Answer: D To calculate the mass of silver deposited during electrolysis, we use the formula: \(m = \frac{Q \cdot M}{n \cdot F}\) Where: \( I = 0.8 \, \text{A} \), \( t = 25 \, \text{minutes} = 25 \times 60 = 1500 \, \text{s} \), \( Q = I \cdot t = 0.8 \, \text{A} \cdot 1500 \, \text{s} = 1200 \, \text{C} \) \( M = 107.87 \, \text{g/mol} \), \( n = 1 \), \( F \approx 96485 \, \text{C/mol} \) Substituting the values: \(m = \frac{1200 \, \text{C} \cdot 107.87 \, \text{g/mol}}{1 \cdot 96485 \, \text{C/mol}} = \frac{129444}{96500} \approx 1.34 \, \text{g}\)