Answer :
There are approximately 144.88 grams of silver in 187 grams of Ag2S.To determine the amount of silver (Ag) present in 187g of Ag2S, we need to consider the molar mass and stoichiometry of the compound.
The molar mass of Ag2S can be calculated as follows:
Ag: 1 atom x 107.87 g/mol = 107.87 g/mol
S: 1 atom x 32.07 g/mol = 32.07 g/mol
Total: 107.87 g/mol + 32.07 g/mol = 139.94 g/mol
The molar mass of Ag2S is 139.94 g/mol, indicating that 1 mole of Ag2S weighs 139.94 grams.
To determine the amount of Ag in 187g of Ag2S, we can set up a proportion:
(187 g Ag2S) / (139.94 g Ag2S) = (x g Ag) / (107.87 g Ag)
Solving this proportion, we find:
x g Ag = (187 g Ag2S) * (107.87 g Ag) / (139.94 g Ag2S) = 144.88 g Ag
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