Bài 1: 

a: \(\Leftrightarrow-10< 3x< -7\)

=>3x=-9

hay x=-3

b: \(\Leftrightarrow-9< 2x< -6\)

=>2x=-8

hay x=-4

a.b.\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3mol\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

   0,3      0,9              0,6           0,9          ( mol )

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

\(V_{H_2}=0,9.22,4=20,16l\)

c.\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

              0,9               1,8     ( mol )

\(V_{O_2}=0,9.22,4=20,16l\)

câu 4

Zn+2Hcl->ZnCl2+H2

0,1--0,2---------------0,1

n Zn=\(\dfrac{6,5}{65}\)=0,1 mol

=>HCl dư

m HCl=0,05.36,5=1,825g

=>VH2=0,1.22,4=2,24l

bài5

4Al+3O2-to>2Al2O3

0,2----0,15

2KMnO4-to>K2MnO4+MnO2+O2

0,3-------------------------------------0,15

n Al=\(\dfrac{5,4}{27}\)=0,2 mol

VO2=0,15.22,4=3,36l

=>m KMnO4=0,3.158=47,4g

Bài 1: 

a, \(\)\(\)\(=>R2//\left[R4nt\left(R3//R5\right)\right]\)

\(=>Rtd=\dfrac{R2\left[R4+\dfrac{R3.R5}{R3+R5}\right]}{R2+R4+\dfrac{R3.R5}{R3+R5}}=\dfrac{1.\left[1+\dfrac{1}{1+1}\right]}{1+1+\dfrac{1}{1+1}}=0,6\left(ôm\right)\)

\(=>I=\dfrac{Uab}{Rtd}=\dfrac{10}{0,6}=\dfrac{50}{3}A=I1\)

\(=>Uab=U2345=10V=U2=U345\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{10}{1}=10A\)

\(=>I345=\dfrac{U345}{R345}=\dfrac{10}{1+\dfrac{1.1}{1+1}}=\dfrac{20}{3}A=I4=I35\)

\(=>U35=I35.R35=\dfrac{20}{3}.\dfrac{1.1}{1+1}=\dfrac{10}{3}V=U3=U5\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{\dfrac{10}{3}}{1}=\dfrac{10}{3}A,\)

\(=>I5=\dfrac{U5}{R5}=\dfrac{10}{3}A\)

b, \(I1=0,1A=Im=I2345\)

\(=>Uab=I2345.R2345=0,1.\dfrac{6\left[8+\dfrac{6.12}{6+12}\right]}{6+8+\dfrac{6.12}{6+12}}=0,4V\)