Answer :
The mass of silver bromide formed when 35.5 ml of 0.184 m silver nitrate is treated with an excess of hydrobromic acid is 9.89 g.
When 35.5 mL of 0.184 M silver nitrate is treated with an excess of hydrobromic acid, the reaction forms silver bromide and a salt containing bromide ions. The mass of silver bromide that is formed can be calculated using the following equation:
Mass = Concentration x Volume x Molecular Weight
Where:
- Mass = Mass of silver bromide
- Concentration = Concentration of silver nitrate (0.184 M)
- Volume = Volume of silver nitrate (35.5 mL)
- Molecular Weight = 187.81 g/mol
Therefore, the mass of silver bromide formed is:
Mass = 0.184 x 35.5 x 187.81 = 9.89 g
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