Bạn tham khảo:
$5/$
Bảo toàn điện tích:
$a+2b=c+d$
$m=23a+40b+61c+62d$
$6/$
$n_{CaCl_2}=0,03(mol)$
$n_{NaCl}=0,04(mol)$
$V_{dd}=300+200=500(ml)=0,5(l)$
$CM_{CaCl_2}=\frac{0,03}{0,5}=0,06M$
$CaCl_2 \to Ca^{2+}+2Cl^{-}$
$[Ca^{2+}]=0,06M$
$[Cl^{-}]=0,06.2=0,12M$
$CM_{NaCl}=\frac{0,04}{0,5}=0,08M$
$NaCl \to Na^{+} +Cl^{-}$
$[Na^{+}]=Cl^{-}=0,08M$
$∑[Cl^{-}]=0,12+0,08=0,2M$
