There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.
Explanation:
39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.
4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.
We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.
So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.
Learn more about:
Avogadro's number
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the balanced equation for the formation of ammonia is
N₂ + 3H₂ ---> 2NH₃
molar ratio of N₂ to NH₃ is 1:2
mass of N₂ reacted is 8.0 g
therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol
according to the molar ratio,
1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant
therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃
therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g
a mass of 9.72 mol of NH₃ is formed
