3,012 tấn

3,012 tấn nhé

= 3,012

=8,06 dm²

8,06

8,06 dm2

không có biết

\(P=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{-2\sqrt{x}}=\dfrac{2x}{-2\sqrt{x}}=-\sqrt{x}\)

\(P=-\sqrt{x}=-\sqrt{4}=-2\left(đpcm\right)\)

Bài 1: \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,04  →  0,04

\(\Rightarrow m_{H_2SO_4}=0,04\cdot98=3,92\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{3,92}{80}\cdot100\%=4,9\%\)

Bài 2: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\uparrow\)

\(\dfrac{1}{15}\)     ←    0,4

\(\Rightarrow m_{Fe_2O_3}=\dfrac{1}{15}\cdot160=\dfrac{32}{3}\left(g\right)\)

a. K mở

\(R_{tđ}=R_1+R_2=9+9=18\Omega\\ I=\dfrac{U}{R_{tđ}}=\dfrac{18}{18}=1A\)

b. K đóng

\(R_{tđ}=\dfrac{\left(R_1+R_2\right).R_3}{R_1+R_2+R_3}=\dfrac{\left(9+9\right)18}{9+9+18}=9\Omega\\ I=\dfrac{U}{R_{tđ}}=\dfrac{18}{9}=2A\)

a: Khi m=3 thì (1): x^2-6x+4=0

=>x^2-6x+9-5=0

=>(x-3)^2=5

=>\(x=3\pm\sqrt{5}\)