11.
$n_{H_2}=\dfrac{0,336}{22,4}=0,015(mol)$
Bảo toàn e: $2n_M=2n_{H_2}$ (M là kim loại)
$\Rightarrow n_M=0,015(mol)$
$M_M=\dfrac{0,6}{0,015}=40(Ca)$
Cấu hinh e: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$
12.
$CaX_2+2AgNO_3\to Ca(NO_3)_2+2AgX$
$n_{CaX_2}=\dfrac{0,2}{40+2X}(mol)$
$\Rightarrow n_{AgX}=\dfrac{0,4}{40+2X}(mol)$
$\Rightarrow \dfrac{0,4}{40+2X}=\dfrac{0,376}{X+108}$
$\Leftrightarrow X=80(Br)$
$\to CaBr_2$
