Question #54069

what is the amount of magnesium that would contain 1.20*10^24 particles? (Mg=24, Avogadro's constant+6.02*10^23)

Expert's answer

The number of moles of magnesium id defined by the equation:

Âµ = N/Na, where N- the number of particles and Na â€“ Avogadro's constant;

Âµ = 1.20Ã—10^24/ 6.02Ã—10^23 mol^-1 = 1.9934 moles.

The mass of this amount of magnesium is:

m = Âµ Ã— M, where M â€“ the atomic weight of Mg

Thus, m = 1.9934 mol Ã— 24 g/mol = 47.84 g

Âµ = N/Na, where N- the number of particles and Na â€“ Avogadro's constant;

Âµ = 1.20Ã—10^24/ 6.02Ã—10^23 mol^-1 = 1.9934 moles.

The mass of this amount of magnesium is:

m = Âµ Ã— M, where M â€“ the atomic weight of Mg

Thus, m = 1.9934 mol Ã— 24 g/mol = 47.84 g

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