Question

How much space would .33 moles of oxygen take up at STP?​

Answer 1

`V = 7.391\,L`

Step-by-step explanation:

Let suppose that oxygen behaves ideally. The equation of state of the ideal gas is:

`P\cdot V = n\cdot R_(u)\cdot T`

The volume is now cleared:

`V = (n\cdot R_(u)\cdot T)/(P)`

`V = ((0.33\,mol)\cdot \left(0.082\,(atm\cdot L)/(mol\cdot K) \right)\cdot (273.15\,K))/(1\,atm)`

`V = 7.391\,L`

Answer 2

0.33 moles of Oxygen at stp would occupy a volume of 7.392 dm³

Step-by-step explanation:

1 mole of every gas at standard temperature and pressure (stp) occupies 22.4 dm³

0.33 moles of Oxygen at stp would occupy a volume of 0.33 × 22.4 dm³ = 7.392 dm³

Hope this Helps!!!