In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al2O3) dissolved in molten cryolite (Na3AlF6), resulting in the reduction of the Al₂O3 to pure aluminum. Suppose a current of 990. A is passed through a Hall-Heroult cell for 68.0 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 0.0 X μ 믐 00

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Chapter17: Electrochemistry
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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al2O3) dissolved in molten cryolite (Na, AIF), resulting in
the reduction of the Al2O3 to pure aluminum.
Suppose a current of 990. A is passed through a Hall-Heroult cell for 68.0 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
0x10
0.0
X
μ
010
5
Transcribed Image Text:In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al2O3) dissolved in molten cryolite (Na, AIF), resulting in the reduction of the Al2O3 to pure aluminum. Suppose a current of 990. A is passed through a Hall-Heroult cell for 68.0 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 0x10 0.0 X μ 010 5
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