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 8900. A is passed through a Hall-Heroult cell for 78.0 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 H 0.0 8 X dh

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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 Al₂O, to pure aluminum.
Suppose a current of 8900. A is passed through a Hall-Heroult cell for 78.0 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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Transcribed Image Text:weks.com/alekscgi/x/isl.exe/1o_u-IgNsikr7j8P3jH-lvTqeviKFP6W0cqJcWJdIACROQwyw24GWHIntCe8TQ2aQJJLTK32H8YI2EB... Reading Schedule 19.6 Reduction Po.... Gibbs Free E... 5.3 Enthalpies of... 18.5 Gibbs Free E... M 2 O ELECTROCHEMISTRY Calculating the mass of an electrolysis product from the applied... 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 Al₂O, to pure aluminum. Suppose a current of 8900. A is passed through a Hall-Heroult cell for 78.0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol. W S 0 Explanation # 3 E D Check $ 4 R 0.0 > LL X F H H % 5 5 Ⓡ tv I T G A 6 T MacBook Pro Y Is & 7 H U * 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility The SOLUTION: The le... 8 J 1 ( 9 K ) 0 Math 115 W-SF O L Lara V P 8 db
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