We will use this two reaction equation:
H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2
HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8
we will use the ICE table for the first equation:
H2SO3 + H2O ↔ H3O+ + HSO3-
initial 0.025 0 0
change -X +X +X
Equ (0.025-X) X X
Ka1 = [H3O+] [HSO3-] / [H2SO3]
1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X
∴ X = 0.0127
when [H3O+] = X
∴[H3O+] = 0.0127 M
and when [HSO3-] = X
∴[HSO3-] = 0.0127 M
and when [H2SO3] = 0.025 - X
∴[H2SO3] = 0.025 - 0.0127
= 0.0123 M
when Kw = [OH-][H3O+]
and Kw = 1.1 x 10^-14 / 0.0127
∴[OH-] = 1.1 x 10^-14 / 0.0127
= 8.66 x 10^-13 M
- by using the ICE table for the second equation:
HSO3- + H2O ↔ H3O+ + SO3 2-
initial 0.0127 0.0127 0
change -X +X +X
Equ (0.0127-X) (0.0127+X) X
when Ka2 = [SO32-] [H3O+] / [HSO3-]
by substitution:
6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)
as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X
6.3 x 10^-8 = 0.0127X /0.0127
∴X = 6.3 x 10^-8
when [SO3 2-] = X
∴[SO32-] = 6.3 x 10^-8
I hate stoichiometry. I will help you but I need to know the question first.
Answer: % error of observation is 4.77%
Explanation:
Given:
Observation value = 415nm
theoretical value= 435.8nm
Percent error of observation = theoretical value- observation value/ theoretical value x 100 %
= 435.8-415/435.8= 0.04772 x 100 = 4.77%
therefore % error of observation is 4.77%
Answer:
A is denser than B as it's volume for the same mass is smaller.
Explanation:
Hello.
In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:
Next, we compute the densities as shown below:
In such a way, A is denser is B as it's volume for the same mass is smaller.
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