What Is The Molecular Weight Of Ag2SeO4?

The molecular weight of Ag2SeO4 (silver selenate) can be calculated by adding the atomic weights of its constituent elements, which are 2 atoms of silver (Ag), 1 atom of selenium (Se), and 4 atoms of oxygen (O).

Using the atomic weights from the periodic table, we have:

2 x atomic weight of Ag = 2 x 107.87 g/mol = 215.74 g/mol

1 x atomic weight of Se = 78.96 g/mol

4 x atomic weight of O = 4 x 15.99 g/mol = 63.96 g/mol

Adding these values together gives us the molecular weight of Ag2SeO4:

215.74 g/mol + 78.96 g/mol + 63.96 g/mol = 358.66 g/mol

Therefore, the molecular weight of Ag2SeO4 is 358.66 g/mol.