n _HNO3=0,2mol

CM_HNO3=\(\dfrac{0,2}{0,5}=0,4M\)

Tích số ion của nước ở 25^oC là: \(\left[H^+\right]\left[OH^-\right]=10^{-14}\left(M^2\right)\)

a) 3-etyl-4-metylhexan

b) 4-clo-3,4-đimetyloctan-6-ol

1) Ket luan nao ( neu duoi day) sai

A. Cac loai dau mo dong vat va thuc vat deu thuoc loai lipit

B. Cac loai dau mo cong nghiep deu thuoc loai lipit

C. Cac loai dau mo cong nghiep khong thuoc loai lipt

D. Cac loai dau mo cong nghiep khong phai la este

        \(NH_4OH⇌NH_4^++OH^-\)

C_o:       0,1

[ ]:       0,1-x            x           x

=> \(\dfrac{x}{0,1}.100\%=1\%\Rightarrow x=0,001\)

=> [OH^-] = 0,001M

=> pH = 11

Gọi \(C_{CH_3COOH}=x\left(mol\right)\)

pH = 4 => [H⁺] = 10^-4M

         \(CH_3COOH⇌CH_3COO^-+H^+\)

C_o:             x

[ ]         x - 10^-4               10^-4           10^-4

=> \(\alpha=\dfrac{10^{-4}}{x}.100\%=1\Rightarrow x=0,01\)

pH = 2,9 => [H⁺] = 10^-2,9M

        \(CH_3COOH⇌CH_3COO^-+H^+\)

C_o:         0,1

[ ]:     0,1-10^-2,9             10^-2,9      10^-2,9

=> \(\alpha=\dfrac{10^{-2,9}}{0,1}.100\%=1,259\%\)

Giả sử có 1 lít dd CH₃COOH 

\(CH_3COOH⇌H^++CH_3COO^-\)

Ta có: \(pH=2,9\)

\(\rightarrow\left[H^+\right]=10^{-2,9}\rightarrow n_{CH_3COOH\left(phân.ly\right)}=10^{-2,9}\left(mol\right)\)

=> \(\alpha=\dfrac{10^{2,9}}{0,1}.100\%=1,26\%\)

đổi 224 ml = 0,224 l 

\(n_{CO_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\\ pthh:CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)  
           0,01           0,02                     0,01 
\(m_{CaCO_3}=0,01.100=1g\\ V_{HCl}=\dfrac{0,02}{0,1}=0,2l\)
        

Đổi 224 ml = 0,224 lít

\(n_{CO_2}=\dfrac{0.224}{22,4}=0,01\left(mol\right)\)

PTHH : CaCO₃ + 2HCl -> CaCl₂ + CO₂ + H₂O

               0,01       0,02

\(m_{CaCO_3}=0,01.100=1\left(g\right)\)

\(V_{HCl}=\dfrac{0.02}{0,1}=0,2\left(l\right)=200\left(ml\right)\)

\(n_{NaOH}=0,2.2,5=0,5\left(mol\right);n_{Ba\left(OH\right)_2}=0,1.1=0,1\left(mol\right)\\ \Rightarrow n_{OH^-}=0,5+0,2.2=0,9\left(mol\right)\)

\(n_{Fe^{2+}}=n_{FeCl_2}=0,5.a=0,5.2b=b\left(mol\right)\\ n_{CuCl_2}=0,5b\left(mol\right)\)

PTHH:
\(Fe^{2+}+2OH^-\rightarrow Fe\left(OH\right)_2\downarrow\)

b------>2b

\(Cu^{2+}+2OH^-\rightarrow Cu\left(OH\right)_2\downarrow\)

0,5b--->b

=> b + 2b = 0,9

=> b = 0,3

dd sau pư có: \(\left\{{}\begin{matrix}Ba^{2+}:0,2\left(mol\right)\\Na^+:0,5\left(mol\right)\\Cl^-:0,5.2.0,3+0,5.0,3=0,45\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{Ba^{2+}}=\dfrac{0,2}{0,2+0,5}=\dfrac{2}{7}M\\C_{Na^+}=\dfrac{0,5}{0,2+0,5}=\dfrac{5}{7}M\\C_{Cl^-}=\dfrac{0,45}{0,2+0,5}=\dfrac{9}{14}M\end{matrix}\right.\)

Mình nhầm sửa lại

\(Cl^-:2.\left(0,5.2.0,3+0,5.0,3\right)=0,9\left(mol\right)\\ C_{Cl^-}=\dfrac{0,9}{0,7}=\dfrac{9}{7}M\)