hóa 9                     

`\color[#094A52 ][@] \color[#0F5132][b] \color[#066839][ơ] ` ` \color[#1E5938][s]\color[#3DAC78 ][e] \color[#57CC99][n] \color[ #74C69D][t]`

$\rm a)$

$-$ các $\rm CTHH$ có thể có của các chất trên là:

$~~~~+$ oxide:

`@` oxide acid:

$\rm P_2O_3: ~ diphosphorus ~ trioxide$

$\rm P_2O_5: ~ diphosphorus ~ pentaoxide$

$\rm N_2O_3: ~ dinitrogen ~ trioxide$

$\rm NO_2: ~ nitrogen ~ dioxide$

$\rm N_2O_5: ~ dinitrogen ~ pentaoxide$

$\rm Cl_2O: ~ dichlor ~ oxide$

$\rm Cl_2O_3: ~ dichlor ~ trioxide$

$\rm Cl_2O_5: ~ dichlor ~ pentaoxide$

$\rm Cl_2O_7: ~ dichlor ~ heptoxide$

`@` oxide base:

$\rm BaO: ~ barium ~ oxide$

$\rm FeO: ~ iron ~ (II) ~ oxide$

$\rm Fe_2O_3: ~ iron ~ (III) ~ oxide$

$\rm Fe_3O_4: ~ iron ~ (II, ~ III) ~ oxide$

`@` oxide lưỡng tính:

$\rm Al_2O_3: ~ aluminium ~ oxide$

$\rm H_2O: ~ nước$

`@` oxide trung tính:

$\rm NO: ~ nitrogen ~ oxide$

$\rm N_2O: ~ dinitrogen oxide$

$~~~+$ acid:

`@` acid có chứa oxygen:

$\rm H_3PO_3: ~ phosphorous ~ acid$

$\rm H_3PO_4: ~ phosphoric ~ acid$

$\rm HNO_2: ~ nitrous ~ acid$

$\rm HNO_3: ~ nitric ~ acid$

$\rm HClO: ~ hypochlorous ~ acid$

$\rm HClO_2: ~ chlorous ~ acid$

$\rm HClO_3: ~ chloric ~ acid$

$\rm HClO_4: ~ perchloric ~ acid$

`@` acid không chứa oxygen:

$\rm HCl: ~ hydrochloric ~ acid$

$-$ base: 

`@` base tan:

$\rm Ba(OH)_2: ~ barium ~ hydroxide$

`@` base không tan:

$\rm Fe(OH)_2: ~ iron ~ (II) ~ hydroxide$

$\rm Fe(OH)_3: ~ iron ~ (III) ~ hydroxide$

`***` $\rm Al(OH)_3: ~ aluminium ~ hydroxide$ đây là hydroxide lưỡng tính có thể viết dưới dạng: $\rm HAlO_2(.H_2O): ~ aluminic ~ acid$ 

$-$ muối:

`@` muối trung hòa:

$\rm Ba_3(PO_3)_2: ~ barium ~ phosphite$

$\rm Ba_3(PO_4)_2: ~ barium ~ phosphate$

$\rm Ba(NO_2)_2: ~ barium ~ nitrite$

$\rm Ba(NO_3)_2: ~ barium ~ nitrate$

$\rm BaCl_2: ~ barium ~ chloride$

$\rm Ba(ClO)_2: ~ barium ~ hypochlorite$

$\rm Ba(ClO_2)_2: ~ barium ~ chlorite$

$\rm Ba(ClO_3)_2: ~ barium ~ chlorate$

$\rm Ba(ClO_4)_2: ~ barium ~ perchlorate$

$\rm Ba(AlO_2)_2: ~ barium ~ aluminate$

$\rm Fe_3(PO_3)_2: ~ iron ~ (II) ~ phosphite$

$\rm Fe_3(PO_4)_2: ~ iron ~ (II) ~ phosphate$

$\rm Fe(NO_2)_2: ~ iron ~ (II) ~ nitrite$

$\rm Fe(NO_3)_2: ~ iron ~ (II) ~ nitrate$

$\rm FeCl_2: ~ iron ~ (II) ~ chloride$

$\rm Fe(ClO)_2: ~ iron ~ (II) ~ hypochlorite$

$\rm Fe(ClO_2)_2: ~ iron ~ (II) ~ chlorite$

$\rm Fe(ClO_3)_2: ~ iron ~ (II) ~ chlorate$

$\rm Fe(ClO_4)_2: ~ iron ~ (II) ~ perchlorate$

$\rm Fe(AlO_2)_2: ~ iron ~ (II) ~ aluminate$

$\rm FePO_3: ~ iron ~ (III) ~ phosphite$

$\rm Fe_3PO_4: ~ iron ~ (III) ~ phosphate$

$\rm Fe(NO_2)_3: ~ iron ~ (III) ~ nitrite$

$\rm Fe(NO_3)_3: ~ iron ~ (III) ~ nitrate$

$\rm FeCl_3: ~ iron ~ (III) ~ chloride$

$\rm Fe(ClO)_3: ~ iron ~ (III) ~ hypochlorite$

$\rm Fe(ClO_2)_3: ~ iron ~ (III) ~ chlorite$

$\rm Fe(ClO_3)_3: ~ iron ~ (III) ~ chlorate$

$\rm Fe(ClO_4)_3: ~ iron ~ (III) ~ perchlorate$

$\rm Fe(AlO_2)_3: ~ iron ~ (III) ~ aluminate$

$\rm AlPO_3: ~ aluminium ~ phosphite$

$\rm AlPO_4: ~ aluminium ~ phosphate$

$\rm Al(NO_2)_3: ~ aluminium ~ nitrite$

$\rm Al(NO_3)_3: ~ aluminium ~ nitrate$

$\rm AlCl_3: ~ aluminium ~ chloride$

$\rm Al(ClO)_3: ~ aluminium ~ hypochlorite$

$\rm Al(ClO_2)_3: ~ aluminium ~ chlorite$

$\rm Al(ClO_3)_3: ~ aluminium ~ chlorate$

$\rm Al(ClO_4)_3: ~ aluminium ~ perchlorate$

`@` muối acid:

$\rm BaHPO_3: ~ barium ~ hydrophosphite$

$\rm Ba(H_2PO_4)_2: ~ barium ~ dihydrophosphate$

$\rm BaHPO_4: ~ barium ~ hydrophosphate$

$\rm FeHPO_3: ~ iron ~ (II) ~ hydrophosphite$

$\rm Fe(H_2PO_4)_2: ~ iron ~ (II) ~ dihydrophosphate$

$\rm FeHPO_4: ~ iron ~ (II) ~ hydrophosphate$

$\rm Fe_2(HPO_3)_3: ~ iron ~ (III) ~ hydrophosphite$

$\rm Fe(H_2PO_4)_3: ~ iron ~ (III) ~ dihydrophosphate$

$\rm Fe(HPO_4)_2: ~ iron ~ (III) ~ hydrophosphate$

$\rm b)$

cho phân tử khối của các nguyên tố, cậu tự tính nhé vì cậu không yêu cầu ý b)

$\rm P ~ = ~ 31$

$\rm N ~ = ~ 14$

$\rm O ~ = ~ 16$

$\rm Al ~ = ~ 27$

$\rm Ba ~ = ~ 137$

$\rm H ~ = ~ 1$

$\rm Fe ~ = ~ 56$

$\rm Cl ~ = ~ 35,5$

$\rm VD:$ tính $\rm PTK$ của $\rm BaCl_2:$

$\rm M_{BaCl_2} ~ = ~ 137 ~ + ~ 35,5 \times ~ 2 ~ = ~ 208$