Ge, Sn, Pb Flashcards

(42 cards)

1
Q

Ge+HCl=

A

Нет реакции

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2
Q

Sn+HCl=

A

Sn+HCl= SnCl2+H2

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3
Q

Pb+HCl(конц)=

A

Pb+HCl(конц)= PbCl2+H2

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4
Q

Sn+HNO3(разб)=

A

Sn+HNO3(разб)= Sn(NO3)2+NO+H2O (Pb также)

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5
Q

Ge+HNO3(конц)=

A

Ge+HNO3(конц)= H2GeO3+NO2+H2O (Sn также)

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6
Q

Ge+H2SO4(конц)=

A

Ge+H2SO4(конц)= Ge(SO4)2+SO2+H2O (Sn также)

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7
Q

Pb+H2SO4(конц)=

A

Pb+H2SO4(конц)=Pb(HSO4)2+SO2+H2O

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8
Q

Ge+KOH=

A

Нет реакции

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9
Q

Sn+KOH+H2O=

A

Sn+KOH+H2O= K[Sn(OH)3]+H2

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10
Q

Sn+KOH(изб)+H2O=

A

Sn+KOH(изб)+H2O= K4[Sn(OH)6]+H2

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11
Q

Mg2Ge+HCl=

A

Mg2Ge+HCl= MgCl2+GeH4

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12
Q

SnCl4+Li[AlH4]=

A

SnCl4+Li[AlH4]= SnH4+LiCl+AlCl3

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13
Q

GeH4=

A

GeH4= Ge+H2 (C, Si, Sn, Pb также)

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14
Q

GeO2+HCl(конц)=

A

GeO2+HCl(конц)= GeCl4+H2O

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15
Q

GeO2+NaOH+H2O=

A

GeO2+NaOH+H2O= Na2[Ge(OH)6]

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16
Q

GeO2+NaOH (расплав)=

A

GeO2+NaOH=Na2GeO3+H2O

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17
Q

SnCl4+NH3+H2O=

A

SnCl4+NH3+H2O= NH4Cl+Sn(OH)4 (α-форма)

18
Q

SnCl4+H2O=

A

SnCl4+H2O= HCl+Sn(OH)4 (α-форма)

19
Q

Sn+HNO3(конц)=

A

Sn+HNO3(конц)= NO2+H2O+H2SnO3 (β-форма)

20
Q

Sn(OH)4+H2SO4(разб)=

A

Sn(OH)4+H2SO4(разб)= Sn(SO4)2+H2O

21
Q

Sn(OH)4+KOH=

A

Sn(OH)4+KOH= K2[Sn(OH)6]

22
Q

Pb(OAc)2+CaOCl2+H2O=

A

Pb(OAc)2+CaOCl2+H2O= PbO2+CaCl2+HOAc

23
Q

PbO2+H2SO4(конц)=

A

PbO2+H2SO4(конц)= Pb(SO4)2+H2O

24
Q

PbO2+HCl=

A

PbO2+HCl= PbCl2+Cl2+H2O

25
PbO2+CaO= (t)
PbO2+CaO= CaPbO3
26
Pb+O2= (200-300°)
Pb+O2= PbO
27
PbO+O2=(400-500°)
PbO+O2= Pb3O4
28
Pb3O4+HNO3=
Pb3O4+HNO3= PbO2+Pb(NO3)2+H2O
29
Pb3O4= (>500°)
Pb3O4= PbO+O2
30
PbCl2+Cl2+KCl=
PbCl2+Cl2+KCl= K2[PbCl6]
31
Sn(OH)2= (t)
Sn(OH)2= SnO+H2O
32
Sn(OH)2+H2SO4=
Sn(OH)2+H2SO4= SnSO4+H2O
33
Sn(OH)2+NaOH=
Sn(OH)2+NaOH= Na[Sn(OH)3]
34
Na[Sn(OH)3]=
Na[Sn(OH)3]= Sn+Na2[Sn(OH)6]
35
SnCl2+HCl(конц)=
SnCl2+HCl(конц)= H[SnCl3]
36
PbO+AcOH=
PbO+AcOH= Pb(OAc)2+H2O
37
Pb(OH)2+HCl=
Pb(OH)2+HCl= PbCl2+H2O
38
Pb(OH)2+KOH(конц)=
Pb(OH)2+KOH(конц)= K2[Pb(OH)4]
39
PbO2+H2S=
PbO2+H2S= PbSO4+PbO+H2O
40
SnS2+Na2S=
SnS2+Na2S= Na2SnS3
41
Na2SnS3+HCl=
Na2SnS3+HCl= SnS2+H2S+NaCl
42
SnS+Na2S2=
SnS+Na2S2= Na2SnS3 (Ge также)