Chemistry: Post your doubts here!

[ABDSyed]

Please Can Anyone Help Me With a Question In PastPaper
October/November 2008 Question 2b

 

[princessanum]

Please Can Anyone Help Me With a Question In PastPaper
October/November 2008 Question 2b

is it p3???

 

[ABDSyed]

is it p3???

Its P4 Sorry I Forgot to Mention

 

[Ahmed Ali Akbar]

help me in mcq 27 , 30 ,34 of oct/nov 10 p12.

 

[Ahmed Ali Akbar]

I can't help you with 16 since I haven't covered energetics yet but for 25, you'll have to learn it. When a haloalkane is converted into CN, the reagents used are KCN in ethanol and the condition is heat under reflux.

To convert the CN atom to CO2H, you will have to use dilute HCL/H2SO4 with heat under reflux. These are some of the reactions of haloalkanes.

what's role of ethanol in KCN..why we can't use KCN in aqueous form....

 

[Doctor Nemo]

Please Can Anyone Help Me With a Question In PastPaper
October/November 2008 Question 2b

The slowest step in the reaction is the rate determining step and therefore only the species involved in that step will determine the overall rate.

If step 1 is slowest then the rate will only depend on the concentrations of H2O2 and I-. Since it is a bimolecular reaction the order will be first order with respect to both of these. Since H+ only enters in the second reaction which is much faster changing its concentration will not affect the rate-increasing H+ will not increase the rate of reaction 1. So the order with respect to H+ is 0.
The values of a, b, c are 1,1,0

If step 2 is the slowest and therefore rate determining then things a a little more complicated. The rate depends on the concentrations of H+ and an intermediate species IO-. One makes the assumption that since the first step is fast that it is in equilibrium and writes an expression for the equilibrium constant. (This is called the steady-state approximation.)

(IO-) =K
(H2O2) (I-)

As always we leave the water out of the equilibrium expression. So we get an expression for the concentration of (IO- ) = K(H2O2) (I-) and substituting this into the rate expression gives

rate = (H+)K(H2O2) (I-)

So the rate is first order for each and the values for a, b and c are 1, 1and 1.

If the third step is the slowest then the rate will depend the concentrations of the three species.

Rate = (HOI)(H+)(I-)

One needs to find an expression for the concentration of the intermediate HOI.

Again one makes the assumption that the first two steps are fast and therefore are in a state of equilibrium


(HOI) = K2 Thus (HOI) = K2(IO-)(H+)
(IO-)(H+)

Substituting the previous expression for IO- gives (HOI) = K2K(H2O2) (I-)(H+)

Substituting in the the rate expression gives

rate = (H+)(I-) K2K(H2O2) (I-)(H+)

or rate = K2K(H+)2(I-)2(H2O2)

So the values of a, b and c are 1,2,2

This is a lot of work for 3 miserable points.

 

[Doctor Nemo]

help me in mcq 27 , 30 ,34 of oct/nov 10 p12.

For 27

The acid will cleave the ester group removing the -OCH3 group. Structural formulas like this can be confusing. I would suggest filling in the carbon atoms so that you can clearly see that this is an ester. The treatment with hydrogen will then reduce all the double bonds so the answer is A.


For 30

The rule is that to have a chiral carbon you have to have 4 different groups on the carbon. It does not matter if the carbon is in a ring. Looking at just the carbon with the -CH3 group we see that for number 1 the carbon is on an axis of symmetry. Draw a line from that carbon to the opposite side and the two halves are the same. This means that the two sides of the ring attached to that carbon are the same and therefore this compound does not have a chiral carbon. If you do that for the other compounds that is not true. The carbon is not on an axis of symmetry and the two sides are not the same. Therefore they do have chiral carbons and the answer is D.


For 34

There are two differences between sulfuric acid and ethanoic acid. Sulfuric acid is a strong acid so it is completely dissociated while ethanoic acid is a weak acid. Sulfuric acid has two protons while ethanoic acid has only one. Since it is a strong acid and it has two protons it will react more rapidly than ethanoic acid producing both a higher temperature and more hydrogen after two minutes.

Since it has two protons it will produce more total hydrogen after the reactions have gone to completion so all three choices are correct so the answer is A.

 

[Ahmed Ali Akbar]

For 27

The acid will cleave the ester group removing the -OCH3 group. Structural formulas like this can be confusing. I would suggest filling in the carbon atoms so that you can clearly see that this is an ester. The treatment with hydrogen will then reduce all the double bonds so the answer is A.


