Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082

- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082
× 273 K
Solving:

n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
A precipitate will form since BaSO₄ is insoluble in water.
Answer:
The mass of P₄O₁₀ (s) formed is 471.03 g
Explanation:
4PH₃(g) + 8O₂(g) → P₄O₁₀ (s) + 6H₂O (g)
This is the ballanced equation.
Every 4 moles of phosphine, 1 mol of tetraphosphorus decaoxide is generated.
Moles of PH₃ = Mass PH₃ / Molar mass PH₃
Moles PH₃ = 225 g / 33.9 g/m
Moles PH₃ = 6.63 moles
So the rule of three, will be:
If 4 moles of PH₃ generate 1 mol P₄O₁₀
Then, 6.63 moles of PH₃ generate (6.63 .1)/4 = 1.66 moles
If we want to know the mass:
Moles . Molar mass = mass
Molar mass P₄O₁₀ = 283.88 g/m
1.66m . 283.88 g/m = 471.03 g
The answer would be B) False cause a line graph only shows ''two'' variables, not three.