How many moles are in 63 grams of na2co3? make sure to show all work and calculations for credit.

0.59 mol Na2CO3

Explanation:

M(Na2CO3) = 2*23.0 + 12.0 + 3*16.0 = 106 g/mol

63 g Na2CO3 * 1 mol Na2CO3/106 g Na2CO3 = 0.59 mol Na2CO3

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explanation:

2b2t

explanation:

2b2t