Răspuns:
Explicație:
4.
nr.moli
300g Al2S3
n= m : M
M=2.27 +3.32=150------->150g/moli
n= 300g : 150g/moli= 2 moli Al2S3
10g CaCO3
40+12+3.16=100--------->100g/moli
n= 10g : 100g/moli =0 ,1 moli CaCO3
80g NaOH
M=23+16+1=40--------->40g/moli
n=80g : 40g/moli=2moli NaOH
5.
g subst.
0,2 moli Na2SO4
m = n . M
M=2.23+ 32 +4.16=142-------->142g/moli
m= 0,2moli . 142g/moli=28,4 g Na2SO4
5moli HNO3
M=1+14+3.16=63-------->63g/moli
m=5moli .63g/moli= 315g HNO3
10moli Ca(OH)2
M=40+ 2.16 +2=74------->74g/moli
m= 10moli . 74g/moli=740g Ca(OH)2
6.
g azot
340g NH3
MNH3=14+3=17-------->17g/moli
17g HNO3---------14gN
340g NH3-----------X
X=340. 14 :17=280 g N
7.
Comp.proc . a K2CO3
M=2.39+ 12 +3.16=138-------->138g/moli
138 g K2CO3--------78g K---------12g C---------48g O
100g---------------------x-------------------y----------------z
x=100.78:138=57,52% K
y=100.12:138 =8,7% C
Z=100.48 : 138=34,78% O
8.
nr. molecule
36g H2O
MH2O=2+16=18------->18g/moli
n= 36g : 18g/moli=2moli H2O
1moli H2O-----------6,023.10²³molec. H2O
2moli---------------------x
x=2 . 6,023.10²³=12,046 .10²³ molec. H2O
9.
nr. neutroni
71 g Cl2
MCl2 = 2 .35,5 =71--------> 71g/moli
Z=17 17p ,17e ,18n
A=35
1mol Cl2 -----------18 . 6,023.10²³ n
