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missladydjoke
@missladydjoke
September 2021
1
15
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Який об'єм кисню потрібний для спалювання 8 л водню?
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Тетянія
×л ₈ л
2H₂ + O₂ → 2H₂O
₄₄.₈ ₂₂.₄
V(H₂)=44.8*8/22.4= 16 л
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Answers & Comments
2H₂ + O₂ → 2H₂O
₄₄.₈ ₂₂.₄
V(H₂)=44.8*8/22.4= 16 л