The partial pressure of ethane over …
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Sia ? 6 years, 5 months ago
Applying the relationship
{tex}\style{font-family:'Times New Roman'}{m\;=K_H\;p\;}{/tex}
In the first case, {tex}\style{font-family:'Times New Roman'}{6.56\;\times\;10^{-2}g\;=\;K_H\;\times\;1\;bar}{/tex}
{tex}\style{font-family:'Times New Roman'}{\;K_H\;=6.56\;\times\;10^{-2}g\;/bar}{/tex}
In second case, {tex}\style{font-family:'Times New Roman'}{\;5.00\;\times\;10^{-2}g\;=6.56\;\times\;10^{-2}g/bar\;\times\;p}{/tex}
{tex}P = \frac{{5.00 \times {{10}^{ - 2}}g}}{{6.56 \times {{10}^{ - 2}}g\,ba{r^{ - 1}}}}{/tex} = 0.7621 bar
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