A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.
We can calculate the total number of moles using the ideal gas equation.

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.
We can calculate the moles of nitrogen using the ideal gas equation.

The mole fraction of nitrogen in the mixture is:

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
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Answer:
B
Explanation:
the gravitational force is a big play.
Answer:
(a) 3:2; (b) 2:1
Explanation:
The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.
That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.
(a) Ammonia and hydrazine.
In ammonia, the mass ratio of H:N is r₁ = 0.2158/1
In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1
The ratio of the ratios is:

(b) Nitrogen oxides
In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1
In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1
The ratio of the ratios is:
