Chemistry: Post your doubts here!

[daredevil]

I was doing mcqs from the redspot chemistry Mcqs book and the questions from the 80s & 90s were extremely difficult and they were from p3 p4 mostly. Are they as level questions or a2??

80's is too far back... have you done all through 2007 - present??
patterns and syllabii have changed continuously over the years. Some of the stuff in the 80's syllabus might not even be in your syllabus. do all variants from 2007 - present THOROUGHLY. then back as far as 2002 and do all papers till then thoroughly until you never forget the questions however they might come in front of you.
i don't think there will be time after all that. but if there still is, do some exercises or practice questions.

 

[Kim Shaw]

hey guys, please help me with this question in equilibria. answer is C. Please explain why..

 

[Kim Shaw]

iron tablets can be bought at chemists to supplement the diet. they contain iron(II)sulfate. which one of the following aqueous reagents could be used under suitable conditions to determine the percentage of iron in the tablets by titration?
A.ethanedioic acid B.Iodine in potassium iodide C.potassium manganate(VII) D.sodium thiosulfate
ans C..plz tell me why C and not the others..thnx

 

[ayanhaider]

Can anyone plz tell me about iodoform test?Im having a problem in it ....

 

[Hassan Ali Abid]

Pls help me with these questions
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_11.pdf Q4, Q5,Q6, Q8
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Q33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_11.pdf Q7

9701_s09_qp_1.pdf Q33 ...the ans is D bcz it says that which reactions have enthalpy change of formation so 2. 2C(s) + O2(g) → 2CO(g) cant be the choice as two moles of CO are formed so the choice left 1. so ans is D.

9701_w10_qp_11.pdf Q7....angle b/w H-C-H=109.5 , C-O-O =104.5 , C-C=O = 117 so order is 3-1-2 .

 

[daredevil]

Can anyone plz tell me about iodoform test?Im having a problem in it ....

It is used to test for alcohols (primary and secondary)
NaOH and I2 is added to the sample that is to be tested. yellow precipitate means test is positive for the sample.
I2 is reduced in the process.

 

[SomeStudent]

HELP!
Can someone please explain these two parts? How do we find out what the order is, like whats the procedure?
Part b in the first file.
Part e in the second file.
its Oct/Nov 08

 

[snowbrood]

HELP!
Can someone please explain these two parts? How do we find out what the order is, like whats the procedure?
Part b in the first file.
Part e in the second file.
its Oct/Nov 08

for part e
zero order reaction changing concentration doesnt affect the rate of reaction
first order reaction rate=k*concentration
second order reaction rate=k*(concentration)^2

we kept the concentration of H+ ions constant for line 1 and line 2
so only concentration of H2O2 will affect the rate of reaction
well if u see in line 1 and line 2 there is change in rate of reaction that means that h2o2 has order of 1 or 2
lets see if it is order 1
0.05=k*1 equation 1
k=0.05/1=0.05
now if we put the value of k for line 2 we get
0.05*1.4=0.07
that means h2o2 is order 1
pretty clear right?
now line 3
again increasing the concentration of h2o2 by 0.02 we would expect a increase of 0.4 in relative rate
however we are also changing the concentration of H+. as the change in conc of H+ didn't affect the rate of reaction so it has zero order

 

[SomeStudent]

for part e
zero order reaction changing concentration doesnt affect the rate of reaction
first order reaction rate=k*concentration
second order reaction rate=k*(concentration)^2

we kept the concentration of H+ ions constant for line 1 and line 2
so only concentration of H2O2 will affect the rate of reaction
well if u see in line 1 and line 2 there is change in rate of reaction that means that h2o2 has order of 1 or 2
lets see if it is order 1
0.05=k*1 equation 1
k=0.05/1=0.05
now if we put the value of k for line 2 we get
0.05*1.4=0.07
that means h2o2 is order 1
pretty clear right?
now line 3
again increasing the concentration of h2o2 by 0.02 we would expect a increase of 0.4 in relative rate
however we are also changing the concentration of H+. as the change in conc of H+ didn't affect the rate of reaction so it has zero order

Right. A bit confused in the beginning. We found k by dividing rate by concentration? Shouldn't it be 1/o.o5 instead?

 

[selena]

Can anyone please solve q9 may/June 2004 9701/1. Do is as fast as you can I have a paper tomorrow

 

[selena]

Also q 18 of the same paper(june04p1)

 

[saifeddin]

how to know if a precipitation dissolves in excess or not?
from the practical papers, the cations and anions questions

 

[saifeddin]

Also q 18 of the same paper(june04p1)

okay so 58.5 kg of NaCl is 1000 moles.. meaning that the rest of all the other reagents are going to have 1000 moles each...
you have 1 H atom, which is 1 g/mol. multiply this by a thousand you will get 1 kg of (H). you also have 1 Cl atom in the equation and it's 35.5 g/mol. multiply this by a thousand you will get 35.5 Kg of (Cl). the next is NaOH, with the same way you will find out that it will have 40 Kg per 1000 moles. that's it
i'm as clueless as you are in the other one xD