For 30

The rule is that to have a chiral carbon you have to have 4 different groups on the carbon. It does not matter if the carbon is in a ring. Looking at just the carbon with the -CH3 group we see that for number 1 the carbon is on an axis of symmetry. Draw a line from that carbon to the opposite side and the two halves are the same. This means that the two sides of the ring attached to that carbon are the same and therefore this compound does not have a chiral carbon. If you do that for the other compounds that is not true. The carbon is not on an axis of symmetry and the two sides are not the same. Therefore they do have chiral carbons and the answer is D.


For 34

There are two differences between sulfuric acid and ethanoic acid. Sulfuric acid is a strong acid so it is completely dissociated while ethanoic acid is a weak acid. Sulfuric acid has two protons while ethanoic acid has only one. Since it is a strong acid and it has two protons it will react more rapidly than ethanoic acid producing both a higher temperature and more hydrogen after two minutes.

Since it has two protons it will produce more total hydrogen after the reactions have gone to completion so all three choices are correct so the answer is A.

thanks..but confused about mcq 34..please explain it in more detail if you could..

 

[abcde]

thanks..but confused about mcq 34..please explain it in more detail if you could..

Sulfuric acid is a stronger acid than ethanoic acid since it dissociates to give a larger proportion of H+ ions than ethanoic acid. It will thus produce more H2 gas than ethanoic acid at any point in time. (So 2 and 3 are correct!). Since it is a stronger acid, it also undergoes a more exothermic reaction, releasing more heat energy than ethanoic acid after 2 mins => higher temp.( 1 is correct, too). Hence, A is the right option.

 

[Noor]

I wanna wake up one day to know that Inorganic chemistry was murdered a slow painful death. Amen.
Anyone knows where can I get Chemistry O/N 2011? /:

 

[Doctor Nemo]

thanks..but confused about mcq 34..please explain it in more detail if you could..

Let´s make it concrete with some numbers.

Suppose you stated with 1 moldm-3 solutions of the two acids. Since ethanoic acid is a weak acid you calculate the concentration of H+ from the acid dissociation constant.

CH3COOH < >------< > H+ + CH3COO- Ka = 1.7 X10^-5

So the concentration of H+ is sqrt (1.7 X 10-5) =4.1 X 10-3

H2SO4 is a strong acid and therefore you can assume that all the protons are released and the concentration of H+ = to 2X that of the acid since there are two protons.

H2SO4 ------->SO4 -2 + 2H+

In the sulfuric acid the concentration of H+ is 2 moldm-3 while in the ethanoic acid it is only 4.1 X 10-3. So the concentration of protons in the sulfuric acid is about 500 X that of ethanoic acid.

Since the concentration of protons is much higher in the sulfuric acid solution the reaction will be much more rapid producing a faster rise in the temperature and a faster production of hydrogen.

At the end of the reaction the sulfuric acid will also have produced more hydrogen because each sulfuric acid has two protons while ethanoic acid has only one. Number 3 would not be correct if the starting acid was another strong acid such as H Cl which like ethanoic acid has only one proton.

 

[Ahmed Ali Akbar]

Let´s make it concrete with some numbers.

Suppose you stated with 1 moldm-3 solutions of the two acids. Since ethanoic acid is a weak acid you calculate the concentration of H+ from the acid dissociation constant.

CH3COOH < >------< > H+ + CH3COO- Ka = 1.7 X10^-5

So the concentration of H+ is sqrt (1.7 X 10-5) =4.1 X 10-3

H2SO4 is a strong acid and therefore you can assume that all the protons are released and the concentration of H+ = to 2X that of the acid since there are two protons.

H2SO4 ------->SO4 -2 + 2H+

In the sulfuric acid the concentration of H+ is 2 moldm-3 while in the ethanoic acid it is only 4.1 X 10-3. So the concentration of protons in the sulfuric acid is about 500 X that of ethanoic acid.

Since the concentration of protons is much higher in the sulfuric acid solution the reaction will be much more rapid producing a faster rise in the temperature and a faster production of hydrogen.

At the end of the reaction the sulfuric acid will also have produced more hydrogen because each sulfuric acid has two protons while ethanoic acid has only one. Number 3 would not be correct if the starting acid was another strong acid such as H Cl which like ethanoic acid has only one proton.

Thanks..

