\(n_Fe=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ a,n_{H_2}=n_{Fe}=0,2(mol)\\ \Rightarrow V_{H_2(đktc)}=0,2.22,4=4,48(l)\\ b,n_{HCl}=2n_{Fe}=0,4(mol)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6(g)\\ c,n_{FeCl_2}=n_{Fe}=0,2(mol)\\ \Rightarrow m_{FeCl_2}=0,2.127=25,4(g)\)

a,\(P\left(x\right)=x^3-2x^4+x^5-\dfrac{1}{2}x^2+=x^5-2x^4+x^3-\dfrac{1}{2}x^2+1\)

bậc :5

b,\(Q\left(x\right)=-x^3+3x^2-5x^4-x^2+3x^3-\dfrac{1}{2}=-5x^4+2x^3+2x^2-\dfrac{1}{2}\)

bậc :4

b,\(P\left(x\right)-Q\left(x\right)=\left(x^5-2x^4+x^3-\dfrac{1}{2}x^2+1\right)-\left(-5x^4+2x^3+2x^2-\dfrac{1}{2}\right)\)

\(=\text{​​}\text{​​}\text{​​}\text{​​}x^5+3x^4-x^3-\dfrac{5}{2}x^2+\dfrac{3}{2}\)

a: Ta có: \(P\left(x\right)=x^3-2x^4+x^5-\dfrac{1}{2}x^2+1\)

\(=x^5-2x^4+x^3-\dfrac{1}{2}x^2+1\)

Bậc là 5

Ta có: \(Q\left(x\right)=-x^3+3x^2-5x^4-x^2+3x^3-\dfrac{1}{2}\)

\(=-5x^4+2x^3+2x^2-\dfrac{1}{2}\)

Bậc là 4

b: Ta có: P(x)-Q(x)

\(=x^5-2x^4+x^3-\dfrac{1}{2}x^2+1+5x^4-2x^3-2x^2+\dfrac{1}{2}\)

\(=x^5+3x^4-x^3-\dfrac{5}{2}x^2+\dfrac{3}{2}\)

1/ x-5=-1

x= (-1)+5

x=4

2: x+30=-4

nên x=-34

1. b

2. a

3. b

4. c

5. c

6. a

7. c

8. b

9. a

10. a

11. b

12. c

13. a

đề đâu

lx

oko

a, \(n_{KClO_3}=\dfrac{36,75}{122,5}=0,3\left(mol\right)\)

PTHH: 2KClO₃ ---t^o→ 2KCl + 3O₂

Mol:        0,3                  0,3      0,45

\(m_{KCl}=0,3.74,5=22,35\left(g\right)\)

\(V_{O_2}=0,45.22,4=10,08\left(l\right)\)

b, 

PTHH: 4Fe + 3O₂ ---t^o→ 2Fe₂O₃

Mol:     0,6     0,45

\(m_{Fe}=0,6.56=33,6\left(g\right)\)