In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al₂O₂) dissolved in molten cryolite (Na, AIF), resulting in the reduction of the Al₂O, to pure aluminum. Suppose a current of 610. A is passed through a Hall-Heroult cell for 84.0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol. 0 D.P 0.0 X H 0 3

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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al₂O,) dissolved in molten cryolite (Na, AIF), resulting in
the reduction of the Al₂O, to pure aluminum.
Suppose a current of 610. A is passed through a Hall-Heroult cell for 84,0 seconds. Calculate the mass of pure aluminum produced.
Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
0
D.P
A.D
X
H
0
3
da
Transcribed Image Text:In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al₂O,) dissolved in molten cryolite (Na, AIF), resulting in the reduction of the Al₂O, to pure aluminum. Suppose a current of 610. A is passed through a Hall-Heroult cell for 84,0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol. 0 D.P A.D X H 0 3 da
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