a chemical reaction produces oxygen gas. you collect 3.00×10^23 oxygen gas molecules at stp .

what volume of oxygen gas did you collect?

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

20 drops hope this