I only knwi one way which is if water evaporates from a mixture
It's basically when a base and an acid of some kind, come together to make H2O or Water.
Answer:
60 cm³ of water
Explanation:
We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:
Concentration of stock solution (C₁) = 17 M
Volume of stock solution (V₁) = 25 cm³
Concentration of diluted solution (C₂) = 5 M
Volume of diluted solution (V₂) =?
C₁V₁ = C₂V₂
17 × 25 = 5 × V₂
425 = 5 × V₂
Divide both side by 5
V₂ = 425 / 5
V₂ = 85 cm³
Thus, the volume of the diluted solution is 85 cm³
Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:
Volume of stock solution (V₁) = 25 cm³
Volume of diluted solution (V₂) = 85 cm³
Volume of water =?
Volume of water = V₂ – V₁
Volume of water = 85 – 25
Volume of water = 60 cm³
Therefore, 60 cm³ of water is needed to dilute the solution.
d, one atom of oxygen and two atoms of hydrogen
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