.


:




:

































 

 

 

 





ω%, , :

ω % = (m (x) / m -) 100%,

m - = V- ρ -,

ω % = (m (x) / V- ρ -) 100%. , 100 .

0,9% , 100 0,9 NaCl 99,1 . .

(), : () = n (x) / V-, /. 1 . , n (x) = m (x) / (), :

() = m (x) / ()V-. ,

() = ω % 10 ρ - / ().

(NaCl) = 0,5 / , 1 0,5 0,5 58,5 / = 29,25 . , . (NaCl) = 0,5 / 0,5 .

(1/ z x), :

(1/ z ) = n (1/ z x) / V-, /.

(z) (, , ) () , , - - - .

, H2SO4 , z(H2SO4) = 2; NaOH , z(NaOH) = 1. , . , : H2SO4 + 2NaOH = Na2SO4 + 2H2O . .

f() = 1/ z, , - - .

:

1.

f() = 1 /

f(HCl) = 1; f(H2SO4) = ½;

f() = 1 /

f(KOH) = 1; f(Cu(OH)2) = 1/2; f(Al(OH)3) = 1/3;

f() = 1 / ( )

f(KCl) = 1; f(K2 SO4) = 1/2; f(Cr2(SO4)3) = 1/6;

f() = 1 / ( )

f(K2O) = 1/2; f(SO3) = 1/6; f(Cr2O3) = 1/6;

2.

- :

H2SO4 + 2NaOH → Na2SO4 + 2H2O

f(H2SO4) = 1/2, , ½ H2SO4; f(NaOH) = 1, .

, : H2SO4 + NaOH → NaHSO4 + H2O Na2B4O7 + 2HCl + 5H2O → 4H3BO3 + 2NaCl : f(H2SO4) = 1; f(NaOH) = 1; f(Na2B4O7) = 1/2; f(HCl) = 1

- :

f() = 1 /

f() = 1 /

2KMnO4+ 5H2O2 + 3H2SO4 → 2MnSO4 + 5O2+ 8H2O + K2SO4

MnO4- + 8H+ +5 → Mn2+ + 4H2O; MnO4- . f(KMnO4)=1/5.

H2O2 - 2 → O2 + 2H+; H2O2 . f(H2O2)=1/2.

, n (1/ z x) = z n (x).

(1/ z ) = 1/ z (), /.

, (1/ z ) = n (1/ z x) / V-, : (1/ z ) = n (x) z / V-

(1/ z ) = m (x) z / () V-.

() = m (x) / () V-, : (1/ z ) = () z;

() = (1/ z ) / z.

1 . 0,1 / , 1 0,1 m(H2SO4) = (1/2 H2SO4) (H2SO4) V- / 2 = 0,1 98 1/2 = 4,9 .

. N. (1/2 H2SO4) = 0,1 / 0,1 0,1 N.

, b (x), - , . b (x) = n (x) / m , /.

- χ, . χ i = ni / Σ ni . , , , , , χ () = n(A) / (n(A)+n(B)+n(C)).

(x), , 1 . - /. :

(x) = m (x) / V-;

(x) = (x) () / 1000;

(x) = (1/ z ) (1/ z ) / 1000.

(l) = 0,00365 / , 1 0,00365 l.

γ() , . γ() = m (x) / V-, /, /.

% - 100 .

0/00 1000 .

: , , , .

(1/ z 1) V1 = (1/ z 2) V2

(1/ z 1) V1 = n (1/ z 1); (1/ z 2) V2 = n (1/ z 2), n (1/ z 1) = n (1/ z 2).

1(1/ z ) V1 = 2(1/ z ) V2





:


: 2018-10-14; !; : 769 |


:

:

, .
==> ...

1704 - | 1588 -


© 2015-2024 lektsii.org - -

: 0.01 .