At STP 32 g of O₂ would occupy by the same volume as 4 g of He
Complete question
At STP 32 g of O₂ would occupy by the same volume as:
Standard Conditions
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So the gas will have the same volume if the number of moles is the same
mol of 32 grams of O₂ :
[tex]\tt \dfrac{32}{32}=1[/tex]
[tex]\tt mol=\dfrac{4}{4}=1[/tex]
[tex]\tt mol=\dfrac{8}{16}=0.5[/tex]
[tex]\tt mol=\dfrac{64}{2}=32[/tex]
[tex]\tt mol=\dfrac{32}{64}=0.5[/tex]
So mol of 4 g He = mol of 32 g O₂
STP is the standard temperature and pressure of the gas to compare with the experimental data. At STP 32 gm of oxygen is equivalent to 4 gm of helium.
STP is the standard temperature and pressure of 0 degrees Celsius and 1 atm. At this temperature and pressure, the volume of the gas is 22.4 L/mol.
So, if the moles of the 32 gm oxygen are the same as the other element then the volume will also be the same. Here, the moles of oxygen are 1 mol.
Moles of 4 gm helium is 1, 8 gm methane is 0.5 mol, 64 gm dihydrogen is 32 mol, and 32 gm sulfur dioxide is 0.5 mol.
Therefore, the mole of 4 gm helium is equivalent to the mole of 32 gm oxygen.
Learn more about STP here:
https://brainly.com/question/25705287
