Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron acceptor. • NO3 + 10e-> N₂ (EO = +0. V) • H*+2e-> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 74 moles of H₂ gas are oxidized to H*, how many moles electrons will be transferred from hydrogen to nitrogen? • Report your answer in standard decimal notation rounded to two decimal places. Include trailing zeros, even if the answer is a whole number oe.g. 18.00, not 18 • Report only the numeric part of your answer oe.g. 18.00, not 18.00 moles e

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Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron
acceptor.
• NO3 + 10e-> N₂ (EO = +0. V)
•
H*+2e-> H₂ (E° = -0.42 V)
If you balance and combine the reactions so that 74 moles of H₂ gas are oxidized to H*, how many moles electrons will be transferred from hydrogen to nitrogen?
• Report your answer in standard decimal notation rounded to two decimal places.
Include trailing zeros, even if the answer is a whole number
oe.g. 18.00, not 18
• Report only the numeric part of your answer
oe.g. 18.00, not 18.00 moles e
Transcribed Image Text:Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron acceptor. • NO3 + 10e-> N₂ (EO = +0. V) • H*+2e-> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 74 moles of H₂ gas are oxidized to H*, how many moles electrons will be transferred from hydrogen to nitrogen? • Report your answer in standard decimal notation rounded to two decimal places. Include trailing zeros, even if the answer is a whole number oe.g. 18.00, not 18 • Report only the numeric part of your answer oe.g. 18.00, not 18.00 moles e
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