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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
Answer:
E = 2.85×10⁻¹⁹ J
Explanation:
Given data:
Wavelength of light = 700 nm
Frequency of light = 4.3×10¹⁴ s⁻¹
What is the energy of light = ?
Solution:
Formula:
E = h × f
h = plancks constant = 6.63×10⁻³⁴ Js
by putting values,
E = 6.63×10⁻³⁴ Js × 4.3×10¹⁴ s⁻¹
E = 28.5×10⁻²⁰ J
E = 2.85×10⁻¹⁹ J
country of origin and chemical makeup
Answer:
two hydrogen atoms and one oxygen atom
Explanation: