What mass of znco3 contains 6.11×1022 o atoms? give your answer correctly to three significant digits?

Answer is: mass of zinc carbonate is 4.241 grams.

N(O) = 6.11·10²²; number of oxygen atoms.

n(O) = N(O) ÷ Na (Avogadro constant).

n(O) = 6.11·10²² ÷ 6.022·10²³ 1/mol.

n(O) = 0.101 mol; amount of oxygen atoms.

One molecule of ZnCO₃ has three oxygen atoms: n(ZnCO₃) : n(O) = 1 : 3.

n(ZnCO₃) = 0.101 mol ÷ 3.

n(ZnCO₃) = 0.034 mol.

m(ZnCO₃) = n(ZnCO₃) · M(ZnCO₃).

m(ZnCO₃) = 0.034 mol · 125.39 g/mol.

m(ZnCO₃) = 4.241 g.

f(x)=-3√ x+5

That is your answer :