Answer:
\[{{w}_{{{H}_{2}}}}=20g;\] \[{{w}_{{{O}_{2}}}}=80g\]
\[{{n}_{{{H}_{2}}}}=\frac{20}{2}=10;\] \[{{n}_{{{O}_{2}}}}=\frac{80}{32}=2.5\]
\[{{x}_{{{H}_{2}}}}=\frac{10}{10+2.5}=0.8;\] \[{{x}_{{{O}_{2}}}}=1-{{x}_{{{H}_{2}}}}=0.2\]
\[{{P}_{{{H}_{2}}}}={{x}_{{{H}_{2}}}}\times P\]
\[=0.8\times 1=0.8bar\]
\[{{P}_{{{O}_{2}}}}=0.2\times
1=0.2bar\]
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