Cho mình xin đáp án và lời giải vs

Cần gấpppp

Tham khảo :

$11,B$

$NaOH \to Na^++OH^-$

`=>`$C_M(NaOH)=C_M(Na^+)=C_M(OH^-)$

$n_{NaOH}=0,4 \cdot 0,5+\dfrac{10 \cdot 1,25 \cdot 20}{40} \cdot 0,1=0,825(mol)$

$V_{hh}=0,4+0,1=0,5(l)$

`=>`$C_M(Na^+)=C_M(OH^-)=\dfrac{0,825}{0,5}=1,65(M)$

$12,C$

$n_{H_2SO_4}=2V \cdot 0,2+3V \cdot 0,5=1,9V$

$V_{hh}=2V+3V=5V$

$C_M(H_2SO_4)=\dfrac{1,9V}{5V}=0,38(M)$

$13,C$

$n_{CaCl_2}=0,3 \cdot 0,3 =0,09(mol)$

$n_{NaCl}=0,2 \cdot 1 =0,2(mol)$

$CaCl_2 \to Ca^{2+}+2Cl^-$

$0,09$  `->`  $0,09$ `->`   $0,18(mol)$

$NaCl \to Na^++Cl^-$

$0,2$  `->`  $0,2$  `->` $0,2(mol)$

$V_{hh}=0,2+0,3=0,5(l)$

$C_M(Na^+)=\dfrac{0,2}{0,5}=0,4(M)$

$C_M(Ca^{2+})=\dfrac{0,09}{0,5}=0,18(M)$

$C_M(Cl^-)=\dfrac{0,18+0,2}{0,5}=0,76(M)$

$14,C$

$n_{K}=1(mol)$

`-` Bảo toàn $e:$

$n_{H_2}=\dfrac{n_{K}}{2}=\dfrac{1}{2}(mol)$

`-` Bảo toàn nguyên tố $K:$

$n_{KOH}=n_K=1(mol)$

`-` Bảo toàn khối lượng:

$m_K+m_{H_2O}=m_{dd}+m_{H_2}$

`=>`$m_{dd}=39+362-\dfrac{1}{2} \cdot 2=400(gam)$

`=>`$C\%=\dfrac{1 \cdot 56}{400} \cdot 100\%=14\%$

$15,C$

$n_{Ba(NO_3)_2}=0,25 \cdot 0,5=0,125(mol)$

$n_{Na_2SO_4}=0,1 \cdot 0,75=0,075(mol)$

$Ba(NO_3)_2 \to Ba^{2+}+2NO_3^{-}$

$0,125$  `->`  $0,125(mol)$

$Na_2SO_4 \to 2Na^++SO_4^{2-}$

$0,075$  `->`  $0,15$   `->`  $0,075(mol)$

$n_{Ba^{+}}>n_{SO_4^{2-}}$`->` Tính theo $SO_4^{2-}.$

$Ba^{2+}+SO_4^{2-} \to BaSO_4$

                  $0,075$  `->` $0,075(mol)$

`=>`$m_{BaSO_4}=0,075 \cdot 233=17,475(gam)$