I have no intention whatsoever of misleading people. This term is used in the books i use too but this is because the AS and A2 books are combined and some information is for AS and some for A2. But it is nowhere mentioned in our syllabus even 'complex ion formation' is not mentioned.. You can check each and every point in the AS part of the syllabus for exams in May/June 2012. I dont know about syllabuses before or after that session it may be in one of them but not in this one. And as far as i know Cambridge RARELY gives thinks that are not in or related to our syllabus.

Page 185 of the latest coursebook mentions the formation of complex ions when ammonia is added to silver halides.

 

[hmlahori]

for the first paper i.e 08 Q3 electronic configuration of Li is 1 S^2 and 2 S^1 and for Cr it is 1S^2, 2S^2, 2P^6,3S^2,3P^6,3D^5 and 4 S^1 so means both Li and Cr have one electron is their S orbital so answer is D for Q31 its not A because permanent dipole occurs in molecules or compounds which have a difference of electronegativity and for Q38 read the following link http://www.chemguide.co.uk/CIE/section103/learningcd.html

For question 31 Statement 3 says: The Kevlar molecule has no permanent dipole.
Then from your explanations it should be correct right?

 

[hmlahori]

yes but Cr and Cu are the two exceptions they have only one electron in the s orbital if you study A2 you would know it better in the topic transition metals

Oh ok i didnt really know that because i dont study A2 so ya thanks thats something i'll try to remember for my upcoming AS exams.

 

[hm12]

For question 31 Statement 3 says: The Kevlar molecule has no permanent dipole.
Then from your explanations it should be correct right?

Carbon and nitrogen do not have difference of electronegativity

 

[hmlahori]

Page 185 of the latest coursebook mentions the formation of complex ions when ammonia is added to silver halides.

Yes but it doesnt say anything about ligands. You are only supposed to know that the complex ion forms
and that the formula of the complex ion is always [Ag(NO3)2]X where X represents the halogen.

 

[UltimateHustla]

october november 2011 paper 12 question 26
help plsssssssssssssssssss

 

[Nikesh]

Fellows need help with these paper 1 questions​

others have explained some of them but im not entirely satisfied.​

​

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf​

http://www.xtremepapers.com/papers/...and AS Level/Chemistry (9701)/9701_w08_er.pdf​

Q3 how do we work out the answer of this question?(which is D)​

Q31 Why is the answer not A?​

Q38 Why is the answer not B?(someone in a previous post said that flourine has to be present in order for a hydrocarbon to be a flame retardant. Can someone confirm that and tell me why?)​

​

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf​

Q9 Why is the answer not A?​

Q31 Why cant the answer be B? The second statement seems correct.​

Q3--> Use box electronic configuration
Q31--> Dipole moment 0 in each part of kelvar molecule
Q38----> Hydrogen bonding is not suitable except in CBrF3

 

[djmohammad]

please tell me ny goodway to revise my concepts for paper 1 AS ..... i try to do paspapers but could only score not more than 20%

 

[Scafalon40]

it cant be D bcoz they r tlkng abt solidification of butane, in which van der waals' forces dnt hv 2 b ovrcum

That question has been answered already, I'm talking about the post in colourful fonts...

 

[Scafalon40]

Fellows need help with these paper 1 questions​

others have explained some of them but im not entirely satisfied.​

​

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf​

http://www.xtremepapers.com/papers/...and AS Level/Chemistry (9701)/9701_w08_er.pdf​

Q3 how do we work out the answer of this question?(which is D)​

Q31 Why is the answer not A?​

Q38 Why is the answer not B?(someone in a previous post said that flourine has to be present in order for a hydrocarbon to be a flame retardant. Can someone confirm that and tell me why?)​

​

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf​

Q9 Why is the answer not A?​

Q31 Why cant the answer be B? The second statement seems correct.​

But in any case you solved something!
So I'll help
Seems alot has already been solved since my post
For Q9
It can't be A because there is 100cm^3 of solution, not 50cm^3 (1cm^3=1g)
For Q 31
Sulphuric Acid does behave as an acid in ethanol: remember that alcohols are very weakly basic!
Further detail: The ions present in that mixture will be HSO4-(the hydrogen sulphate ion)
and CH3CH2O^+H2(i.e the alcohol has gained a proton and H2SO4 has lost a proton)
By the way, the hydrogen sulphate ion has a negative charge and the alcohol has a positive charge. You can thank the admin for not letting me include proper scientific symbols in this post, i.e a proper power symbols instead of ^+....
If any body needs more help in return for answering my questions below in my signature(the questions in coloured fonts), feel free to ask...

