STT CTHH Phân loại Tên gọi tương CT axit ứng tương CT bazơ ứng hộ mik theo thứ tự này nhé

Đáp án `+` Giải thích các bước giải:

`16-NaHSO_4-` muối axit `-H_2SO_4-NaOH`

`17-K_2CO_3-` muối trung hòa `-H_2CO_3-KOH`

`18-Ba(HCO_3)_2-` muối axit `-H_2CO_3-Ba(OH)_2`

`19-NO_2-` oxit axit `-HNO_3-` không có

`20-SO_3-` oxit axit `-H_2SO_4-` không có

`21-Zn(OH)_2-` bazơ không tan `- H_2ZnO_2 - Zn(OH)_2`

`22-FeHPO_4-` muối axit `-H_3PO_4-Fe(OH)_2`

`23-Fe_2(HPO_4)_3-` muối axit `-H_3PO_4-Fe(OH)_3`

`24-SnO-` oxit lưỡng tính `-H_2SnO_2 - Sn(OH)_2`

`25-PbO_2 -` oxit lưỡng tính `- H_2PbO_3 - Pb(OH)_4`

`26-Fe_2O_3-` oxit bazơ `-` không có `-Fe(OH)_3`

\begin{array}{|c|c|c|}\hline \text{STT}&\text{CTHH}&\text{Phân loại}&\text{Tên gọi}&\text{CT Acid tương ứng}&\text{CT base tương ứng}\\\hline\text{16}&\text{$NaHSO_4$}&\text{Muối trung acid}&\text{Sodium monohydrogen sulfate}&\text{$H_2SO_4$}&\text{$NaOH$}\\\hline \text{17}&\text{$K_2CO_3$}&\text{Muối trung hòa}&\text{Potassium carbonate}&\text{$H_2CO_3$}&\text{$KOH$}\\\hline \text{18}&\text{$Ba(HCO_3)_2$}&\text{Muối trung acid}&\text{Barium monohydrogen carbonate}&\text{$H_2CO_3$}&\text{Ba(OH)_2}\\\hline \text{19}&\text{$NO_2$}&\text{oxide acid}&\text{Nito monoxide}&\text{$HNO_3$}&\text{-}\\\hline \text{20}&\text{$SO_3$}&\text{oxide acid}&\text{Lưu huỳnh trioxide}&\text{$H_2SO_4$}&\text{-}\\\hline\text{21}&\text{$Zn(OH)_2$}&\text{Hydroxide lưỡng tính}&\text{Kẽm hydroxide }&\text{$H_2ZnO_2$}&\text{$Zn(OH)_2$}\\\hline\text{22}&\text{$FeHPO_4$}&\text{Muối acid}&\text{Sắt(II) monohydrogen phosphate}&\text{$H_3PO_4$}&\text{$Fe(OH)_2$}\\\hline\text{23}&\text{$Fe_2(HPO_4)_3$}&\text{Muối acid}&\text{Sắt (III) monohydrogen phosphate}&\text{$H_3PO_4$}&\text{$Fe(OH)_3$}\\\hline\text{24}&\text{$SnO$}&\text{oxide lưỡng tính}&\text{Thiếc (II) oxide}&\text{$H_2SnO_2$}&\text{$Sn(OH)_2$}\\\hline\text{25}&\text{$PbO_2$}&\text{Oxide lưỡng tính}&\text{Chì(IV) oxide}&\text{$H_2PbO_3$}&\text{$Pb(OH)_4$}\\\hline\text{25}&\text{$Fe_3O_4$}&\text{Oxide base }&\text{oxide sắt từ}&\text{-}&\text{$Fe(OH)_3,Fe(OH)_2$}\\\hline\end{array}