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
Answer:
1 gram
Explanation:
Half life = 25 years
Starting mass = 16 grams
Time = 100 years
Number of half lives = Time / Duration of Half life = 100 / 25 = 4
After first Half life;
Remaining mass = 16 / 2 = 8 g
After Second Half life;
Remaining mass = 8 / 2 = 4 g
After Third Half life;
Remaining mass = 4 / 2 = 2 g
After Fourth Half life;
Remaining mass = 2 / 2 = 1 g
Answer:
a) 32.09 kPa
b) 32.09 kPa
Explanation:
Given data:
rate constant 
initial pressure is = 32.1 kPa
half life of A is calculated as
for calculating pressure we have follwing expression
a) 
b) 
HELPPPPPPPPPPPPPPPPPPPPPPPf it is a good idea to
