The same amount of electricity was passed through electrolyzers with aqueous solutions of mercury (II)

The same amount of electricity was passed through electrolyzers with aqueous solutions of mercury (II) nitrate and silver nitrate. At the same time, mercury with a mass of 401.2 g was released. What is the mass of the released silver? Give the equations of electrode processes.

Given:
Hg (NO3) 2
AgNO3
m (Hg) = 401.2 g

To find:
m (Ag) -?

Decision:
1) Using the periodic system of D.I.Mendeleev, calculate the molar masses of Hg and Ag:
M (Hg) = Mr (Hg) = Ar (Hg) = 201 g / mol;
M (Ag) = Mr (Ag) = Ar (Ag) = 108 g / mol;
2) Calculate the mass of Ag:
m (substances) = (I * t / F) * (M / z);
I1 * t1 / F = I2 * t2 / F;
m (Hg) * z (Hg) / M (Hg) = m (Ag) * z (Ag) / M (Ag);
m (Ag) = m (Hg) * z (Hg) * M (Ag) / (M (Hg) * z (Ag)) = 401.2 * 2 * 108 / (201 * 1) = 431.1 g …

Answer: The mass of Ag is 431.1 g. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.