 

[daredevil]

how to know if a precipitation dissolves in excess or not?
from the practical papers, the cations and anions questions

u need to learn it for all the salts in the syllabus.... in paper 3 data sheet for that is given....
it's basically the tests for anions and cations. if u dont find it or dont get it then tell me and i'll elaborate

 

[aaaamfa]

2) In the preparation of soft margarine, glyceryl trieleostearate
is suitably hydrogenated so that, on average, one of its side-chains is converted into the
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2 residue and two side-chains are converted into the
CH3(CH2)7CH=CH(CH2)7CO2 residue.
How many moles of hydrogen are required to convert one mole of glyceryl trieleostearate into the
soft margarine?
A 4 B 5 C 6 D 9

need help with the explaination how do we get to the ans in such questions

Section A question no. 6 i dont know what to do and how to do

section A
question no. 9
question no. 10
question no. 11
Section B
question no. 33

hello ppl waiting for ur reply :::
any explainations would be appreciated

 

[Kim Shaw]

2) In the preparation of soft margarine, glyceryl trieleostearate
is suitably hydrogenated so that, on average, one of its side-chains is converted into the
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2 residue and two side-chains are converted into the
CH3(CH2)7CH=CH(CH2)7CO2 residue.
How many moles of hydrogen are required to convert one mole of glyceryl trieleostearate into the
soft margarine?
A 4 B 5 C 6 D 9

need help with the explaination how do we get to the ans in such questions

the diagram for glyceryl trieleostearate is given...plz write the full question WITH diagrams next time...http://2.bp.blogspot.com/-BvWESUAvw...BmBo9gYuGM/s1600/glyceryl+trieleostearate.jpg
http://2.bp.blogspot.com/-BvWESUAvw...BmBo9gYuGM/s1600/glyceryl+trieleostearate.jpg
Step 1: Count the hydrogen atoms in the side-chains.
Upper side-chain and lower side-chain: 2 x (3+6+6+14) = 58
Middle side-chain: 3 + 6 + 6 + 14 = 29
Step 2: Count the hydrogen atoms in each residue
First residue: 3 + 8 + 6 + 14 = 31
Second residue x 2 (because two side-chains are converted): 2 x (3 + 14 + 2 + 14) = 66
Step 3: Calculate how many hydrogen atoms had to be added to each side-chain in order to form the residues.
Upper and lower side-chain residues have 66 - 58 extra hydrogens = 8 extra hydrogens
Middle side-chain residue has 31-29 extra hydrogens = 2 extra hydrogens
Step 4: You know that 10 hydrogen atoms have been added to the original molecule. Each hydrogen molecule contains two hydrogen atoms, so there are 10/2 = 5 hydrogen molecules ie. 5 moles of hydrogen

 

[Kim Shaw]

Section A question no. 6 i dont know what to do and how to do

the question asked for 'the most accurate value of Mr'. I is also said that the compound is gaseous. Hence, the most accurate value is given when gas is in ideal conditions, that is it is an ideal gas. This occurs as high temperature and low pressure. thus, C is correct answer

 

[Kim Shaw]

section A
question no. 9
question no. 10
question no. 11
Section B
question no. 33

9. answer is B because nitric acid is HNO3. It dissociates in water to give 1H+ ion and 1 NO3- ion. It is the only one which dissociates to give 1 mole H+, like HCl(which dissociates to give 1 H+ and 1 Cl-)

10. answer is D. This is found in Equilibria chapter. You have to know le Chatelier’s principle to do that. Basically, on RHS of equa, there is OH-. By making solution more alkaline, u are adding more OH-. The system will thus shift to LHS so as to oppose this change. Thus LHS will be favoured and mor V2+ will be formed.

11. answer is D. This is again an Equilibria question. If u study the graph given, u will see that, as pressure is increased and temperature is decreased, %yield of Z increases. To understand the pressure part, in equation there r more moles on LHS so increasing pressure will increase Z as RHS will be favoured. Equilibrium constant is a CONSTANT(as its name suggests)... so it does NOT change. Now, coming back to graph, if by decreasing temperature %yield of Z is increasing, this means that forward reaction is favoured by low temp, so forward is exothermic(this is understood better if u have knowledge on Chatelier’s principle).

33. ans is C. Catalyst merely lowers activation energy. So it increases BOTH forward and backward reaction. Kinetic energy is the SAME, with or without enzyme. As far as i know, kinetic energy rises ONLY when temp rises, u have to confirm that with some1 else..but i’m 100& sure that catalyst does NOT in ANY WAY increase kinetic energy. So, only statement 2 and 3 are correct...hence C is correct ans.

If anything is unclear, just msg me and I’ll clarify it if i can

 

[Kim Shaw]

Hi guys, PLZ PLZ PLZ reply to my questions...its #6814 and #6815. thnx in advance

 

[rockmoinul]

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