 

[histephenson007]

october november 2011 paper 12 question 26
help plsssssssssssssssssss

Ethanol + Oxidation agent -----(immediate distillation)----> Ethanal + [(hydrogen gas or water...but that is insignificant)]

n (ethanol) = 2.3/ (24+16+6) = 0.05 mol
n (ethanol) = n (ethanal) = 0.05 mol

therefore, m of ethanal = n (Mr) = 0.05(24+16+4) = 2.2 g

But, the yield is 70%. so 0.7(2.2) = 1.54 g
answer is A?!

 

[Nikesh]

Can anyone please help me out with this?? November 2008, Paper 4, Question 2(b).
View attachment 7680

to find values of a,b and c
1st) if eqn 1 is slowest rexn then rate only depends upon the concentration of [H2O2] and [I-] that is value of c is automaticall 0. and since the molar ratio in the given equation is 1:1 your values will be a=b=1.
2nd) If equation 2 is the slowest rexn. then u have to take notice of 1st eqn too as 2nd equation has reactant IO- which is the product of 1st equation......i.e now Rate depends upon 1 concentration of H2O2, I- (from equation 1) and H+ (from equation 2) i.e. a=b=c=1
3rd) If eqn 3 is d slowest rexn. then u have to take notice of all three given equations. Check it you will find Rate now depends upon 1 concentration of H2O2 and 2 concentration of I- and H+ each.....
Remember they are just assumption....

 

[histephenson007]

Somebody solve these:

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

There were so many questions it looks intimidating! I'm not sure about the answers, but this is all the stuff I know. Next time try posting one question at once.

First, Oct/Nov 2011 Paper 11 ,

Q8. I was wrong
Q19. Answer is A. Bond-making is exothermic, bond-breaking is endothermic. Only in A, bonds are broken.
Q27. Either C or D. dunno about the ratios. Homolytic is right, cuz both atoms get one electron after the fission.

Q11. Basic information from the text : AlxCy + H2O -----> H2O + CO2
m ( AlxCy) = 0.144g
V ( CO2) = 72 cm^3
--> Note that it says 72 cm^3 ONLY

Solution (mine's kinda confusing): n (CO2) = V / 24 = (72*10^-3) / 24 = 0.003 mol

Because it said ONLY, we can know that n(CO2) should be equal to the n of Carbon in AlxCy. Hence, we can the mass of the carbon atoms in AlxCy.
(or)
n( Carbon in AlxCy) = n(CO2) = 0.003 mol
m (Carbon in AlxCy) = n(Mr) = 0.003(12) = 0.036 g
m( Aluminium in AlxCy) = 0.144 - 0.036 = 0.108
So, n(Al) = 0.108/27 = 0.004 mol
So, ratio of Al : C is 0.004 : 0.003
Q17. I don't like inorganic chemistry. So, not doing it.
Q6. I was wrong
Q4. In SO2, S has an oxidation number of +4 (because Oxygen's oxidation number is -2). When SO2 is oxidised by I2 (Each Iodine atom has a single negative charge). So, finally, in SO2I2, S will have +6 charge.

 

[hmlahori]

Can someone give explanations to the following paper 1 questions...
Thanks.....

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w02_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Chemistry (9701)/9701_w02_er.pdf
Q3 How do we work out the answer?
Q16 and Q26?

 

[shan5674]

Can someone please explain chemical energetics and how exactly to calculate the enthaply questions with examples of each type of calculation pleaseee