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Chemistry: Post your doubts here!

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upload_2015-6-7_15-41-32-png.54755

Why would the ratio be 3:1??
upload_2015-6-7_15-42-47-png.54756

Why is 3 wrong? If something can form H-bonds then it will mix right?
upload_2015-6-7_15-43-6-png.54757

Why would 3 also occur?
 
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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
 
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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

no shaz
SO3 + NaOH -----> Na2SO4 not when SO2 rects wid NaOH
check notes buddy :p
 
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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.
 
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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:
11414951_389508844590379_443543328_o.jpg

It behaves as an ideal gas so shouldnt the particles be I instead of I2?
11126037_389508841257046_229252897_n.jpg

HCl is polar,while Ccl4 is not so shouldnt it have higher mp?
 
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Thanks,mind explaining these as well:
11414951_389508844590379_443543328_o.jpg

It behaves as an ideal gas so shouldnt the particles be I instead of I2?
11126037_389508841257046_229252897_n.jpg

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
 
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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
 
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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 :/
 
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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.
 
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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
 
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