Chemistry: Post your doubts here!

[Thampi4]

Hey guys,

i dont get the overlap of orbitals.
could someone explain it pls?

much appreciated

 

[shimeka]

I get many problem in attempting questions including hess law.. could someone help me.. I know the concept but i am unable to do most question related to that... If someone would explain that to me again...

 

[Ahmed Ali Akbar]

paper 11 or 12..??

Paper 11

 

[soumayya]

i need help in may june 11 P1 ...mcq 22, 6, 35..

6) only ze addition of some more reactants would cause more products to be formed, given that the graph had levelled off in X...adding water would only decrease the conc. and the no. of reactant molecules will be the same as would be the no. of product...this reaction is not reversible..so, altering temp. would not affect conc. of products....
manganese (IV) oxide is only a catalyst which will affect only rate and not conc. of product...

 

[soumayya]

[QUOTEi need help in may june 11 P1 ...mcq 22, 6, 35..][/QUOTE]
22)


a radical contains an unpaired electron...
an electrophile is an electron deficient reagent..it can be a positive ion..
a nuleophile is an electron rich reagent..it can be a negative ion...

 

[soumayya]

CaO +H2O → Ca(OH)2
Na(OH)2 + H2O → NaOH PH=13
SrO + H2O → SR(OH)2

All of these oxides dissolve in water to form alkalis..SrO will dissolve more readily in water than CaO because of its lower lattice energy and hence form a more alkaline solution...

 

[hassam]

Nickel often occurs in ores along with iron. After the initial reduction of the ore with coke,
a nickel-iron alloy is formed.
Use data from the Data Booklet to explain why nickel can be purified by a similar
electrolysis technique to that used for copper, using an impure nickel anode, a pure
nickel cathode, and nickel sulfate as the electrolyte. Explain what would happen to the
iron during this process.

 

[Ahmed Ali Akbar]

6) only ze addition of some more reactants would cause more products to be formed, given that the graph had levelled off in X...adding water would only decrease the conc. and the no. of reactant molecules will be the same as would be the no. of product...this reaction is not reversible..so, altering temp. would not affect conc. of products....
manganese (IV) oxide is only a catalyst which will affect only rate and not conc. of product...

Thanks .. a lot.could you explain me effect of adding water again?but in more detail..And will addition of water affects effective collisions between reactants..

 

[Ahmed Ali Akbar]

may june 07 paper 01,,,Mcq 16 and 25....why in mcq 16 answer is not B ?

 

[CaptainDanger]

9701_w01_qp_3.pdf

9701_w01_ms_3.pdf

Q2 e part... Where they say filter the mixture... Explain observation and Deduction to that part?

 

[Pals_1010]

may june 07 paper 01,,,Mcq 16 and 25....why in mcq 16 answer is not B ?

The activation energy does not affect the enthalpy change.

See? Finally the enthalpy change is unaffected by whether the activation energy is high or not.

 

[Ahmed Ali Akbar]

The activation energy does not affect the enthalpy change.

See? Finally the enthalpy change is unaffected by whether the activation energy is high or not.[/.
Then how bond energies of HI and HCL explains the reason for enthalpy change of formation?

 

[soumayya]

Thanks .. a lot.could you explain me effect of adding water again?but in more detail..And will addition of water affects effective collisions between reactants..

normally the addition of water will cause a solution to become more dilute..so particles of reactant will become further apart..hence reducing the chance of effective collision..in zis case there would be less chance of a reactant molecule to collide with the catalyst...but since the no. of H2O2 in the solution is constant , there wouldn't be an increase in the volume of O2 produced (as is the case in the graph)...

