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How many grams are in 4.23E44 particles of PbO2

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Answer:

1.679 × 10²³ g of PbO₂ is present in 4.23×10⁴⁴ particles of PbO₂.

Step-by-step explanation:

First the number of particles is converted into moles by dividing it by the Avogadro's number and then moles multiplied by the molar mass, we will get the mass of PbO₂ in grams.

4.23×10⁴⁴/ 6.022×10²³ = 7.02×10²⁰ moles

Now we have to multiply this by the molar mass of PbO₂, that is 239.2 g/mol, we will get the mass in grams.

7.02×10²⁰ moles × 239.2 g/mol = 1.679 × 10²³ g of PbO₂

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