In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (a¹₂03) dissolved in molten cryolite (Na3AlF6), resulting in the reduction of the A¹₂o to pure aluminum. Suppose a current of 830. A is passed through a Hall-Heroult cell for 38.0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

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In the Hall-Heroult process, a large electric current is passed through a
solution of aluminum oxide (a¹₂03) dissolved in molten cryolite (Na3AlF6),
resulting in the reduction of the A¹₂o, to pure aluminum.
Suppose a current of 830. A is passed through a Hall-Heroult cell for 38.0
seconds. Calculate the mass of pure aluminum produced.
Round your answer to 3 significant digits. Also, be sure your answer
contains a unit symbol.
Transcribed Image Text:In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (a¹₂03) dissolved in molten cryolite (Na3AlF6), resulting in the reduction of the A¹₂o, to pure aluminum. Suppose a current of 830. A is passed through a Hall-Heroult cell for 38.0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
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