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hodyreva [135]
3 years ago
14

How many atoms are present in 179.0 g of iridium

Chemistry
1 answer:
nika2105 [10]3 years ago
8 0
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We do as follows:

 </span>179.0 g of iridium (1 mol / 192.217 g) ( 6.022 x 10^23 atoms /  1 mol ) = 5.61 x 10^23 atoms of iridium 

Hope this answers the question.
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