III. 

A. 

1. B

2. C

3. B

4. A

5. D

B. 

1. C

2. B

3. B

4. A

5. C

1 b c b a d

2 c b b a c nha

1 The meeting was canceled 3 days ago

2 She told me she was watching a film with her sister then

3 I admire the guitarist who is perfroming on the stage

4 Had it not been for Pauline's interest, the project would have been abandoned 

IV. 

1. The meeting was canceled 3 days ago.

2. She told me shewas watching a film with her sister then

3. I admire the guitarist who is performing on the stage

4. Had it not been for Pauline's interest, the project was abandoned. (chắc thế)

Ta có: \(\left\{{}\begin{matrix}p+e+n=52\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=17\\n=18\end{matrix}\right.\)

A là đáp án đúng

Câu 2 :

a) PTHH : \(2Al+3S-->Al_2S_3\)

               \(Zn+S-->ZnS\)

m_S = 29,55 - 15,15 = 14,4 (g)

b) nS = 14,4/32 = 0,45 (mol)

PTHH : \(Al_2S_3+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2S\)

                \(ZnS+H_2SO_4-->ZnSO_4+H_2S\)

               \(H_2S+2AgNO_3-->Ag_2S+2HNO_3\)

Bảo toàn S : nAg₂S = nH2S = nS = 0,45 (mol)

=> m_ktAg2S = 111,6 (g)

c) PTHH : \(H_2S+CuCl_2-->CuS+2HCl\)

nH2S = 0,45/2 = 0,225 (mol)

Theo pthh : nCuS = nH2S = 0,225 (mol)

=> mCuS = 0,225.96 = 21,6 (g)

a)

Gọi số mol Fe, Al là a, b (mol)

=> 56a + 27b = 8,94 (1)

\(n_{H_2}=\dfrac{0,36}{2}=0,18\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

              a------------->a----->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

             b---------------->b------>1,5b

=> a + 1,5b = 0,18 (2)

(1)(2) => a = 0,15 (mol); b = 0,02 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{8,94}.100\%=93,96\%\\\%m_{Al}=\dfrac{0,02.27}{8,94}.100\%=6,04\%\end{matrix}\right.\)

b) 

m_FeCl2 = 0,15.127 = 19,05 (g)

m_AlCl3 = 0,02.133,5 = 2,67 (g)

Gọi số mol Mg, Zn là a, b (mol)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

            a---------------->a----->a

            Zn + 2HCl --> ZnCl₂ + H₂

             b--------------->b---->b

=> a + b = 0,3 (1)

Và 95a + 136b = 36,7 (2)

(1)(2) => a = 0,1 (mol); b = 0,2 (mol)

\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{0,1.24+0,2.65}.100\%=15,6\%\\\%m_{Zn}=\dfrac{0,2.65}{0,1.24+0,2.65}.100\%=84,4\%\end{matrix}\right.\)

a, Hiện tượng: mất dần màu vàng lục của Clo, có khí không màu thoát ra

\(K_2CO_3+Cl_2\rightarrow KCl+KClO+CO_2\uparrow\)

b, \(F_2+H_2\xrightarrow[\text{nhiệt độ âm}]{\text{bóng tối}}2HF\)

\(Cl_2+H_2\underrightarrow{as}2HCl\)

\(Br_2+H_2\underrightarrow{t^o}2HBr\\ I_2+H_2\underrightarrow{t^o,xt}2HI\)