Chemistry: Post your doubts here!

[histephenson007]

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

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

Q3. According to the definition of enthalpy change of ionisation, an electron should be removed from a gaseous atom for first ionisation energy. Two electrons for second ionisation energy. Then, you just have to use the Data Booklet. Adding the first and second ionisation energies of Aluminium gives us 2397. In the four options available, only Cobalt has the same sum of first and second ionisation energies.

Q16. CIE just likes giving questions about CaCO3. lol. I don't know the answer.

Q26. This one is about reaction of alkenes with cold dilute solution of Potassium Manganate. But since the answer given is C, and C has phenyl groups, I am not really sure if this one is a part of the syllabus. You may ask your teacher for further details about this one.

 

[pratikdahal]

Can anyone kindly explain how to sketch in this graph question..and explain how it tends to be so..

 

[hmlahori]

Q3. According to the definition of enthalpy change of ionisation, an electron should be removed from a gaseous atom for first ionisation energy. Two electrons for second ionisation energy. Then, you just have to use the Data Booklet. Adding the first and second ionisation energies of Aluminium gives us 2397. In the four options available, only Cobalt has the same sum of first and second ionisation energies.

Q16. CIE just likes giving questions about CaCO3. lol. I don't know the answer.

Q26. This one is about reaction of alkenes with cold dilute solution of Potassium Manganate. But since the answer given is C, and C has phenyl groups, I am not really sure if this one is a part of the syllabus. You may ask your teacher for further details about this one.

Well ok thanks...

 

[Scafalon40]

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.

First of all, thanks
Secondly, I did post a few questions at the beginning, the list just got bigger and bigger because they were not being answered
But again, thankyou
Comments:
For Q19
Bonds are being broken and made in all the four reactions
The difference between the the bond making and breaking energies is what will tell us if a reaction is exothermic or not..
I think I figured out that this MCQ tests our general knowledge of chemistry rather than enthalpy changes!
For Q 27
I think this follows Markinov's rule...(I think it's spelled like that!), but that is for alkenes, plus the result goes against the rule

All I can think of is that 1 chloropropane is formed more easily because it is the main reaction, 2 chloropronae is more of a side reaction and is formed by different free radicals
For Q 8
Still thinking..

 

[Scafalon40]

For Q 11...
WELL DONE!
For Q17
Reaction D is one of the only one I could find in the syllabus which actually mentions that H2SO4 behaves as a strong acid and oxidising agent i.e special reference to it by the examiners!

For Q6
Tricky one: it actually requires us to know about some melting points: MgO and NaCl
NaF has a slightly higher melting point than NaCl, the latter has a melting point of 883
MgO has a melting point of about 2500
NH3 is soluble, not insoluble
Hence B
Finally Q4
Clever deduction here: requires us to actually know some common oxidation states( mentioned in the syllabus)
SO2 has an oxidation state of +4
If it's oxidised, the only possible higher oxidation number of sulpher is +6 not +5

 

[histephenson007]

Can anyone kindly explain how to sketch in this graph question..and explain how it tends to be so.. View attachment 7833

For competitive inhibition, there is a competition between substrate molecules and the competitive inhibitors for the active site on the enzyme. However, when the substrate concentration increases, there is more probability that a substrate molecule enters the active site. When the substrate concentration increases to a very high level, the competitive inhibitors can be neglected. Hence, Vmax can be achieved. Its just that it takes a longer time to be achieved because of a slower initial rate (some active sites being occupied by the competitive inhibitor).

For non-competitive inhibition, once the inhibitor binds to the enzyme, it never lets go. The inhibitor binds to it irreversibly. So, some enzymes are inhibited forever. Hence, it can never achieve Vmax (where all enzymes are catalysing a reaction in their active site.)

I am only able to solve these because I do Biology. I don't know how they expect others to know this stuff or expect others to even care about this stuff.

