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14 October, 21:39

How many grams are in 4.00 x 1023 particles of alcl3?

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  1. 14 October, 21:42
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    Answer is: 88,9 g of AlCl₃.

    N (AlCl₃) = 4,00 · 10²³.

    n (AlCl₃) = N (AlCl₃) : Na.

    n (AlCl₃) = 4 · 10²³ : 6·10²³ 1/mol

    n (AlCl₃) = 0,66 mol.

    m (AlCl₃) = n (AlCl₃) · M (AlCl₃)

    m (AlCl₃) = 0,66 mol · 133,35 g/mol = 88,9 g.

    Na - Avogadro number.

    M - molar mass.

    n - amount of substance
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