$l=\dfrac{R.S}{\rho}=\dfrac{2.5,1.10^{-6}}{1,7.10^{-8}}=600m$

QTHT: x.III=y.I

$\Rightarrow \dfrac{x}{y}=\dfrac{1}{3}$

$\Rightarrow AlCl_3$

$2Al+6HCl\to 2AlCl_3+3H_2$

HCl: axit clohidric

H2SO4: axit sunfuric

HNO3: axit nitric

H2CO3: axit cacbonic

H2S: axit sunfuhidric

H2SO3: axit sunfuro

$S=\dfrac{R.l}{\rho}=\dfrac{100.50}{0,4.10^{-6}}=1,25.10^{10}m$

a/ $R_{td}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{30}+\dfrac{1}{15}+\dfrac{1}{25}}=\dfrac{50}{7}\Omega$

b/ $U_3=I_3R_3=0,75.25=18,75V$

$U=U_3=18,75V$

$I=\dfrac{U}{R_{td}}=\dfrac{18,75}{\dfrac{50}{7}}=2,625A$

c/ $U_2=U_3=18,75V$

$I_2=\dfrac{U_2}{R_2}=\dfrac{18,75}{15}=1,25A$

$0,34mm^2=0,34.10^{-6}m^2$

$R=\rho\dfrac{l}{S}=1,7.10^{-8}.\dfrac{2}{0,34.10^{-6}}=0,1m$

Pt $\Leftrightarrow (x-4)^3=0\\\Leftrightarrow x-4=0\\\Leftrightarrow x=4$

$\dfrac{1}{R_{td}}=\dfrac{1}{80}+\dfrac{1}{65}+\dfrac{1}{45}=\dfrac{469}{9360}\\\Rightarrow R_{td}=\dfrac{9360}{469}\Omega$

Gọi phân tử muối Sodium Carbonate là $Na_2CO_x$

$PTK_{Na_2CO_x}=58+16x(g/mol)$

Vì PTK của muối sodium carbonate là 106

=> 58+16x=106

<=> 16x=48

<=> x=3

=> Có 3 nguyên tử Oxigen trong muối

PTK SO₃=32+48=80 đvC

PTK NaHCO₃=23+1+12+48=84 đvC

PTK Al₂(SO₄)₃=54+96+192=342 đvC

PTK PbSO₄=207+32+64=303 đvC

PTK CH₄=12+4=16 đvC