Câu 1 :
$m_{Na} = (11 + 12)\ amu = 23\ amu = 23.1,66.10^{-24} (gam) = 38,18.10^{-24}(gam) = 38,18.10^{-27}(kg)$
Câu 2 :
$1,97A^o = 1,97.10^{-8}cm$
$V_{nguyên\ tử\ Ca} = \dfrac{4}{3}.\pi.(1,97.10^{-8})^3 = 3,2.10^{-23}(cm^3)$
$m_{Ca} = 40.1,66.10^{-27} = 66,4.10^{-27}(kg)$
$\Rightarrow D_{Ca} = \dfrac{m}{V} = \dfrac{66,4.10^{-27}}{3,2.10^{-23}} = 2,075.10^{-3}(kg/cm^3)$