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30 tháng 3 2022
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 --to--> 2FeCl3
0,025-->0,0375
Cu + Cl2 --to--> 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
=> %mFeCl2 = 100%
13 tháng 1 2022
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
30 tháng 11 2021
\(2,ĐK:\left\{{}\begin{matrix}\dfrac{2}{a+5}\ge0\\a+5\ne0\end{matrix}\right.\Leftrightarrow a+5>0\Leftrightarrow a>-5\left(C\right)\\ 3,M=2\sqrt{3}=\sqrt{12}< \sqrt{15}=N\left(C\right)\\ 4,=\left|3-\sqrt{3}\right|=3-\sqrt{3}\left(A\right)\\ 5,=\dfrac{3\sqrt{5}-3\sqrt{3}+3\sqrt{5}+3\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{6\sqrt{5}}{2}=3\sqrt{5}\left(C\right)\)
19 tháng 3 2022
\(I=100mA=0,1A\)
\(l=200cm=2m\)
Lực từ tác dụng lên dây dẫn:
\(F=B\cdot I\cdot l\cdot sin\alpha\)
\(\Rightarrow F=4\cdot10^{-2}\cdot0,1\cdot2=8\cdot10^{-3}N\)
18 tháng 9 2021
\(-x-2=\dfrac{5}{4}\)
\(\Leftrightarrow x=-2-\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{13}{4}\)
câu 1
a)\(=-\dfrac{5}{3}+\dfrac{4}{6}=-\dfrac{10}{6}+\dfrac{4}{6}=-\dfrac{6}{6}=-1\)
b)\(=\left(\dfrac{-3}{7}+\dfrac{10}{7}\right)+\left(-\dfrac{4}{9}-\dfrac{23}{9}\right)+\dfrac{2}{3}=1-3+\dfrac{2}{3}=-2+\dfrac{2}{3}=-\dfrac{4}{3}\)
c)\(=\dfrac{9}{4}\cdot\dfrac{7}{2}+\dfrac{9}{4}\cdot\dfrac{7}{16}=\dfrac{9}{4}\cdot\left(\dfrac{7}{2}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\left(\dfrac{56}{16}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\dfrac{63}{16}=\dfrac{567}{64}\)
cây 2
a)\(=>x\left(\dfrac{3}{5}-1\right)=2-\dfrac{1}{3}=\dfrac{6}{3}-\dfrac{1}{3}=\dfrac{5}{5}\)
\(=>x=\dfrac{5}{3}:\left(\dfrac{3}{5}-1\right)=\dfrac{5}{3}:\left(\dfrac{3}{5}-\dfrac{5}{5}\right)=\dfrac{5}{3}:\dfrac{-2}{5}=-\dfrac{25}{6}\)
b)\(x=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{7}{14}-\dfrac{2}{14}=\dfrac{5}{14}\)
c)\(=>\left[{}\begin{matrix}4x^2=25\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{25}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\\x=-3\end{matrix}\right.\)