The volume is 6,19 L.
moles = mass in grams / atomic weight So moles in 37.9 g or argon = 37.9 / 39.948 = 0.948 moles
The mass number of Argon is 39.948 so it is 119.98/39.948=3.0034 moles of Argon in 119.98 g of Argon or approximately 3 moles of Argon is in 119.98 g this gas.
the equation is Xg multiplied by the moles/grams of X = moles of X (the grams cancel leaving you with moles) 607g Ar x 1 mole/ 39.95g = 15.19 moles
The number of moles is 0,579.
Approx 0.223 moles.
In a container the volume remain constant but the pressure increase.
moles = mass in grams / atomic weight So moles in 37.9 g or argon = 37.9 / 39.948 = 0.948 moles
The mass number of Argon is 39.948 so it is 119.98/39.948=3.0034 moles of Argon in 119.98 g of Argon or approximately 3 moles of Argon is in 119.98 g this gas.
the equation is Xg multiplied by the moles/grams of X = moles of X (the grams cancel leaving you with moles) 607g Ar x 1 mole/ 39.95g = 15.19 moles
The number of moles is 0,579.
Approx 0.223 moles.
Mass of 1mole of Argon is 39.95g. So, The mass of 7 moles of Argon is 39.95x7= 279.65g.
Moles = weight in g / atomic weight. So moles in 24.7 g of Ar = 24.7 / 39.948 = 0.62 moles
1,67.1024 argon atoms is equal to 2,773 moles.
Have: 607gAr Need: Moles of Argon From the periodic table we know that there are 39.948gAr per every 1 mole of Argon. 607g/39.948 = your answer.
This is another calculation. there are 0.123 moles inn this volume.
0.25 moles, approx.