STT CTHH Phân loại Tên gọi CT axit tương ứng , CT tương bazơ ứng mn giúp theo thứ tự vs ạ

Đáp án `+` Giải thích các bước giải:

`41-KClO_3-` muối trung hòa `-HClO_3-KOH`

`42-Ba(OH)_2-` bazơ tan `-` không có `- Ba(OH)_2`

`43-CuOH-` bazơ không tan `-` không có `- CuOH`

`44-Na_2S-` muối trung hòa `- H_2S-NaOH`

`45-CuBr_2-` muối trung hòa `-HBr - Cu(OH)_2`

`46-CaCO_3 -` muối trung hòa `- H_2CO_3 - Ca(OH)_2`

`47-CuSO_4-` muối trung hòa `-H_2SO_4-Cu(OH)_2`

`48-ZnCl_2-` muối trung hòa `-HCl-Zn(OH)_2`

`49-AlCl_3-` muối trung hòa `-HCl-Al(OH)_3`

`50-FeBr_2-` muối trung hòa `HBr-Fe(OH)_2`

\begin{array}{|c|c|c|}\hline \text{41}&\text{KClO3}&\text{muối trung hoà}&\text{kali clorat}&\text{HClO3}&\text{KOH}\\\hline \text{42}&\text{Ba(OH)2}&\text{bazơ tan}&\text{bari hiđroxit}&\text{x}&\text{Ba(OH)2}\\\hline \text{43}&\text{CuOH}&\text{bazơ không tan}&\text{đồng (I) hiđroxit}&\text{x}&\text{CuOH}\\\hline \text{44}&\text{Na2S}&\text{muối trung hoà}&\text{natri sunfua}&\text{H2S}&\text{NaOH}\\\hline \text{45}&\text{CuBr2}&\text{muối trung hoà}&\text{đồng (II) bromua}&\text{HBr}&\text{Cu(OH)2}\\\hline \text{46}&\text{CaCO3}&\text{muối trung hoà}&\text{canxi cacbonat}&\text{H2CO3}&\text{Ca(OH)2}\\\hline \text{47}&\text{CuSO4}&\text{muối trung hoà}&\text{đồng (II) sunfat}&\text{H2SO4}&\text{Cu(OH)2}\\\hline \text{48}&\text{ZnCl2}&\text{muối trung hoà}&\text{kẽm clorua}&\text{HCl}&\text{Zn(OH)2}\\\hline \text{49}&\text{AlCl3}&\text{muối trung hoà}&\text{nhôm clorua}&\text{HCl}&\text{Al(OH)3}\\\hline \text{50}&\text{FeBr2}&\text{muối trung hoà}&\text{sắt (II) bromua}&\text{HBr}&\text{Fe(OH)2}\\\hline\end{array}