 

[Doctor Nemo]

I wanna wake up one day to know that Inorganic chemistry was murdered a slow painful death. Amen.
Anyone knows where can I get Chemistry O/N 2011? /:

A wished death never comes.

Love thy enemies. That will really bother and upset them.

Remember that both beauty and ugliness are only skin deep. Try to see beyond the repulsive surface of inorganic chemistry, beyond the trivia, useless and random information and seek the inner beauty and lovable nature of inorganic chemistry.

While you are on that mission impossible, wandering through the deep, dark and damp passageways with no more light or company than that of a birthday candle, remind yourself why you are there.

It is because subjects like inorganic chemistry are what separates an A* from just an A....

 

[Ahmed Khider]

I found this question in paper 63 oct-nov 2011>>>>Seawater contains sodium chloride and other salts.
Plan an experiment to fi nd the mass of salts in 1 dm3 of seawater.
You will be provided with a small bottle of seawater.
You should include details of the method and any apparatus used.
(1 dm3 = 1000 cm3). How do i solve this question? I thought of heating it and collecting the salts but they are soluble in water!

 

[Noor]

A wished death never comes.

Love thy enemies. That will really bother and upset them.

Remember that both beauty and ugliness are only skin deep. Try to see beyond the repulsive surface of inorganic chemistry, beyond the trivia, useless and random information and seek the inner beauty and lovable nature of inorganic chemistry.

While you are on that mission impossible, wandering through the deep, dark and damp passageways with no more light or company than that of a birthday candle, remind yourself why you are there.

It is because subjects like inorganic chemistry are what separates an A* from just an A....

._. That's deep, man! XD I'm tryina do that but paper one and those many reactions really do get in the way! :c
Ikr? It won't take just couple of days to let it sink in my brain, I'll try to give it more time. XD

 

[soumayya]

I found this question in paper 63 oct-nov 2011>>>>Seawater contains sodium chloride and other salts.
Plan an experiment to fi nd the mass of salts in 1 dm3 of seawater.
You will be provided with a small bottle of seawater.
You should include details of the method and any apparatus used.
(1 dm3 = 1000 cm3). How do i solve this question? I thought of heating it and collecting the salts but they are soluble in water!

u just hve to measure the mass of salt after crystallisation...it's solubility has no connection here...

 

[MEGUSTA_xD]

I found this question in paper 63 oct-nov 2011>>>>Seawater contains sodium chloride and other salts.
Plan an experiment to fi nd the mass of salts in 1 dm3 of seawater.
You will be provided with a small bottle of seawater.
You should include details of the method and any apparatus used.
(1 dm3 = 1000 cm3). How do i solve this question? I thought of heating it and collecting the salts but they are soluble in water!

When you heat it to crystaLisaTion point ( no water ) you will then measure the mass.
You need to note down the mass of the beaker where the experiment you are going to do, and then minus that mass with the final mass.

 

[Doctor Nemo]

I found this question in paper 63 oct-nov 2011>>>>Seawater contains sodium chloride and other salts.
Plan an experiment to fi nd the mass of salts in 1 dm3 of seawater.
You will be provided with a small bottle of seawater.
You should include details of the method and any apparatus used.
(1 dm3 = 1000 cm3). How do i solve this question? I thought of heating it and collecting the salts but they are soluble in water!

One could try to make it more complicated but basically that is it-evaporating
off the water by heating it to leave a solid.

So details

label 3 beakers 100 cm3 volume, add a few boiling stones to each and then weigh them.

Use a volumetric pipette to measure out 10.0 cm3 of seawater to each beaker

Gently heat the beakers on a bunsen burner or electric hot plate so that the liquid does not boil too vigorously and cause liquid to splash out of beaker.

Handle hot beakers with metal tongs (always try to think of some safety feature to add to method).

Heat to dryness, then let beakers cool to room temperature and then weigh them again.

Subtract weight of beaker + boiling stones from weight of beaker after heating to obtain weight of salt from 10.0 cm3 sea water.

Average the 3 values

Multiply the value for 10 cm3 by 100 to obtain the amount in 1 dm3 sea water.

 

[Ahmed Ali Akbar]

help needed in mcq 6,8,27 of may june 08....

 

[Just call me MJ]

Ok. a really annoying question that is repeated many times over is Question 6, paper 1, may/june 2002.....
I memorized it but I don't understand why O__o could somebody explain it please?!?!?!