AP Chemistryhardmcq1 pt
A 3.00 A current passes through AgNO₃ solution for 1930 s. What mass of silver deposits? (Ag = 107.9 g/mol)
A.6.47 g
B.3.24 g
C.12.9 g
D.0.647 g
Explanation
Core Concept
Charge = 3.00 × 1930 = 5790 C; moles e⁻ = 5790/96,485 = 0.0600 mol = moles Ag (1:1); mass = 0.0600 × 107.9 = 6.47 g.
Correct Answer
A6.47 g
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