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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ + 6e-> NH4* (E° = +0.34 volts) • O₂ + 4e -> 2H₂O (E° = +0.82 volts) If you balance and combine the reactions so that 34 moles of NH4* are oxidized to NO₂, how many moles of H* will be produced?

Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ + 6e-> NH4* (E° = +0.34 volts) • O₂ + 4e -> 2H₂O (E° = +0.82 volts) If you balance and combine the reactions so that 34 moles of NH4* are oxidized to NO₂, how many moles of H* will be produced?

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Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ + 6e-> NH4+ (E° = +0.34 volts) O₂ +4e-> 2H₂O (E° = +0.82 volts) If you balance and combine the reactions so that 34 moles of NH4* are oxidized to NO₂, how many moles of H* will be produced?
Transcribed Image Text:Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ + 6e-> NH4+ (E° = +0.34 volts) O₂ +4e-> 2H₂O (E° = +0.82 volts) If you balance and combine the reactions so that 34 moles of NH4* are oxidized to NO₂, how many moles of H* will be produced?
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