HOC24
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Sửa đoạn: \(\dfrac{8}{5}-\dfrac{11}{18}\) là: \(\dfrac{8}{15}-\dfrac{11}{18}\) nhé
\(\dfrac{8}{15}-x:\dfrac{3}{4}=\dfrac{11}{18}\)
\(x:\dfrac{3}{4}=\dfrac{8}{5}-\dfrac{11}{18}\)
\(x:\dfrac{3}{4}=-\dfrac{7}{90}\)
\(x=-\dfrac{7}{90}\cdot\dfrac{3}{4}\)
\(x=-\dfrac{7}{120}\)
(1) \(Cl_2+H_2⇌HCl+HClO\)
(2) \(MnO_2+4HCl\left(\text{đặc}\right)\xrightarrow[]{}MnCl_2+Cl_2\uparrow+H_2O\)
(3) \(Cl_2+2NaOH\xrightarrow[]{}NaCl+NaClO+H_2O\)
(4) \(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2\uparrow+H_2\)
1. \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(2Mg+O_2\xrightarrow[]{t^\circ}2MgO\)
0,2 → 0,1
\(\Rightarrow V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\)
2. \(n_{P_2O_5}=\dfrac{28,2}{142}=\dfrac{141}{710}\left(mol\right)\)
\(4P+5O_2\xrightarrow[]{t^\circ}2P_2O_5\)
\(\dfrac{141}{355}\) ← \(\dfrac{141}{710}\)
\(\Rightarrow m_P=\dfrac{141}{355}\cdot31\approx12,312\left(g\right)\)
Check lại đi bạn.
\(\sqrt{2}x^2+6x=0\)
\(\Leftrightarrow x\left(\sqrt{2}x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{2}x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\sqrt{2}\end{matrix}\right.\)
Vậy \(S=\left\{0;-3\sqrt{2}\right\}\)
Hurry up! We are watting for you.
\(\dfrac{3}{8}+\dfrac{4}{10}=\dfrac{30}{80}+\dfrac{32}{80}=\dfrac{62}{80}=\dfrac{31}{40}\)
\(-\dfrac{3}{4}+\dfrac{5}{7}+\dfrac{7}{2}\)
\(=-\dfrac{21}{28}+\dfrac{20}{28}+\dfrac{98}{28}\)
\(=\dfrac{97}{28}\)
4% của 15 là:
\(15\times\dfrac{4}{100}=0,6\)