\(PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Đặt:a=n_{CH_4}\left(mol\right);b=n_{C_2H_4}\left(mol\right)\left(a,b>0\right)\\ Có:\left\{{}\begin{matrix}22,4a+22,4b=8,96\\22,4a+44,8b=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\\ \%V_{CH_4}=\dfrac{a}{a+b}.100\%=\dfrac{0,3}{0,3+0,1}.100\%=75\%;\%V_{C_2H_4}=100\%-75\%=25\%\)

\(a,C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\\ b,\%V_{C_2H_4}=\dfrac{0,1.22,4}{3,36}.100\%\approx66,667\%\\ \%V_{CH_4}\approx100\%-66,667\%\approx33,333\%\)

\(n_{NaOH}=0,2.0,5=0,1\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaCl}=n_{NaOH}=0,1\left(mol\right)\\ a,C_{MddHCl}=\dfrac{0,1}{0,05}=2\left(M\right)\\ b,C_{MddNaCl}=\dfrac{0,1}{0,05+0,2}=0,4\left(M\right)\)

Gọi CT của X là C_nH_2nO₂

X no, đơn chức, mạch hở → n_CO2 = n_H2O

\(\Rightarrow\dfrac{m-0,25}{44}=\dfrac{m-3,5}{18}\)

⇒ m = 5,75 (g)

\(\Rightarrow n_X=\dfrac{5,75}{14n+32}\left(mol\right)\)

\(n_{CO_2}=\dfrac{5,75-0,25}{44}=0,125\left(mol\right)\)

Có: n_CO2 = n.n_X

\(\Rightarrow0,125=\dfrac{5,75n}{14n+32}\)

⇒ n = 1

Vậy: X là CH₂O₂.

2Fr + 2H₂O --> 2FrOH + H₂↑

Tuy nhiên Fr không có đồng vị bền (cực hiếm) nhưng nó vẫn là kim loại kiềm !

\(a)n_K=\dfrac{7,8}{39}=0,2mol\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{H_2}=2n_K=0,4mol\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96l\\ V_{H_2\left(đkc\right)}=0,4.24,79=9,916g\\ b)n_{KOH}=n_K=0,2mol\\ 2KOH+Cu\left(NO_3\right)_2\rightarrow2KNO_3+Cu\left(OH\right)_2\\ n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1mol\\ m_{\downarrow}=m_{Cu\left(OH\right)_2}=0,1.98=9,8g\\ c)Cu\left(OH\right)_2\xrightarrow[]{t^0}CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=0,1mol\\ m_{CuO}=0,1.80=8g\)

Ta có: n_NaOH = 0,2.0,25 = 0,05 (mol)

Gọi: V_H2O = a (l)

\(\Rightarrow\dfrac{0,05}{a+0,2}=0,1\) \(\Rightarrow a=0,3\left(l\right)=300\left(ml\right)\)

 \(V_{H_2O} = x(l) \\ 200ml=0,2l \\ n_{NaOH}=0,2 . 0,25=0,05(mol) \\ \Rightarrow C_{M\ NaOH\ 0,1M}=\frac{0,05}{a+0,2}=0,1(M) \\ Giải\ phương\ trình: x=0,3(l)=300(ml) \)

Gọi: n_CuSO4.5H2O = n_CuSO4 = a (mol)

⇒ m_CuSO4.5H2O = 250a (g)

\(\Rightarrow\dfrac{160a}{250a+200}=0,16\)

⇒ a = 4/15 (mol)

⇒ m_CuSO4.5H2O = 4/15.250 = 200/3 (g)