Рабочая карточка по теме «Определение валентности ХЭ в сложных соединениях» (Химия, 8 класс)

V II

N2O5

I III II

HNO2

I VI II

H2S2O7

II II I

Cu(OH)2

III VI II

Al2(SO4)3

II∙5 = 10

10:2=V

1∙1 = 1

II∙2 = 4

I∙2 = 2

II∙7 = 14

I∙2 = 2

II∙2 = 4

III∙2 = 6

II∙4∙3 = 24

4-1= III

14-2 = 12

12:2 = VI

24 – 6 = 18

18:3 = VI

4-2 = II

SeO3

HPO3

H2Cr2O7

CuOH

Ca3(PO­4)2

SiH4

H2SiO3

Na2S2O3

Ni(OH)2

Ba3(BO3)2

K3N

H3PO4

H4P2O7

Fe(OH)3

Al2(CO3)3

Al4C3

H5IO3

H5P3O10

Cr(OH)6

* Al2(Cr2O7)3

V II

N2O5

I III II

HNO2

I VI II

H2S2O7

II II I

Cu(OH)2

III VI II

Al2(SO4)3

II∙5 = 10

10:2=V

1∙1 = 1

II∙2 = 4

I∙2 = 2

II∙7 = 14

I∙2 = 2

II∙2 = 4

III∙2 = 6

II∙4∙3 = 24

4-1= III

14-2 = 12

12:2 = VI

24 – 6 = 18

18:3 = VI

4-2 = II

SeO3

HPO3

H2Cr2O7

CuOH

Ca3(PO­4)2

SiH4

H2SiO3

Na2S2O3

Ni(OH)2

Ba3(BO3)2

K3N

H3PO4

H4P2O7

Fe(OH)3

Al2(CO3)3

Al4C3

H5IO3

H5P3O10

Cr(OH)6

* Al2(Cr2O7)3

V II

N2O5

I III II

HNO2

I VI II

H2S2O7

II II I

Cu(OH)2

III VI II

Al2(SO4)3

II∙5 = 10

10:2=V

1∙1 = 1

II∙2 = 4

I∙2 = 2

II∙7 = 14

I∙2 = 2

II∙2 = 4

III∙2 = 6

II∙4∙3 = 24

4-1= III

14-2 = 12

12:2 = VI

24 – 6 = 18

18:3 = VI

4-2 = II

SeO3

HPO3

H2Cr2O7

CuOH

Ca3(PO­4)2

SiH4

H2SiO3

Na2S2O3

Ni(OH)2

Ba3(BO3)2

K3N

H3PO4

H4P2O7

Fe(OH)3

Al2(CO3)3

Al4C3

H5IO3

H5P3O10

Cr(OH)6

* Al2(Cr2O7)3

V II

N2O5

I III II

HNO2

I VI II

H2S2O7

II II I

Cu(OH)2

III VI II

Al2(SO4)3

II∙5 = 10

10:2=V

1∙1 = 1

II∙2 = 4

I∙2 = 2

II∙7 = 14

I∙2 = 2

II∙2 = 4

III∙2 = 6

II∙4∙3 = 24

4-1= III

14-2 = 12

12:2 = VI

24 – 6 = 18

18:3 = VI

4-2 = II

SeO3

HPO3

H2Cr2O7

CuOH

Ca3(PO­4)2

SiH4

H2SiO3

Na2S2O3

Ni(OH)2

Ba3(BO3)2

K3N

H3PO4

H4P2O7

Fe(OH)3

Al2(CO3)3

Al4C3

H5IO3

H5P3O10

Cr(OH)6

* Al2(Cr2O7)3