How many atoms of oxygen are present in 10.6g of na2co3
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Answered by
71
formula weight of Na2CO3 is 106g/mol
hence 106g of Na2CO3≡1 mol Na2CO3≡3g-atom O≡3*6.022*10∧23atom of O
∴10.6g of Na2CO3≡3*6.022*10∧22atom of O
hence 106g of Na2CO3≡1 mol Na2CO3≡3g-atom O≡3*6.022*10∧23atom of O
∴10.6g of Na2CO3≡3*6.022*10∧22atom of O
Answered by
192
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 x 10²³
= 1.8 x 10²³ atoms
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 x 10²³
= 1.8 x 10²³ atoms
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