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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. NO26eNH4 (-0.41 volts) 02+ 4e-> 2H20 (+0.82 volts) If you balance and combine the reactions so that 78 molecules of NH4 are oxidized to NO2, how many molecules 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. NO26eNH4 (-0.41 volts) 02+ 4e-> 2H20 (+0.82 volts) If you balance and combine the reactions so that 78 molecules of NH4 are oxidized to NO2, how many molecules of H will be produced?

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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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. NO26eNH4 (-0.41 volts) 02+ 4e-> 2H20 (+0.82 volts) If you balance and combine the reactions so that 78 molecules of NH4 are oxidized to NO2, how many molecules 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. NO26eNH4 (-0.41 volts) 02+ 4e-> 2H20 (+0.82 volts) If you balance and combine the reactions so that 78 molecules of NH4 are oxidized to NO2, how many molecules of H will be produced?
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