Hitunglah banyaknya mol kapur CaCO3 yang mengandung 3,612 × 10^23 molekul CaCO3!
nicholasns
Jumlah Molekul = mol x 6.02 x 10^23 3,612 x 10^23 = mol CaCO3 x 6,02 x 10^23 mol CaCO3 = 3,612 x 10^23 / 6,02 x 10^23 mol CaCO3 = 3/5 mol = 0.6 mol
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AC88
Jumlah molekul = n x L 3,612 x 10^23 = n x 6,02 x 10^23 n = 0,6 mol
3,612 x 10^23 = mol CaCO3 x 6,02 x 10^23
mol CaCO3 = 3,612 x 10^23 / 6,02 x 10^23
mol CaCO3 = 3/5 mol = 0.6 mol
3,612 x 10^23 = n x 6,02 x 10^23
n = 0,6 mol
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