A chemist requires 0.422 mil Na2CO3 for a reaction. How many grams does this correspond to?

Question

Question

Answer 1

Answer #1

Moles of Na₂CO₃ = 0.422

Mass of Na₂CO₃ = number of moles * molar mass of Na₂CO₃ = 0.422 * (2*23 + 12 + 3 * 16) = 48.952 g

Answer 2

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