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FAQ

⇐ 42434445464748495051



1. . :

2 6 4 ( )

N2 + 3H2 ↔ 2NH3

1 2 4 ( )

 

-

[N2]= 1 \

[H2]=2 \

[NH3]=4 \

N2 H2,

 

:

[N2] = 1 \ +2 \ = 3 \

[H2] = 2 \ +6 \ = 8 \

 

 

2. (HClO) 0,001 \.

, , =10-7

 

α= = 10-2 1%. = -lg = 5. [H+] = = 10-5

2. , NH4OH, 0,001 \, =10-5

 

α= = 10-1 10%. = -lg = 4.

[H-] = = 10-4

 

= 14 = 14 -4 =10 [H+] = 10-10 /

3. D :

N2H4 + O2 = N2 + 2H2O

() () () ()

D0, \, N2H4()= -80 H2O()= -242

DS , ( )

 

DrH0298 = ∑ v i D f H0298, i - ∑ vj D f H0298, j = 0 + 2(-242) (-80 + 0) =-404

. .

DrH0298

DS (DrS>0), 2, 3, .. . D G = D - D S Dr G <0,

 

4. 500 4,5 C6H12O6 (R = 0,082 ./.)

 

- . = RT . = nRT/V, n =m/M (C6H12O6) = 6 12 + 121 + 616 = 180, V = 0,5

: . = 1,22 .

 

 

5. 250 60 CO(NH2)2. , =1,86.

 

 

: Δ. = . m,

Δ. , m ,

, :

Cm = = m/M . m = 4 /

 

Δ. = . m = 4.1,86 = 7,440

 

 

6. , - , MnO4 -. MnO4 -

= 0 + (0,059/n) . lg ([Ox]/[Red])

:

KMnO4 + KNO2 + H2SO4 = MnSO4 + K2SO4 + KNO3 + H2O

MnO4- + 8H+ + 5e = Mn2+ + 4H2O 2

NO2- + H2O - 2e = NO3- + 2H+ 5

, 2, , 5

2MnO4- + 16H+ + 10e +5NO2- + 5H2O - 10e = 2Mn2+ + 8H2O + 5NO3- + 10H+

:

2MnO4- + 6H+ +5NO2- = 2Mn2+ + 3H2O + 5NO3-

2KMnO4 + 5KNO2 + 3H2SO4 = 2MnSO4 + K2SO4 + 5KNO3 + 3H2O 0 MnO4- / Mn2+ = 1,51

:

KMnO4 + K2SO3 + H2O = MnO2 + K2SO4 + H

2KMnO4 + 3K2SO3 + H2O = 2MnO2 + 3K2SO4 + 2H 0 MnO4- / MnO2 = 0,60

:

) KMnO4 + K2SO3 + KOH = K2MnO4 + K2SO4 + H2O 0 MnO4- / MnO4-2 = 0,56

 

) KMnO4 +3 +KOH =2MnO4+ K2O3 + H2O

) 2KMnO4 + K2SO3 + 2KOH = 2K2MnO4 + K2SO4 + H2O

) 6KMnO4 +3 +8KOH =62MnO4+ K2O3 + 6H2O

 

 

7. -

(),,

:

Cl2+2KF ↔ 2KCl+F2

 

Cl2+2e- ↔ 2Cl-

2F- -2e- ↔ F2

: E Cl2/Cl- = 1,36 B < E F2/F- = 2,87 B. .

 

8. ,

BeCl2, F3 F3,,

 

M BeCl2 F3 , .

( ) F3, - sp3 , , NH3, NF3. : ,

 

9. 0,001 , . . (CH3COOH) = 10-5

 

 

CH3COOK + H2O ↔ CH3COOH + KOH

 

α= = = 10-3

 
 


= - lg = 6

 
 


pH = 14 = 14 + lg = 8

 

10. 0,1 : 1)NH4Cl, . (NH4OH) = 10-5.

