Chemistry: Post your doubts here!

[qwertypoiu]

Help please........ qwertypoiu can you please explain me this question, like why cant the Answer be C? is there anything else behind the unbalanced part ... as i have even asked this question before but didnt get a good response ..

Which of these equations represents the reaction of sulphur dioxide with an excess of aqueous

sodium hydroxide?

A SO2 + NaOH ----> NaHSO3

B SO2 + 2NaOH -------> Na2SO3 + H2O

C SO2 + 2NaOH ------> Na2SO4 + H2O

D SO2 + 2NaOH -----> Na2SO4 + H2

As fantastic girl has already said,
NaOH, when reacting with SO2, initially produces Na2SO3.
It may later get oxidised to Na2SO4 however.

 

[shazmina]

no shaz
SO3 + NaOH -----> Na2SO4 not when SO2 rects wid NaOH
check notes buddy

As fantastic girl has already said,
NaOH, when reacting with SO2, initially produces Na2SO3.
It may later get oxidised to Na2SO4 however.

Thanks both of you

 

[fantastic girl]

Thanks both of you

Ma pleasure

 

[asadalam]

Can anyone explain this?
shazmina

 

[fantastic girl]

Can anyone explain this?
shazmina

balance equation 2 to get 4e- nd this will give 2 moles of I2
so the ratio will be 1:2

 

[shazmina]

Can anyone explain this?
shazmina

First of all we Balance the 2nd equation inorder to cancel the electrons from both the Eqns...
Therefore we multiply 2nd eqn by 2 to get 4 electrons which can be cancelled ... then the final eqn would have 1 mole of IO3- and 2 moles of I2

 

[asadalam]

balance equation 2 to get 4e- nd this will give 2 moles of I2
so the ratio will be 1:2

First of all we Balance the 2nd equation inorder to cancel the electrons from both the Eqns...
Therefore we multiply 2nd eqn by 2 to get 4 electrons which can be cancelled ... then the final eqn would have 1 mole of IO3- and 2 moles of I2

Thanks,mind explaining these as well:

It behaves as an ideal gas so shouldnt the particles be I instead of I2?

HCl is polar,while Ccl4 is not so shouldnt it have higher mp?

 

[fantastic girl]

Thanks,mind explaining these as well:

It behaves as an ideal gas so shouldnt the particles be I instead of I2?

HCl is polar,while Ccl4 is not so shouldnt it have higher mp?

Ur welcome

Mr of iodine is 127x2 = 254
PV=nRT
1x10^5 x 1.247x10^-3 = 6.35/254 x 8.31 x T
T= 600 K

 

[asadalam]

Ur welcome

Mr of iodine is 127x2 = 254
PV=nRT
1x10^5 x 1.247x10^-3 = 6.35/254 x 8.31 x T
T= 600 K

I got this but it says it behaves like an ideal gas,so shouldnt it have separate atoms of I instead of atoms that have intermolecular forces like I2

 

[fantastic girl]

I got this but it says it behaves like an ideal gas,so shouldnt it have separate atoms of I instead of atoms that have intermolecular forces like I2

i didnt think abt this :/
they said iodine vapour so i went wid the formula I2 :/

 

[The Sarcastic Retard]

I got this but it says it behaves like an ideal gas,so shouldnt it have separate atoms of I instead of atoms that have intermolecular forces like I2

Nobel gases have weak van der waals force of attraacton b/w themselves

 

[asadalam]

i didnt think abt this :/
they said iodine vapour so i went wid the formula I2 :/

And what about the second one?HCl and Ccl4 one

 

[qwertypoiu]

And what about the second one?HCl and Ccl4 one

Number of electrons in hydrogen chloride molecule = 18
Number of electrons in tetrachloromethane molecule = 74.
The van der Waal's forces in the second molecule is much greater.
Polarity should be used to justify higher melting point of a molecule over another only if their Mr is same or very similar.
In huge differences like this, it shouldn't be done.

 

[fantastic girl]

And what about the second one?HCl and Ccl4 one

ummm iam not really sure
HCl is a polar molecule so we wud expect it to hve higher melting point
However HCl is a much smaller molecule than CCl4 and CCl4 has more attractive forces between them since it has a greater number of electrons

 

[asadalam]

Number of electrons in hydrogen chloride molecule = 18
Number of electrons in tetrachloromethane molecule = 74.
The van der Waal's forces in the second molecule is much greater.
Polarity should be used to justify higher melting point of a molecule over another only if their Mr is same or very similar.
In huge differences like this, it shouldn't be done.

Thanks a lot,can you explain this as well?

 

[fantastic girl]

Thanks a lot,can you explain this as well?

1- u can calculate the values if u want
or see , H-Cl bond is stronger than H-Br so its less likely that H-Cl bond breaks , thats means enthalphy change will be higher for H-Cl
2 - HBr is more likely to form than HI due to stronger H-Br bond than H-I nd Kp is proportional to the products so a higher value of Kp for H-Br since hydrogen bromide can form more easily than HI
3 - Cl atom is smaller than I so more force of attraction from nucleus so its not easy to remove an electron from a Cl atom.....so it has higher Ionization energy

 

[asadalam]

1- u can calculate the values if u want
or see , H-Cl bond is stronger than H-Br so its less likely that H-Cl bond breaks , thats means enthalphy change will be higher for H-Cl
2 - HBr is more likely to form than HI due to stronger H-Br bond than H-I nd Kp is proportional to the products so a higher value of Kp for H-Br since hydrogen bromide can form more easily than HI
3 - Cl atom is smaller than I so more force of attraction from nucleus so its not easy to remove an electron from a Cl atom.....so it has higher Ionization energy

Thanks!Mind explaining this one as well:

 

[fantastic girl]

Thanks!Mind explaining this one as well:

A nd B is wrong since its an electrophilic addition reaction which is undergone by alkenes
D- alkyl halide cnt form hydrogen bonds
C - its the only right one - when both Br2 in first compound nd one Br2 in second is removed it gives the same product

 

[qwertypoiu]

Thanks!Mind explaining this one as well:

These are halogenoalkanes. They undergo nuclephilic substitution.
The first compound has no hydrogen bonds with water.
When halogenoalkanes undergo hydrolysis, the halogen atoms are replaced by OH groups.
The first compound will have both its bromine atoms replaced, and the second one will have the one replaced. Both will end up with the same result.

EDIT: maybe second compound should not be called a halogenoalkane because it has OH group as well. The answer remains the same though.

 

[asadalam]

These are halogenoalkanes. They undergo nuclephilic substitution.
The first compound has no hydrogen bonds with water.
When halogenoalkanes undergo hydrolysis, the halogen atoms are replaced by OH groups.
The first compound will have both its bromine atoms replaced, and the second one will have the one replaced. Both will end up with the same result.

EDIT: maybe second compound should not be called a halogenoalkane because it has OH group as well. The answer remains the same though.

A nd B is wrong since its an electrophilic addition reaction which is undergone by alkenes
D- alkyl halide cnt form hydrogen bonds
C - its the only right one - when both Br2 in first compound nd one Br2 in second is removed it gives the same product

Thanks a lot bro/sis!Please help out in these as well: