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Question #fbc22

1 Answer

Sep 29, 2016

Because both processes involve the same reaction:

$H_{2} + (\frac{1}{2}) O_{2} \to H_{2} O$ $"H"_2+(1/2)"O"_2 \rarr "H"_2"O"$

Explanation:

Formation of water: Hydrogen combined with oxygen

$H_{2} + (\frac{1}{2}) O_{2} \to H_{2} O$ $"H"_2+(1/2)"O"_2 \rarr "H"_2"O"$

Combustion of hydrogen: Hydrogen combined with oxygen, the usual oxidizer for combustion, to nake the most stable product(s):

$H_{2} + (\frac{1}{2}) O_{2} \to H_{2} O$ $"H"_2+(1/2)"O"_2 \rarr "H"_2"O"$

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