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:
KE = 1/2*m*v^2
KE = 1/2*150kg*(20 m/s)^2
KE = 75kg * 400m²/s²
KE = 30,000 kg*m²/s²
KE = 30,000 N*m
KE = 30,000 J
Explanation:
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Iron bending is just a physical change because the iron has only changed shape but there is no new substance. Iron rusting is a chemical change because there is a new substance which is rust.
The Ideal Gas Equation
The term pVnRT is also called the compression factor and is a measure of the ideality of the gas. An ideal gas will always equal 1 when plugged into this equation……