The SI unit of time in seconds
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
Answer:
Partial pressure of CO₂ is 406.9 mmHg
Explanation:
To solve the question we should apply the concept of the mole fraction.
Mole fraction = Moles of gas / Total moles
We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)
Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles
To determiine the partial pressure of CO₂ we apply
Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P
Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure
We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg
