96,485 C will plate 1/3 mol Cr^3+ or 0.333 mol Cr^3+. You want 0.25. You will need 96,485 C x 0.25/0.333 = ?
C = A x sec
?C = A x 8 hr x (60min/hr) x (60sec/min)
Solve for A.
What amperage is required to plate out .250 mol Cr from a Cr+3 solution in a period of 8.00 h?
1 answer