An oxygen gas has volume of 10 L and contains .568 moles. What will be the new volume of the container if .450 moles will be used to fill in the adjustable container?*2 puntos5.92 L6.92 L7.92 L8.92 l
Question
An oxygen gas has volume of 10 L and contains .568 moles. What will be the new volume of the container if .450 moles will be used to fill in the adjustable container?*2 puntos5.92 L6.92 L7.92 L8.92 l
Solution
This problem can be solved using the ideal gas law, which states that the volume of a gas is directly proportional to the number of moles of the gas if the temperature and pressure are held constant.
The formula for the ideal gas law is: V1/n1 = V2/n2
Where: V1 = initial volume = 10 L n1 = initial number of moles = 0.568 moles V2 = final volume n2 = final number of moles = 0.450 moles
We can rearrange the formula to solve for V2:
V2 = (V1 * n2) / n1
Substituting the given values:
V2 = (10 L * 0.450 moles) / 0.568 moles
V2 = 7.92 L
So, the new volume of the container will be 7.92 L.
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