Question

Calculate the weight of 6.022 x 1023 formula units of CaCO3.​

Answer 1

Answer:

Solution — Molar mass (Molecular mass in gram) of CaCO3 = 40+12+3×16 = 100 g No. of moles of CaCO3 = No. of molecules/Avogadro constant = 6.022 × 1023/ 6.022 × 1023 = 1 mole Mass of CaCO3 = No. of moles × molar mass = 1 × 100 g = 100 g.

Answer 2

Answer:

6160.50 will be your answer

Explanation:

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