How many oxygen atoms are present in 0.500 mol carbon dioxide
1 answer:
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Here we have to get the effect of addition of 0.25 moles of gas C on the mole fraction of gas A in a mixture of gas having constant pressure.
On addition of 0.25 moles of C gas, the mole fraction of gas A will be .
The partial pressure of gas A can be written as =
×P (where
is the mole fraction of gas A present in the mixture and P is the total pressure of the mixture.
The mole fraction of gas A in a mixture of gas A and C is = and
respectively.
Thus on addition of 0.25 moles of C gas, the mole fraction of gas A will be .
Which is different from the initial state.
Answer:
T2 = 133.333°K
Explanation:
Using Combined Gas Laws:
(600 torr)(10L)/500°K = (200 torr)(8L)/x°K
Cross multiply:
x°K (600 torr)(10L) = 500°K(200 torr)(8L)
Divide:
x°K = (500°K(200 torr)(8L))/(600 torr)(10L)
x = 400/3°K or 133.333°K