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
Although the models are not provided, I was able to find them and the beakers with solid present in them are:
1C
2A
2C
3A
3C
This is determined by the fact that the beakers all have a piece of closely packed substance laying at the bottom. This closely packed lattice is characteristic of solid substances, and the fact that they exist in the solution in the solid states indicates that they are insoluble.
Answer:
0.032 L or 32 mL
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 9.0 M
V1 = This is what we're looking for.
M2 = 0.145 M
V2 = 2 L
Solve for V1 --> V1 = M2V2/M1
V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L
Answer:
What happens when it is squeezed is that its volume increases, the pressure of the material increases.
Explanation:
This is due to the fact that the elastic modulus of the sponge is high and withstands broad forces without deforming its structure, since the force is made within the proportional limit of its particles without modifying or permanently deforming them, that is why when stopping doing pressure or force on it its shape returns to being the original, this also happens due to the phenomenon of resilience
