1. She asked us not to use too much hot water.

2. She invited me to come to her party.

3. She told them not to do it again.

4. She asked me if Mr.Brown had sent the potatoes to me.

5. She told boys not to get their shoes dirty.

1. Nam has been written a letter for a month.

2. Songs were being sung by the children this time.

3. A new bike is going to be bought by my brother tomorrow.

 \(CO+CuO\underrightarrow{t^0}Cu+CO_2\)

Hỗn hợp A gồm Cu và CuO dư sau phản ứng

\(Cu+2H_2SO_{4\left(đ,t^0\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(CuO+H_2SO_{4\left(đ,t^0\right)}\rightarrow CuSO_4+H_2O\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(x^2=\dfrac{4}{9}\)

\(\Rightarrow x=\dfrac{2}{3}\)

1.B

2.A

3.C

4.C

5.A

6.D

7.D

8.C

9.A

10.B

11.D

12.C

13.A

14.D

15.D

          \(R_xO_y+yCO\underrightarrow{t^0}xR+yCO_2\)

TĐB:  \(\dfrac{0,07}{y}\) - \(0,07\)  - \(\dfrac{0,07x}{y}\) - \(0,07\)   (mol)

           \(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)

TĐB:   0,07        0,07            0,07        0,07   (mol)

          Gọi n là hóa trị của kim loại R

            \(2R+2nHCl\rightarrow2RCl_n+nH_2\uparrow\)

TĐB: \(\dfrac{0,105}{n}\) - \(0,105\)                      \(0,0525\)    (mol)

\(n_{BaCO_3}=\dfrac{m}{M}=\dfrac{13,79}{197}=0,07\left(mol\right)\)

\(n_{H_2}=\dfrac{m}{M}=\dfrac{0,105}{2}=0,0525\left(mol\right)\)

\(m_{CO_2}=n.M=0,07.44=3,08\left(g\right)\)

\(m_{CO}=n.M=0,07.28=1,96\left(g\right)\)

Áp dụng định luật bảo toàn khối lượng

     \(m_{R_xO_y}+m_{CO}=m_R+m_{CO_2}\)

\(\Leftrightarrow4,06+1,96=m_R+m_{CO_2}\)

\(\Leftrightarrow6,02=m_R+3,08\)

\(\Leftrightarrow m_R=2,94\left(g\right)\)

    \(m_R=n.M\)

\(\Leftrightarrow2,94=\dfrac{0,105}{n}.R\)

\(\Leftrightarrow\dfrac{2,94}{0,105}=\dfrac{R}{n}\)

\(\Leftrightarrow28n=R\)

Biện luận

Nếu n=1 \(\Rightarrow R=28\) (loại)

        n=2 \(\Rightarrow R=56\) (nhận)

        n=3 \(\Rightarrow R=84\) (loại)

Vậy kim loại R là Fe

      \(m_{Fe_xO_y}=n.M\)

\(\Leftrightarrow4,06=\dfrac{0,07}{y}.\left(56x+16y\right)\)

\(\Leftrightarrow4,06y=3,92x+1,12y\)

\(\Leftrightarrow4,06y-1,12y=3,92x\)

\(\Leftrightarrow2,94y=3,92x\)

\(\Leftrightarrow\dfrac{2,94}{3,92}=\dfrac{x}{y}\)

\(\Leftrightarrow\dfrac{3}{4}=\dfrac{x}{y}\)

Vậy CTHH là \(Fe_3O_4\)

b)   

        \(2Fe_3O_4+10H_2SO_{4\left(đ,t^o\right)}\rightarrow3Fe_2\left(SO_4\right)_3+SO_2\uparrow+10H_2O\)

TĐB:0,0175       0,0875                                                                 (mol)

\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{4,06}{232}=0,0175\left(mol\right)\)

\(m_{H_2SO_4}=n.M=0,0875.98=8,575\left(g\right)\)

\(m_{ddH_2SO_4}=\dfrac{m_{ct}.100\%}{C\%}=\dfrac{8,575.100\%}{98\%}=8,75\left(g\right)\)