 

[histephenson007]

For Q 11...
WELL DONE!
For Q17
Reaction D is one of the only one I could find in the syllabus which actually mentions that H2SO4 behaves as a strong acid and oxidising agent i.e special reference to it by the examiners!

glad u understand it

and I hate inorganic chemistry. I don't care about it ;p

 

[pratikdahal]

For competitive inhibition, there is a competition between substrate molecules and the competitive inhibitors for the active site on the enzyme. However, when the substrate concentration increases, there is more probability that a substrate molecule enters the active site. When the substrate concentration increases to a very high level, the competitive inhibitors can be neglected. Hence, Vmax can be achieved. Its just that it takes a longer time to be achieved because of a slower initial rate (some active sites being occupied by the competitive inhibitor).

For non-competitive inhibition, once the inhibitor binds to the enzyme, it never lets go. The inhibitor binds to it irreversibly. So, some enzymes are inhibited forever. Hence, it can never achieve Vmax (where all enzymes are catalysing a reaction in their active site.)

I am only able to solve these because I do Biology. I don't know how they expect others to know this stuff or expect others to even care about this stuff.

and can you tell me how the two graphs would look like??

 

[histephenson007]

and can you tell me how the two graphs would look like??

I already uploaded a file. I'm not sure if I uploaded it correctly. How do I upload pictures?

 

[histephenson007]

For Q 8
Still thinking..

A. In the reverse reaction of the first reaction, HSO3- acts as a OH- donor. Hence it might be called a base.
B. The reaction is SO2 - > SO3- , Sulphur actually got oxidised from +4 to +6. So, its a reducing agent, not an oxidising agent.
C. SO3 (2-) accepts a proton in the reverse reaction of second reaction. So, it acts as a base. Not an acid.
D. The second reaction isn't a redox reaction at all. So, SO3 (2-) can't be any agent.

 

[ousamah112]

For competitive inhibition, there is a competition between substrate molecules and the competitive inhibitors for the active site on the enzyme. However, when the substrate concentration increases, there is more probability that a substrate molecule enters the active site. When the substrate concentration increases to a very high level, the competitive inhibitors can be neglected. Hence, Vmax can be achieved. Its just that it takes a longer time to be achieved because of a slower initial rate (some active sites being occupied by the competitive inhibitor).

For non-competitive inhibition, once the inhibitor binds to the enzyme, it never lets go. The inhibitor binds to it irreversibly. So, some enzymes are inhibited forever. Hence, it can never achieve Vmax (where all enzymes are catalysing a reaction in their active site.)

I am only able to solve these because I do Biology. I don't know how they expect others to know this stuff or expect others to even care about this stuff.

Yes seriously only bio students knw this. There isnt anything abt enzymes graph in chem syllabus. :S

 

[geek101]

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

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

histephenson007 already answered 3 and 26 so ill chip in to the 16. If something is being precipitated then it must be insoluble, of all the options CaCO3 is the only insoluble salt, or at least thats what i think.

 

[Student12]

Is that about enzymes for A2 right ?

 

[geek101]

can anyone please post a link for the paper 1s' of 2001 and 2002 both june and november! thanx! even if uv got any...just link it

 

[histephenson007]

Is that about enzymes for A2 right ?

yea, its for A2. Don't worry if your just doing AS.

 

[pratikdahal]

I already uploaded a file. I'm not sure if I uploaded it correctly. How do I upload pictures?

there is this "Upload a File" down to the text box in which we type comment.. click on it and browse to the required destination of "your" file..and simply click 'OPEN'...it will get uploaded i guess

 

[histephenson007]

there is this "Upload a File" down to the text box in which we type comment.. click on it and browse to the required destination of "your" file..and simply click 'OPEN'...it will get uploaded i guess

I already did that. And I can see my uploaded file in my original answer under attached files.
Anyway, its just the same as in the answer sheet.

 

[pratikdahal]

I already did that. And I can see my uploaded file in my original answer under attached files.
Anyway, its just the same as in the answer sheet.

Yeah..got it..thank you!

 

[smartangel]

please tell me how is the answer D?

 

[CaptainDanger]

please tell me how is the answer D?

Water is given out as H2 from NH2 and O from Ketone combines together leaving behind the structure in D...