#3 What size container would 67.3mol CO2 need at STP?

The volume of the container would be 1507 L

Explanation:

These question can be solved in two different ways:

Let's do the short road.

We know that 1 mol of anything at STP conditions occupies 22.4 L

By a rule of three:

1 mol contains 22.4L at STP

Then, 67.3 mol of gas would be contained in (67.3 . 22.4) /1 = 1507.5 L

We assume CO₂ as an Ideal Gas, so we can apply P . V = n . R . T

At STP, T is 273.15K and P = 1 atm. We replace:

1 atm . V = 67.3 mol . 0.082 . 273.15K

V = (67.3 mol . 0.082 . 273.15K) / 1 atm → 1507 L

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.