A sample of 0.300 moles of nitrogen occupies 0.600 L. Under the same conditions, what number of moles occupies 1.200L

0.600 moles has volume 1.200L

Explanation:

The volume is numerically double the moles

0.600 mol occupies 1.200L

Mathematically:

V1/moles1 = V2/moles2

1.200L / moles1 = 0.600L/ 0.300mol

Moles 0.600L = 1.200L * 0.300mol

Moles = 1.2000L *0.300 mol / 0.600L

moles = 0.600 moles.

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