SUBIECTUL E:
1.
a.
+5 0 +3 0
KNO3 + Fe --> Fe2O3 + N2 + K2O
2N⁺⁵__+10e-__> N2⁰ reducere (ag. ox)
2Fe⁰ __-6e-___> Fe2³⁺ oxidare (ag. red)
b.
KNO3
2.
6KNO3 + 10Fe --> 5Fe2O3 + 3N2 + 3K2O
6x 2N⁺⁵__+10e-__> N2⁰ reducere (ag. ox)
10x 2Fe⁰ __-6e-___> Fe2³⁺ oxidare (ag. red)
3.
ms1 = 50 g , c% = 10%
ms2 = ? g , c% = 60%
ms.final = ? g , c%.final = 20%
metoda aplicarii regulei dreptunghiului:
c1%=10% ----------------------> 60-20= 40 parti masa sol. 10%
c.%final=20%
c2%= 60% -------------------> 20-10= 10 parti masa sol 60%
------------------------------------------------
total = 40+10 = 50 parti msa sol. finala 20%
=>
50 parti sol.finala ......... 40 parti sol 10% ............ 10 part sol. 60%
ms.final ........................... ms1=50 g .......................... ms2 g
= 62,5 g sol.finala 20%
metoda algebrica:
stim ca c% = mdx100/ms
=> md1 = ms1xc1%/100 = 50x10/100 = 5 g
md.tot = md1 + md2 = 5+md2
ms.final = ms1 + ms2 = 50 + ms2
md2 = ms2xc2%/100 = 0,6ms2
=> md.tot = 5 + 0,6ms2
ms.final = md.totx100/c%.final
= (5+0,6ms2)100/20
=> 50+ms2 = (5+0,6ms2)100/20
=> ms2 = 500/40 = 12,5 g
=> ms.final = 50 + 12,5 = 62,5 g
4.
a.
Zn + CuSO4 --> ZnSO4 + Cu
b.
m g 1,6 g
Zn + CuSO4 --> ZnSO4 + Cu
65 64
=> m = 1,625 g Zn
10 g Zn ................................... 100%
1,625 g Zn consumat ........... a% = 16,25% Zn transformat
5.
a.
pila Daniell
b.
Zn este anodul acestei pile
