Answer:
The chemical term in the equation for the precipitate of AgCl(s) is n=3.54*10^-3
Step-by-step explanation:
the quantity of AgCl(s) in moles is:
n = 0.508g / 143.32 g/mol = 3.54*10^-3 mol
to verify it the mass of AgNO3 involved in the reaction should be
n AgNO3 required = n = 3.54*10^-3 mol
the mass of n involved should be higher than n AgNO3
n existing = V*N = 0.523 mol/L * 35*10^-3 L = 18.305*10^-3 mol