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.