You know 1 mole occupies 22.4 L @ STP. So there must be 37.8/22.4 = ?mols.
mass Ar = grams x atomic mas Ar = ?
Determine the number of moles and the mass of argon occupying 37.8L at stp
4 answers
56.7 atm
n=?
w=?
P=1atm
V=37.8L
R=0.0821
T=273K
m=39.948g
PV=nRT
PV=(w÷m)RT
w=(PVm)÷(RT)
w=(1×37.8×39.948)÷(0.0821×273)
w=1510.03÷22.41
w=67.37g
n=w÷m
n=67.37÷39.948
n=1.68 mole
Really be posting ur hw here.
btw idk if it's right this is what I wrote
w=?
P=1atm
V=37.8L
R=0.0821
T=273K
m=39.948g
PV=nRT
PV=(w÷m)RT
w=(PVm)÷(RT)
w=(1×37.8×39.948)÷(0.0821×273)
w=1510.03÷22.41
w=67.37g
n=w÷m
n=67.37÷39.948
n=1.68 mole
Really be posting ur hw here.
btw idk if it's right this is what I wrote
Sir, how did u got m=39.94g