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
Answer:
The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.
Explanation:
Mass of compound A in a mixture = 119 mg
Mass of compound A after re-crystallization = 83 mg
Percent recovery from re-crystallization :

Percent recovery of compound A:

Mass of compound B in a mixture = 97 mg
Mass of compound B after re-crystallization = 79 mg
Percent recovery of compound B:

In a food chain, energy is passed through one link to another. When a herbivore eats only a certain fraction of the energy, (which comes from the food) it becomes new body mass; the rest of the energy is lost as waste or used up by the herbivore in order to carry out its life processes (ex. movement, digestion, reproduction). It doesn’t necessarily threaten the plants survival, there’s also a benefit. When a animals poops out the fruit (defecate) in another area those seeds get carried to new places with the help of a dab of fertilizer and a little bit of moisture. They also help supply nutrients when they die and decompose.
The balanced chemical
reaction will be:
2H2O = 2H2 + O2
<span>We are given the amount of water used in the decomposition reaction. This will be our
starting point.</span>
<span>17.0 g H2O</span> (1 mol H2O/ 18.02 g H2O) (1 mol O2/2
mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2
Percent yield = actual yield / theoretical yield x 100
<span>Percent yield =10.2 g / 15.09 g
x 100</span>
Percent yield = 67.58%