AP Chemistryhardmcq1 pt
How long must a 5.00 A current run to deposit 10.0 g of silver? (Ag = 107.9 g/mol)
A.1790 s
B.894 s
C.3580 s
D.447 s
Explanation
Core Concept
Moles Ag = 10.0 ÷ 107.9 = 0.0927 mol needs 0.0927 mol e⁻ (1:1). Charge = 0.0927 × 96,485 = 8944 C. t = 8944 ÷ 5.00 = 1789 s ≈ 1790 s.
Correct Answer
A1790 s
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