Answer :

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Americium has a molar mass of 243 g/mol.
There are 6.022*10^23 atoms in a mol (Avogadro's number).

6 atoms / 6.022 * 10^23 atoms/mol gives us:

9.963*10^-24 mols

Lastly: 9.963*10^-24 mols * 243 g/mol = 2.42*10^-21 g

Answer: [tex]243\times 10^{-23}[/tex]  grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles. and weighs equal to the molecular mass.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given atoms}}{\text {Avogadro's number}}=\frac{6}{6.023\times 10^{23}}=1.0\times 10^{-23}moles[/tex]

1 mole of Americium weighs = 243 g

Thus [tex]1.0\times 10^{-23}moles[/tex] of Americium weigh=[tex]\frac{243}{1}\times 1.0\times 10^{-23}=243\times 10^{-23}g[/tex]

Thus the mass  mass in grams of a sample of Americium containing six atoms is [tex]243\times 10^{-23}[/tex]

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