0.02 g of calcium carbonate (CaCO₃)
Explanation:
To solve the question we use the Avogadro's number using the following reasoning:
if 1 moles of calcium carbonate contains 6.022 × 10²³ units of calcium carbonate
then X moles of calcium carbonate contains 12.044 × 10²³ units of calcium carbonate
X = (1 × 12.044 × 10²³) / 6.022 × 10²³ = 2 moles of calcium carbonate
mass = number of moles × molecular weight
mass of calcium carbonate (CaCO₃) = 2 / 100 = 0.02 g
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Avogadro's number
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Mole=mass/molecular mass (Mr)
mole=volume/ 22.4
Molecular amount: Mole=amount atoms/ Avogadro's number
Concentration: mole=volume x M
The answer would be D.) convex
