The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Hey there!:
Given that;
SO2 (g ) = (-296.8 kJ/mol) and SO3 (g )= (-395.7 Kj/mol)
2 SO2(g) + O2(g) --------------> 2 SO3(g)
ΔH reaction = ΔH of product - ΔH of reactant
We know that ΔH O2 = 0 kj
ΔH reaction = [2 x - 395.7] - [2 x - 296.8 +0.0 KJ]
= - 791.4 - (- 593.6)
= - 791.4 + 593.6
= - 197.8 kJ
Because there are 2 mole of SO2 coverted into 2 mole of SO3 :
=- 197.8 kJ/2 mol
ΔH°rxn = - 98.9 kJ/ mol
Hope that helps!
Mater means mom
The mater has kept the house in London
If you meant matter
There is not a lot of matter in an apple compared to the matter in a person.
N -3
Ba +2
Sr +2
F -1
I -1
Ca +2
Mg +2
S -2
S -2
Al +3
//
Ba3N2
SrF2
CaI2
MgS
Al2S3
//
I don't really understand 2.
For an ideal gas, the ideal gas equation can be used to relate the changes in the conditions of the system. The equation is expressed as:
PV = nRT
Therefore, the following are classified as:
P and T = directly proportional
V and n = directly proportional
T and V = <span>directly proportional</span>
P and V = inversely proportional
P and n = directly proportional