The number of molecules involved in the reaction, in this case, the number of Ag atoms involved.
Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is 
Explanation:
Dissociation of weak acid is represented as:
cM 0 0
So dissociation constant will be:
Give c= 0.188 M and 
= ?
Putting in the values we get:
Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is
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The empirical and molecular formula : C₂₁H₂₂N₂O₂
<h3>Further explanation</h3>
Given
The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen,
8.38 % Nitrogen, and 9.57 % Oxygen
Required
The empirical and molecular formula
Solution
C : 75.42 : 12 = 6.285
H : 6.63 : 1 = 6.63
N : 8.38 : 14 = 0.599
O : 9.57 : 16 = 0.598
Divide by 0.598
C : H : N : O = 10.5 : 11 : 1 : 1 = 21 : 22 : 2 : 2
The empirical formula : C₂₁H₂₂N₂O₂
(C₂₁H₂₂N₂O₂)n = 334
(334)n=334
n = 1
