Giải gấp hộ em ạ Em cảm ơn

`a)`

Đặt \(\%^{24}Mg=x\%\)

\(\to\%^{25}Mg=(100-x)\%\)

`->\overline{M_{Mg}}={24x+25(100-x)}/{100}=24,4`

`->x=60`

\(\to\%^{24}Mg=60\%;\%^{25}Mg=40\%\)

`n_{MgCl_2}={9,54}/{24,4+71}=0,1(mol)`

`->n_{Mg}=n_{MgCl_2}=0,1(mol)`

\(n_{^{24}Mg}=0,1.60\%=0,06(mol)\)

Số ngtu \(^{24}Mg\) là `0,06.6,022.10^{23}=3,6132.10^{22}` ngtu.

\(n_{^{25}Mg}=0,1-0,06=0,04(mol)\)

Số ngtu \(^{25}Mg\) là `0,04.6,022.10^{23}=2,4088.10^{22}` ngtu.

`b)` \(\%m_{^{25}Mg}=\dfrac{0,04.25}{9,54}.100\%\approx 10,48\%\)

`c)`

`MgCl_2+2NaOH->Mg(OH)_2+2NaCl`

`0,1->0,2->0,1(mol)`

`m=m_{Mg(OH)_2}=0,1.58=5,8(g)`

`V_{dd\ NaOH}={0,2}/2=0,1(L)=100(mL)`