\(b,n_{CaCO_3}=\dfrac{m}{M}=\dfrac{50}{100}=0,5(mol)\\ \Rightarrow n_O=3n_{CaCO_3}=1,5(mol)\\ \Rightarrow \text{Số nguyên tử oxi là: }A_O=1,5.6.10^{23}=9.10^{23}\)
Ta có: \(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\)
\(\Rightarrow n_O=0,5.3=1,5\left(mol\right)\)
Vậy số nguyên tử oxi là: \(SNT_O=1,5.6.10^{23}=9,10^{23}\left(phân.tử\right)\)
\(M_{CaCO_3}=40+12+16.3=100\left(\dfrac{g}{mol}\right)\)
\(b,n_{CaCO_3}=\dfrac{m}{M}.3=\dfrac{50}{100}.3=1,5\left(mol\right)\)
Số nguyên tử oxi là: \(n.6.10^{23}=1,5.6.10^{23}=9.10^{23}\)(nguyên tử)