 

[Sabcore]

Predict the group in the periodic table in which an element with the following ionization energies would most likely be found.
1st IE = 786 kJ/mol
2nd IE = 1577
3rd IE = 3232
4th IE = 4355
5th IE = 16,091
6th IE = 19,784

My teacher didn't explain this well. I am doing third yr for A-level and got E in chem last yr. I want to go for an A* now and I am hitting hard on learning every titbit.
Someone please explains to me with all the details how to solve the above stuff. I tried some but i end up getting something different.
jazakallah

 

[Doctor Nemo]

Predict the group in the periodic table in which an element with the following ionization energies would most likely be found.
1st IE = 786 kJ/mol
2nd IE = 1577
3rd IE = 3232
4th IE = 4355
5th IE = 16,091
6th IE = 19,784

My teacher didn't explain this well. I am doing third yr for A-level and got E in chem last yr. I want to go for an A* now and I am hitting hard on learning every titbit.
Someone please explains to me with all the details how to solve the above stuff. I tried some but i end up getting something different.
jazakallah

Ionisation energies are a measure of the energy of an electron in an atom and these energies can be compared to the electron configurations of the atoms. In group I the amount of energy needed to remove the 2nd electron is much greater than that needed to remove the 1st electron. This is because the 2nd electron is in a filled shell that is closer to the nucleus than that of the 1st electron. The relative values of the first and second ionisation energies explains why the group I elements form ions with +1 charge and not ions with +2 charge.


Generally, it is the difference in ionisation energy that one analyses in identifying elements on the basis of these ionisation energies.
One looks for a large change in this difference that would indicate that an electron is being removed from a lower shell than the others.
Predict the group in the periodic table in which an element with the following ionization energies would most likely be found.

1st IE = 786 kJ/mol
2nd IE = 1577
3rd IE = 3232
4th IE = 4355
5th IE = 16,091
6th IE = 19,784

difference 1 &2 = 791

2 & 3=1655

3 & 4=1123

4 & 5=11736

For the first 4 electrons the difference is about the same. However it takes much more energy to remove the 5th electron so it much be in a lower shell then the first 4 so this is an element from group 4.

 

[Sandhya Mahat]

Anyone can u plz tell me where can i get the data booklet for chemistry? I cn't prepare without it... i wud b soo grateful 4 ur help!!

 

[XPFMember]

Anyone can u plz tell me where can i get the data booklet for chemistry? I cn't prepare without it... i wud b soo grateful 4 ur help!!

It's there towards the end of the syllabus btw..

anyway check the attachment!

 

[Hinafatima]

i also go with SOCl2 because then the product would be SO2 which is a gas along with chloroalkane

 

[Sabcore]

Ionisation energies are a measure of the energy of an electron in an atom and these energies can be compared to the electron configurations of the atoms. In group I the amount of energy needed to remove the 2nd electron is much greater than that needed to remove the 1st electron. This is because the 2nd electron is in a filled shell that is closer to the nucleus than that of the 1st electron. The relative values of the first and second ionisation energies explains why the group I elements form ions with +1 charge and not ions with +2 charge.


Generally, it is the difference in ionisation energy that one analyses in identifying elements on the basis of these ionisation energies.
One looks for a large change in this difference that would indicate that an electron is being removed from a lower shell than the others.
Predict the group in the periodic table in which an element with the following ionization energies would most likely be found.

1st IE = 786 kJ/mol
2nd IE = 1577
3rd IE = 3232
4th IE = 4355
5th IE = 16,091
6th IE = 19,784

difference 1 &2 = 791

2 & 3=1655

3 & 4=1123

4 & 5=11736

For the first 4 electrons the difference is about the same. However it takes much more energy to remove the 5th electron so it much be in a lower shell then the first 4 so this is an element from group 4.

Ahaha Thanks a lots I understood it well now
I hope I will be able to solve my problems, i I can't I will just post here again. BTW do you have msn or something? I would like to talk to you when I encounter problems I need to get an A* from my E of last year in 7 months, so I am doing my "bestest"

Thanks again buddy.

 

[leosco1995]

may june 07 paper 01,,,Mcq 16 and 25....why in mcq 16 answer is not B ?

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.