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.
Learn more: brainly.com/question/2060778
Glucose + Oxygen > Carbon Dioxide
C(6)H(12)0(6)+ 0(2) >CO(2)
For the atom to be neutral, it must have the same number of electrons and protons. The aromic number indicates the number of protons. Since Carbon has 6 protons, it would have to have 6 electrons for it to be neutral
Corresponding in size or amount to something else.
Answer:
B. They both contain three atoms around the central atom.
Explanation:
Do the unit test on edg20 and got it right!