\(\begin{array}{l}
4)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
nMg = \dfrac{{4,8}}{{24}} = 0,2\,mol\\
 \Rightarrow nHCl = 2nMg = 0,4\,mol\\
C\% HCl = \dfrac{{0,4 \times 36,5}}{{100}} \times 100\%  = 14,6\% \\
5)\\
Cu + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2Ag\\
nAgN{O_3} = a\,mol\\
 \Rightarrow nCu\,pu = 0,5a\,mol\\
nAg = nAgN{O_3} = a\,mol\\
108a - 64 \times 0,5a = 3,04\\
 \Rightarrow a = 0,04\,mol\\
CMAgN{O_3} = \dfrac{{0,04}}{{0,4}} = 0,1M\\
6)\\
nNaOH = 0,2 \times 1 = 0,2\,mol\\
nHCl = 0,4 \times 1 = 0,4\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
\dfrac{{0,2}}{1} < \dfrac{{0,4}}{1}\\
 \Rightarrow nHCl\,spu = 0,4 - 0,2 = 0,2\,mol\\
n{H^ + } = nHCl = 0,2\,mol\\
V{\rm{dd}}spu = 0,2 + 0,4 = 0,6l\\
CM{H^ + } = \dfrac{{0,2}}{{0,6}} = \dfrac{1}{3}M\\
pH =  - \log ({\rm{[}}{H^ + }{\rm{]) =  - log(}}\dfrac{1}{3}) = 0,477\\
7)\\
Dr = \dfrac{{Vr}}{{V{\rm{dd}}}} \times 100\\
 \Rightarrow Vr = \dfrac{{V{\rm{dd}} \times Dr}}{{100}} = \dfrac{{400 \times 40}}{{100}} = 160ml\\
V{H_2}O = V{\rm{dd}} - Vr = 400 - 160 = 240ml
\end{array}\)