Respuesta :
Answer is: volume of nitrogen is 34.72 liters.
m(N₂) = 43.5 g.
n(N₂) = m(N₂) ÷ M(N₂).
n(N₂) = 43.5 g ÷ 28 g/mol.
n(N₂) = 1.55 mol, amount of substance.
T = 273 K, standard temperature.
p = 1 atm, standard pressure.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
m(N₂) = 43.5 g.
n(N₂) = m(N₂) ÷ M(N₂).
n(N₂) = 43.5 g ÷ 28 g/mol.
n(N₂) = 1.55 mol, amount of substance.
T = 273 K, standard temperature.
p = 1 atm, standard pressure.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
V =
n·R·T / p.
V(N₂) = 1.55 mol · 0.08206 L·atm/mol·K · 273 K /
1.00 atm.
V(O₂) = 34.72 L.
we are going to use this formula:
PV= n RT
when at STP
p is the pressure = 1 atm
and n is the moles = mass /molar mass = 43.5 g /28g/mol
= 1.56 moles
R is the ideal gas constant = 0.0821
and T is a temperature in Kelvin = 273 K
by substitution, we will get V (the volume)
1 atm * V = 1.56 moles* 0.0821 *273K
∴ V = 34.96 L
PV= n RT
when at STP
p is the pressure = 1 atm
and n is the moles = mass /molar mass = 43.5 g /28g/mol
= 1.56 moles
R is the ideal gas constant = 0.0821
and T is a temperature in Kelvin = 273 K
by substitution, we will get V (the volume)
1 atm * V = 1.56 moles* 0.0821 *273K
∴ V = 34.96 L