For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:
Kd = c </span>× α²
α = √(Kd/c) × 100%
Kd = 6.0×10⁻⁷
c(HA) = 0.1M
α = √(6.0×10⁻⁷/0.1) × 100% = 0.23%
So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>
Answer:
Explanation:
The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.
The correct answer is 124 ✌
Answer:
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