Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
Here we go ~
Energy difference btween the two electronic states can be expressed as :
[ h = planks constant,
= frequency ]
These oppositely charged compounds are strongly held by electrostatic forces of attraction as these be together for a long time the rise in temparature occurs so that the the melting points rises in them.
See , from the equation we can see that for forming two mole of H2O 2Mole of H2 has to react.
Mass of 2 Mole H2O is 18*2gm or 36gm.
So for forming 36 gm H2O 2×2 I.e. 4 gm H2 has to take part in reaction.
Therefore, to form 1 gm H2O 4÷36 gm of H2 has to take part.
So, for forming 47gm H2O (4÷36)×47 gm H2 has to take part
I.e. 5.22 gm of H2 has to take part
So, ans is 5.22 gm of hydrogen.
Hope it helps!!!
Atomic mass is the quantity of protons that have in a atom. Which atom have different numbers of protons, being able to be identified by them.
