What quantity of aluminium is deposited when a current of 10A is passed through a solution of aluminium salt for 1930s? (All=27,
F=96500cmol)
1 answer:
Answer:
AL*3+ + 3e- =AL.
（10A×1930）÷96500=0.2mole e-
0.2÷3×27=1.8g（AL）
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