Answer:
We colect 11.2 L of gas
Explanation:
Let's apply the Ideal Gases Law to solve this:
P . V = n . R .T
where P and T in STP are 1 atm and 273°K
The thing is n which means the number of moles for the gas.
As we know, 1 mol of anything has 6.02x10²³ particles so
6.02x10²³ are occupied in 1 mol of gas
3.00x10²³ are occupied in (3.00x10² .1) / NA = 0.500 moles
So let's go to the formula:
1 atm . V = 0.500m . 0.082 . 273K
V = (0.500m . 0.082 . 273K) / 1atm
V = 11.2L
There is a rule, which says that 1 mol of gas in STP, occupies 22.4L so, since we have half a mole, it will occupy half the volume
