\(n_A=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1(mol)\\ \Rightarrow M_A=\dfrac{5,6}{0,1}=56(g/mol)\)
Vậy A là sắt (Fe)
\(n_A=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\)
=> \(M_A=\dfrac{5,6}{0,1}=56\left(g/mol\right)\)
=> A là Fe(sắt)
\(n_A=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\\ \Rightarrow M_A=\dfrac{m}{n}=\dfrac{5,6}{0,1}=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow A\text{là}\text{F}e\)