Fresh day bazuunelson, let's solve. Step 1: Write the half-reaction for copper deposition. During electrolysis, copper(II) ions gain electrons to form solid copper. $$Cu^{2+}(aq) + 2e^- \to Cu(s)$$ Step 2: Calculate the moles of copper deposited. The molar mass of copper (Cu) is $63.5 \text{ g/mol}$. $$\text{Moles of Cu} = \frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}$$ $$\text{Moles of Cu} = \frac{1.48 \text{ g}}{63.5 \text{ g/mol}}$$ $$\text{Moles of Cu} \approx 0.023307 \text{ mol}$$ Step 3: Calculate the moles of electrons transferred. From the half-reaction, $1 \text{ mole of Cu}$ requires $2 \text{ moles of electrons}$. $$\text{Moles of electrons} = \text{Moles of Cu} \times 2$$ $$\text{Moles of electrons} = 0.023307 \text{ mol} \times 2$$ $$\text{Moles of electrons} \approx 0.046614 \text{ mol}$$ Step 4: Calculate the total charge (Q) passed. Faraday's constant is $96500 \text{ C/mol e}^-$. $$\text{Charge (Q)} = \text{Moles of electrons} \times \text{Faraday's constant}$$ $$\text{Charge (Q)} = 0.046614 \text{ mol} \times 96500 \text{ C/mol}$$ $$\text{Charge (Q)} \approx 4500.85 \text{ C}$$ Step 5: Convert time to seconds. The time given is $2.5 \text{ hours}$. $$\text{Time (t)} = 2.5 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min}$$ $$\text{Time (t)} = 9000 \text{ s}$$ Step 6: Calculate the amount of current (I) passed. The relationship between charge, current, and time is $Q = I \times t$. $$I = \frac{Q}{t}$$ $$I = \frac{4500.85 \text{ C}}{9000 \text{ s}}$$ $$I \approx 0.500 \text{ A}$$ The amount of current passed is approximately $\boxed{\text{0.500 A}}$. Send me the next one 📸