Number of atoms of oxygen present in 10.6 grams of naco3 will be
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1 mole of Na₂CO₃ = 106 g moles in 10.6 g of Na₂CO₃ = 10.6 /106 = 0.1 mole 106 g contain atoms of oxygen = 3 x 6.022 x 10²³
10.6 contain atoms = 0.1 x 3 x 6.022*10^23
=1.8*10^23 atoms
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10.6 contain atoms = 0.1 x 3 x 6.022*10^23
=1.8*10^23 atoms
hope it's helps
pls give it brainliest and follow me
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