Question

What is the percent ionization of H₂NNH₂ in a solution with a concentration of a 0.620 M? (Kb = 1.3 × 10⁻⁶)

Answer 1

Answer: hi

Explanation: bye

Answer 2

Answer:

$\alpha =\sqrt{\frac{Kb}{c} } =\sqrt{\frac{1.3\ X\ 10^{-6} }{0.620} } = 1.01 X 10^{-2}=$

1.01%