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: pH = 14
Explanation: Please see the attachments below
Formula units in 450 g of
is 1.93 × 10²⁴ formula units.
<u>Explanation:</u>
First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.
3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.
So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.
Your closes answer would be a.10