a) \(n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

\(n_{Cl_2}=\dfrac{1,4}{22,4}=0,0625\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,025<------0,025<--0,025

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

         0,025-->0,0375

             Cu + Cl₂ --t^o--> CuCl₂

         0,025<-0,025

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,025.56}{0,025.56+0,025.64}.100\%=46,67\%\\\%m_{Cu}=\dfrac{0,025.64}{0,025.56+0,025.64}.100\%=53,33\%\end{matrix}\right.\)

\(\%n_{Fe}=\%n_{Cu}=\dfrac{0,025}{0,025+0,025}.100\%=50\%\)

b) X chỉ chứa FeCl₂

=> %m_FeCl2 = 100%

a: =>3x-2x=-3+2

=>x=-1

bL =>2u+27=4u+27

=>u=0

c: =>5-x+6=12-8x

=>-x+8x=12-6-5=1

=>7x=1

hay x=1/7

2. C
3A
4A
5C

\(2,ĐK:\left\{{}\begin{matrix}\dfrac{2}{a+5}\ge0\\a+5\ne0\end{matrix}\right.\Leftrightarrow a+5>0\Leftrightarrow a>-5\left(C\right)\\ 3,M=2\sqrt{3}=\sqrt{12}< \sqrt{15}=N\left(C\right)\\ 4,=\left|3-\sqrt{3}\right|=3-\sqrt{3}\left(A\right)\\ 5,=\dfrac{3\sqrt{5}-3\sqrt{3}+3\sqrt{5}+3\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{6\sqrt{5}}{2}=3\sqrt{5}\left(C\right)\)

\(I=100mA=0,1A\)

\(l=200cm=2m\)

Lực từ tác dụng lên dây dẫn:

\(F=B\cdot I\cdot l\cdot sin\alpha\)

\(\Rightarrow F=4\cdot10^{-2}\cdot0,1\cdot2=8\cdot10^{-3}N\)

Giúp j bn

đây ạ

giúp mk vs ạ

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

\(\Leftrightarrow x=-2-\dfrac{5}{4}\)

\(\Leftrightarrow x=-\dfrac{13}{4}\)

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

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

\(-3x=\dfrac{13}{4}\)

\(x=-\dfrac{13}{12}\)