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:

\(15\times\dfrac{4}{100}=0,6\)