1. What will you do?
I will organize free tutoring classes, donate books and school supplies, and raise funds for students in need.

2. Why do you choose to do them?
Because education is very important, and I want to help students have better opportunities to learn and succeed.

3. Do you think the students in need will receive benefits? Why?
Yes, they will. They can improve their knowledge, have enough materials to study, and feel more confident about their future.

4. How can you make more people know about your voluntary activities?
I can share information on social media, invite friends to join, and cooperate with schools or local organizations to spread the message.

Câu 2

\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)

\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2SO_4}=0,1.2,5=0,25\left(mol\right)\)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,2       0,2             0,2          0,2

Ta có: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\) ⇒ Zn hết, H₂SO₄ dư

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c) \(m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)

  \(m_{H_2SO_4\left(dư\right)}=\left(0,25-0,2\right).98=4,9\left(g\right)\)

Bài 2 : 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

100ml = 0,1l

\(n_{H2SO4}=2,5.0,1=0,25\left(mol\right)\)

a) Pt : \(Zn+2H_2SO_4\rightarrow ZnSO_4+H_2|\)

            1           1             1           1

          0,2        0,25        0,2         0,2

b) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,25}{2}\)

                     ⇒ Zn phản ứng hết , H₂SO₄ dư

                    ⇒ Tính toán dựa vào số mol của Zn

\(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

 Chúc bạn học tốt

Bài 5:

a: \(A=\frac{-3\left(x+1\right)}{x^2-x-6}\)

\(=\frac{-3\left(x+1\right)}{x^2-3x+2x-6}\)

\(=\frac{-3\left(x+1\right)}{\left(x-3\right)\left(x+2\right)}\)

\(x^2-4=0\)

=>(x-2)(x+2)=0

=>x=2(nhận) hoặc x=-2(loại)

Khi x=2 thì \(A=\frac{-3\cdot\left(2+1\right)}{\left(2-3\right)\left(2+2\right)}=\frac{-3\cdot3}{\left(-1\right)\cdot4}=\frac94\)

b: \(B=\frac{2x}{x+3}-\frac{x}{3-x}-\frac{3x^2+9}{x^2-9}\)

\(=\frac{2x}{x+3}+\frac{x}{x-3}-\frac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x\left(x-3\right)+x\left(x+3\right)-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\frac{2x^2-6x+x^2+3x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-3x-9}{\left(x-3\right)\left(x+3\right)}=\frac{-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3}{x-3}\)

c: P=B:A

\(=-\frac{3}{x-3}:\frac{-3\left(x+1\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{3}{x-3}\cdot\frac{\left(x-3\right)\left(x+2\right)}{3\left(x+1\right)}=\frac{x+2}{x+1}\)

Để P nguyên thì x+2⋮x+1

=>x+1+1⋮x+1

=>1⋮x+1

=>x+1∈{1;-1}

=>x∈{0;-2}

mà x là số tự nhiên

nên x=0

VIII

1 Have you ever been

2 Did you go

3 went

4 got up

5 have never got 

6 got 

7 Have you ever seen

D

I

1 B

2 A

3 D

4 A

5 C

Em cảm monw ah