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28 August, 05:38

Em 150 gramas de alumínio, quantos átomos deste elemento estão presentes?

Dados:MM (AI) = 27 gmol-¹

Número de avogadro = 6,02 x 10²³

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  1. 28 August, 06:51
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    The question in English:

    In 150 grams of aluminum, how many atoms of this element are present?

    dа ta: MM (AI) = 27 g mol-1

    Number of avogadro = 6.02 x 10²³

    Answer:

    Moles (mol) = mass (g) / molar mass (g/mol)

    Mass of Al is given as 150 g.

    Hence, moles of Al = 150 g / 27 g/mol = 5.56 mol

    According to the Avogadro number, 1 mole of substance has 6.02 x 10²³ of particles.

    Hence, 1 mole of Al = 6.02 x 10²³ of Al atoms

    Then, atoms in 5.56 mol of Al = 6.02 x 10²³ mol⁻¹ x 5.56 mol

    = 3.34712 x 10²⁴

    Hence, 150 g of Al have 3.34712 x 10²⁴ of atoms.
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