2)Na2CO3 ( ), 1 (2CO3) = 10-7, 2 (2CO3) = 10-11

 

NH4Cl →NH4+ + Cl-

NH4+ + ↔ NH4OH + + ,

pH = -lg = = -lg = 5

Na2CO3 → 2Na+ + CO32-

CO32- + ↔ CO3- + - :

       
 
   
 


1 = α1 = = = 10-1= 0,1

 

CO3- + ↔ 2CO3 + - :

 
 


2 = α2 = = = 10-4

 

, , :

 
 


= - lg = 2

 
 


pH = 14 = 14 + lg =12

 

11. Al(OH)3, (Al(OH)3) = 27∙10-8. .

 

x x 3x

Al(OH)3 → Al+3 + 3OH-

(Al(OH)3)= [Al+3]∙[OH-]3 = x ∙(3x)3 = 27x 4 = 27∙10-8, x = 10-2 , [OH-] = 3∙10-2

= 1,52 = 14 = 14 1,52 = 12,48

 

12. DS0 . D0,

, , ,

( ) :

a) 2N2 + O2 = 2N2O; D0 >0 ΔS < 0, ΔG > 0,

) N2 + O2 = 2NO; D0 >0 ΔS = 0, ΔG > 0,

) 2NO + O2 = 2NO2; D0 <0 ΔS < 0,

) NO + NO2 = N2O3; D0 <0 ΔS = 0, ΔG < 0

13. () , 35%, 1,20/.

 

, : = . V = 1, = n = , , : m = ω.m = ω.ρ.1000, ω = , : m = ρ.1000, ρ /. : = ω. ρ.1000/M = = /

C. = Z.CM =8,56 ../

 

14. 5,2 14,7 ( ). .

 

 

: ,

.. . .. :

 

15. .

 

-

Li+H-, Be+2H2-, B+3H-3, CH4, N-3H3+, H2+O-2, H+F-

 

Li+H- + N-H3+ = LiNH2 + H2 Li+H- + H2+O-2= LiOH + H2 Na0 + N-H3+ = NaNH2 + H2

N-3H3+ + H+ = [NH4]+ H2+O-2+ H+ = [H3O]+

. , α- :

3--3 + Li+H- = Li+ -2--3 + 2

:

Li+H- +Al+3H3- = Li+[AlH4]-

B+3H-3 + Na+H- = Na+[BH4]-

3--3 + Na+[BH4]-→ 3--3

 

15. K2[CoCl4]

.

 

: K2[CoCl4] → 2K++[CoCl4]2-

. :

[CoCl4]2- ↔ [CoCl3]-+ Cl-

[CoCl3]- ↔ [CoCl2]0 + Cl-

[CoCl2]0 ↔ [CoCl]+ + Cl-

[CoCl]+ ↔ Co2+ + Cl-

() :

[CoCl4]2- ↔ Co2+ + 4Cl-

16. :

3PO4 CuSO4 Ca(NO3)2

.

 

3PO4 → 3+ + PO4-3

PO4-3 + HOH ↔ HPO4-2 + OH- ( )

CuSO4 → Cu+2 + SO4-2

Cu+2 + HOH ↔ (CuOH) + + H+ ( )

Ca(NO3)2 → Ca+2 + 2NO3- ( )

 

17. ,

KAl(SO4)2 K3AlF6?

 

:

KAl(SO4)2 → K+ + [Al(SO4)2]- ( , SO42- - , 4)

[Al(SO4)2]-↔ Al+3 + 2SO4)2-2 ( )

 

K3AlF6 → 3K+ + [AlF6]-3 ( , Al+3 6)

 

[AlF6]-3 ↔ Al+3 + 6F- ( )

18. , ( ):

SiH4, CH3OH, H2S, CH3COOH, HBr, NH3, F3, HF, CH3Cl, H2O, 3--3

 

, - , =N, , F, Cl, Br, I .., , , , ,

SiH4, CH3OH, H2S, CH3COOH, HBr, NH3, F3, HF, CH3Cl, H2O, 3--3

CH3OH, H2S, CH3COOH, HBr, NH3, HF, H2O

SiH4, F3, CH3Cl, 3--3

 

19. , , HCl 0,0001 / NaOH 0,001 /

 

