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19-02-2021
Chemistry

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PLEASE HELP ME!! CHEMISTRY
139. How many formula units are present in 500.0 g of lead (II) chloride?

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19-02-2021

Answer:

(500.0 g PbCl2) / (278.1164 g PbCl2/mol) × (6.022 × 10^23 f.u./mol) =

1.083 × 10^24 formula units

Explanation:

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