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:

µ = N/Na, where N- the number of particles and Na – Avogadro's constant;

µ = 1.20×10^24/ 6.02×10^23 mol^-1 = 1.9934 moles.

The mass of this amount of magnesium is:

m = µ × M, where M – the atomic weight of Mg

Thus, m = 1.9934 mol × 24 g/mol = 47.84 g

µ = N/Na, where N- the number of particles and Na – Avogadro's constant;

µ = 1.20×10^24/ 6.02×10^23 mol^-1 = 1.9934 moles.

The mass of this amount of magnesium is:

m = µ × M, where M – the atomic weight of Mg

Thus, m = 1.9934 mol × 24 g/mol = 47.84 g

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