1)
a)
Ta có:
\([{H^ + }] = {C_{M{\text{ }}{\text{HCl}}}} = 0,2M\)
\( \to pH = - \log [{H^ + }] = 0,699M\)
b)
Ta có:
\([O{H^ - }] = 2{C_{M{\text{ Ca(OH}}{{\text{)}}_2}}} = 0,4.2 = 0,8M\)
\( \to pOH = - \log [O{H^ - }] = 0,097 \to pH = 14 - pOH = 13,903\)
2)
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{16,8}}{{22,4}} = 0,75{\text{ mol = }}{{\text{n}}_C}\)
\({n_{{H_2}O}} = \frac{{13,5}}{{18}} = 0,75{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 1,5{\text{ mol}}\)
\( \to {n_O} = \frac{{18,5 - 0,75.12 - 1,5.1}}{{16}} = 0,5{\text{ mol}}\)
\( \to {n_C}:{n_H}:{n_O} = 0,75:1,5:0,5 = 3:6:2\)
Vậy \(A\) có dạng \((C_3H_6O_2)_n\)
\( \to {M_A} = (12.3 + 6 + 16.2).n = 37{M_{{H_2}}} = 37.2 = 74 \to n = 1\)
Vậy \(A\) là \(C_3H_6O_2\)
3)
Gọi số mol \(Al;Fe\) lần lượt là \(x;y\)
\( \to 27x + 56y = 6,95{\text{ gam}}\)
Ta có:
\({n_{N{O_2}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol}}\)
Bảo toàn e:
\(3{n_{Al}} + 3{n_{Fe}} = {n_{N{O_2}}} = 0,45 \to 3x + 3y = 0,45\)
Giải được: \(x=0,05;y=0,1\)
\( \to {m_{Al}} = 0,05.27 = 1,35{\text{ gam}}\)
\({m_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)