HCl ko phải là khí nha, nó chỉ là ở dạng khí thôi, thực ra đó là HCl đặc

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

\(FeS+2HCl--->FeCl_2+H_2S\uparrow\)

\(Zn+2HCl--->ZnCl_2+H_2\uparrow\)

\(MnO_2+4HCl--->MnCl_2+Cl_2\uparrow+2H_2O\)

\(2H_2S+3O_2\overset{t^o}{--->}2SO_2\uparrow+2H_2O\)

\(2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\)

Loại pứ anh giống mấy bn ở trên nhé

\(a.\left(1\right)4K+O_2\overset{t^o}{--->}2K_2O\)

\(\left(2\right)K_2O+H_2O--->2KOH\)

\(b.\left(1\right)4P+5O_2\overset{t^o}{--->}2P_2O_5\)

\(\left(2\right)P_2O_5+3H_2O--->2H_3PO_4\)

\(c.\left(1\right)4Na+O_2\overset{t^o}{--->}2Na_2O\)

\(\left(2\right)Na_2O+H_2O--->2NaOH\)

\(d.\left(1\right)2Cu+O_2\overset{t^o}{--->}2CuO\)

\(\left(2\right)CuO+H_2SO_4--->CuSO_4+H_2O\)

\(\left(3\right)CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

\(e.\left(1\right)2H_2+O_2\overset{tia.lửa.điện}{--->}2H_2O\)

\(\left(2\right)H_2O+SO_3--->H_2SO_4\)

\(\left(3\right)H_2SO_4+Fe--->FeSO_4+H_2\uparrow\)

Khí nặng hơn H₂ là tất cả các khí có trong đề bài

Đ: x² - 25 = (x + 5)(x - 5)

S: \(x^2\ge0\) nên \(x^2+25\ne0\)

\(a.\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{14}{15}\)

\(b.\dfrac{-2}{5}.\dfrac{5}{8}+\dfrac{5}{8}.\dfrac{3}{5}=\dfrac{5}{8}.\left(\dfrac{-2}{5}+\dfrac{3}{5}\right)=\dfrac{5}{8}.\dfrac{1}{5}=\dfrac{1}{8}\)

\(c.25\%-1\dfrac{1}{2}+0,5.\dfrac{12}{5}=\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{6}{5}=-\dfrac{1}{20}\)

C

Ta có: \(n_{KMnO_2}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

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

a. Theo pt: \(n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,05.22,4=1,12\left(lít\right)\)

b. Theo pt: \(n_{MnO_2}=n_{O_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{MnO_2}=0,05.87=4,35\left(g\right)\)