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

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

mn làm giải thích giùm e luôn ạ

8. invention

9. discovery

R=1/2CD=a

h=AD=2a

S1=Sxq=2*pi*r*h=2*pi*a*2a=4*pi*a^2

S2=Stp=2*pi*r^2+2*pi*r*h

=2*pi*a^2+2*pi*a*2a

=6*pi*a^2

>S1/S2=2/3

\(P=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\left(x\ge0,x\ne4\right)\)

\(=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

c) \(P=\dfrac{4}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{4}{3}\Rightarrow3\sqrt{x}=4\sqrt{x}-8\Rightarrow\sqrt{x}=8\Rightarrow x=64\)

13.

\(\lim\limits_{x\rightarrow1}\left(3x^2-2x-1\right)=3.1^2-2.1-1=0\)

14.

\(\lim\limits_{x\rightarrow9}\sqrt{x+16}=\sqrt{9+16}=5\)

15.

\(\lim\limits_{x\rightarrow+\infty}x^{2021}=+\infty\)

16.

\(\lim\limits_{x\rightarrow-\infty}\left(x-x^3+1\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)\)

Do \(\lim\limits_{x\rightarrow-\infty}x^3=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=-1< 0\)

\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=+\infty\)

17.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+5x-3}{x^2+6x+3}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{5}{x}-\dfrac{3}{x^2}\right)}{x^2\left(1+\dfrac{6}{x}+\dfrac{3}{x^2}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{5}{x}-\dfrac{3}{x^2}}{1+\dfrac{6}{x}+\dfrac{3}{x^2}}=\dfrac{2+0-0}{1+0+0}=2\)

18.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-x-1}}{x+1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left|x\right|\sqrt{4-\dfrac{1}{x}-\dfrac{1}{x^2}}}{x+1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{4-\dfrac{1}{x}-\dfrac{1}{x^2}}}{x\left(1+\dfrac{1}{x}\right)}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4-\dfrac{1}{x}-\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}\)

\(=\dfrac{-\sqrt{4}}{1}=-2\)

19.

\(\lim\limits_{x\rightarrow-\infty}\left(2x^3-x^2\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(2-\dfrac{1}{x}\right)\)

Do \(\lim\limits_{x\rightarrow-\infty}x^3=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\left(2-\dfrac{1}{x}\right)=2>0\)

\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x^3\left(2-\dfrac{1}{x}\right)=-\infty\)

20.

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{x^2+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(\dfrac{1}{x}\right)}{x\left(\sqrt{1+\dfrac{1}{x^2}}+1\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{0}{1+1}=0\)

C

C

A

Cái này sao học muộn thế :V

C

C

A