HCl→ H+ + Cl- pH = -lg[H+] = -lg, ., N . =-lg10-4 =4
NaOH→Na+ + OH- = - lg[OH-] = - lgo, o ., N = 14 - = 14 + lg[OH-] = 14 + lgo = 14 -3 = 11

 

 

20. - : Mg2[Fe(CN)6], Na2[Co(NH3)2(NO2)4], [Co(NH3)5(H2O)]Br3

 

CN-, (NH3)0, NO2- , (H2O)0

+2, 6 +2, 6 +3, 6

Mg2[Fe(CN)4 (N2)2], Na2[Co(NH3)2(N)4], [Co(NH3)5(H2O)]Br3

+4 +2 -4 -2 +2 +2 0 -4 +3 0 0 -3

 

21. , :

3H2 () + N2() = 2NH3 ()

NH4NO2 () = N2 () + 2H2O()

Na2CO3() + 2HCl () = 2NaCl() + H2O() + CO2()

 

3H2 () + N2() = 2NH3 () ∆S<0

NH4NO2 () = N2 () + 2H2O() ∆S>0

Na2CO3() + 2HCl () = 2NaCl() + H2O() + CO2() ∆S>0

 

22. ()

2NO() + O2() = 2NO2()

2 ?

 

 

23. , :

SO4-2, NO2-, Cu2+, Sn2+, F-, IO4-, Al3+, I-, Fe3+, SO32-

 

SO4-2, NO2-, Cu2+, Sn2+, F-, JO4-, Al3+, I-, Fe3+, SO32-

SO4-2 (S+6), Cu2+, F-, IO4- (I+7), Al3+, Fe3+ -

NO2- (N+3), Sn2+, I-, SO32- (S+4)

 

24. Dr0

2C2H6() + 7O2() = 4CO2() + 6H2O ()

DflH0298 :

D0(CO2()) = -393,5 \

D0(H2O ()) = -241,8 \

D0(CH4()) = -89,7 \

 

: DrH0298 = ∑ v i D f H0298, i - ∑ vj D f H0298, j=

. .

,

C2H6() + 7/2O2() = 2CO2() + 3H2O ()

DflH0298 =DrH0298/2 = -2845,4/2 = -1422,7 /

 

25. 3. 600 900 ?

 

 

 

26. (), , :

1) MnO4- + Cl- + H+ ↔ MnO2 + Cl2 + H2O

2) MnO4- + Br-+ H+ ↔ MnO2 + Br2 + H2O

3) MnO4- +F- + H+ ↔ MnO2 + F2 + H2O

 

2MnO4- + 6Cl- + 8H+ ↔ 2MnO2 + 3Cl2 + 4H2O

MnO4- + 4H+ + 3e- ↔ MnO2 + 2H2O E0= 1,70 B

2Cl- - 2e- ↔ Cl2 E0= 1,36 B

2MnO4- + 6Br-+ 8H+ ↔ 2MnO2 + 3Br2 + 4H2O

MnO4- + 4H+ + 3e- ↔ MnO2 + 2H2O E0= 1,70 B

2Br- - 2e- ↔ Br2 E0= 1,07 B

2MnO4- +6F- + 8H+ ↔ 2MnO2 + 3F2 + 4H2O

MnO4- + 4H+ + 3e- ↔ MnO2 + 2H2O E0= 1,70 B

2F- - 2e- ↔ F2 E0= 2,87 B

 

27. :

2NO () + O2 ↔ 2NO2()

:

[NO]=10 /

[O2]=12,5 /.

, 50% (II), NO. NO, NO2 , O2;

5 2,5

2NO () + O2 ↔ 2NO2()

5 10 5

 

 

28. 10% , 250 0,1 ? ?

 

ω% = 10%, ω=0,1; V = 250 = 0,25

= .

ω =

C = Z∙CM; = / Z = 0,05 /

29. 50%- ( ρ =1,4 /) , 2 0,1 ? ?

 

ω% = 50%, ω=0,5; V = 2000 = 2

= .

ω =

V2=

C = Z∙CM; = / Z = 0,05 /

 

 

 

⇐ 42434445464748495051


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