Chemistry: Post your doubts here!

[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...

 

[Firebender]

Guys, I'm really panicking right now: http://www.xtremepapers.com/papers/...and AS Level/Chemistry (9701)/9701_y12_sy.pdf

I was looking through the AS level chemistry syllabus and I came across the learning outcomes of the chapter Electrochemistry(Page 19 and 20 in the link). I don't know any of that stuff. AT ALL. I don't remember this being discussed in class or appearing in any exams. And I scored well on my chemistry mock. What's going on?

 

[histephenson007]

please tell me how is the answer D?

2,4- DNP ( or phenylhydrazine in this case) is used as a test for Carbonyl groups. The test happens because the hydrogens on the hydrazine group (from the NH2) undergo a condensation reaction with the Carbonyl group (-C=O )to form a water molecule. In compound P, the carbonyl group is the one on the second Carbon from the left ( note that DNP doesn't react with esters which is given in the middle of the molecule to confuse you).

So now, just connect the Carbonyl group to the hydrazine group, you should be able to see that D is the only option available.

 

[histephenson007]

Guys, I'm really panicking right now: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y12_sy.pdf

I was looking through the AS level chemistry syllabus and I came across the learning outcomes of the chapter Electrochemistry(Page 19 and 20 in the link). I don't know any of that stuff. AT ALL. I don't remember this being discussed in class or appearing in any exams. And I scored well on my chemistry mock. What's going on?

FYI, all the text in bold are for A2 level.

 

[CaptainDanger]

Guys, I'm really panicking right now: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y12_sy.pdf

I was looking through the AS level chemistry syllabus and I came across the learning outcomes of the chapter Electrochemistry(Page 19 and 20 in the link). I don't know any of that stuff. AT ALL. I don't remember this being discussed in class or appearing in any exams. And I scored well on my chemistry mock. What's going on?

Talking about the stuff written in BOLD?