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
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 20 g
m (final mass after time T) = 5 g
x (number of periods elapsed) = ?
P (Half-life) = ? (in minutes)
T (Elapsed time for sample reduction) = 8 minutes
Let's find the number of periods elapsed (x), let us see:
Now, let's find the half-life (P) of the radioactive sample, let's see:
I Hope this helps, greetings ... DexteR! =)
To dilute a solute in a solution, it is necessary to add a proper solvent for the reaction to occur, and adding more solvent will cause the solution to dilute even more, therefore the best answer will be letter B
Answer: N2
blah blah blah blah blah blah
Answer:
It comes from the chemical reaction of The wax and the gas because the wax is made up of carbon-based chemicals so the gas and carbon make the candle light =)
Explanation:
