Giải thích ra nhé em

\(pthh:2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)

\(n_{KClO_3}=\dfrac{122,5}{122,5}=1\left(mol\right)\)

Theo pt: \(n_{O_2}=\dfrac{3}{2}.1=1,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(lít\right)\)

Chọn A

Giải thích ra nha em

\(pthh:S+O_2\overset{t^o}{--->}SO_2\)

\(n_S=\dfrac{32}{32}=1\left(mol\right)\)

Theo pt: \(n_{O_2}=n_S=1\left(mol\right)\)

\(\Rightarrow V_{kk}=1.22,4.5=112\left(lít\right)\)

Chọn A

\(pthh:C+O_2\overset{t^o}{--->}CO_2\left(1\right)\)

\(S+O_2\overset{t^o}{--->}SO_2\left(2\right)\)

Ta có: \(\left\{{}\begin{matrix}n_C=\dfrac{6}{12}=0,5\left(mol\right)\\n_S=\dfrac{8}{32}=0,25\left(mol\right)\end{matrix}\right.\)

Theo pt₍₁₎: \(n_{O_2}=n_C=0,5\left(mol\right)\)

Theo pt₍₂₎: \(n_{O_2}=n_S=0,25\left(mol\right)\)

\(\Rightarrow m_{O_{2_{cần.dùng}}}=0,5.32+0,25.32=24\left(g\right)\)

Chọn B

\(\dfrac{x-1}{2}-\dfrac{x-2}{3}=x-\dfrac{x-3}{4}\)

\(\Leftrightarrow\dfrac{6x-6}{12}-\dfrac{4x-8}{12}=\dfrac{12x}{12}-\dfrac{3x-9}{4}\)

\(\Rightarrow6x-6-4x+8=12x-3x+9\)

\(\Leftrightarrow-7x=7\)

\(\Leftrightarrow x=-1\)

\(a.6x-4=5x\)

\(\Leftrightarrow x=4\)

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

\(\Leftrightarrow\dfrac{4x+6}{6}=\dfrac{15-12x}{6}\)

\(\Leftrightarrow4x+6=15-12x\)

\(\Leftrightarrow16x=9\)

\(\Leftrightarrow x=\dfrac{9}{16}\)

\(c.\left(x+7\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

\(d.\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\left(đk:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x+6}{x^2-9}+\dfrac{3x-9}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(\Rightarrow2x+6+3x-9=3x+5\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\left(tm\right)\)

\(pthh:2KMnO_4\overset{t^o}{--->}K_2MnO_4+MnO_2+O_2\uparrow\)

\(n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

Theo pt: \(n_{O_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,05.32=1,6\left(g\right)\)

Diện tích HCN trên bản đồ: 8 x 5 = 40 (cm²)

Diện tích HCN thực tế: 40 x 200 = 8000 (cm²)

\(\dfrac{2x}{3}+\dfrac{2x-1}{6}=\dfrac{x-1}{2}\)

\(\Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}=\dfrac{3x-3}{6}\)

\(\Leftrightarrow4x+2x-1=3x-3\)

\(\Leftrightarrow3x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)