Răspuns:
Explicație:
NH4 NO3
a. M= 2 . 14 + 4 + 3.16=80--------> 80g/moli
b. r.a. = 2 : 4 : 3
r.m.= 28 : 4 : 48 = 7 : 1 : 12 ( s-a simplif. cu 7 )
c. Comp. proc.
80g NH4NO3-------28g N---------4gH---------48g O
100-------------------------X----------------Y--------------Z
X= 100 . 28 : 80=35%
Y= 100 . 4 : 80=5%
Z= 100 . 8 : 80=60%
d. nr. moli din 16 g subst.
n= m : M
n= 16g : 80g/moli=0,2 moli NH4NO3
e. masa din 2 kmoli subst.
m = n . M
m= 2000 moli . 80g/moli= 160000g= 160 kg NH4NO3
f. nr. de atomi de azot din 3,2g subst.
n= 3,2g : 80g/moli= 0,04 moli NH4NO3
1mol NH4NO3---------2 . 6,023.10²³ atomi N
0,04 moli-------------------------x
x= 0,04 . 2 . 6,023.10²³=0,48184 .10²³atomi N
11.
- nr. molecule si nr. atomi
a. 1,5 moli NH3
1 mol NH3------------6,023.10 ²³molecule NH3
1,5 moli------------------------x
x= 1,5 . 6,023.10²³molecule NH3 =
1mol NH3----------------4 . 6,023.10²³atomi
1,5 moliNH3-------------------y
y= 1,5 . 4 . 6,023.10²³ atomi =
b. 14g CO
MCO= 12 + 16= 28-------> 28g/moli
n= 14 g : 28g/moli= 0,5 moli CO
1mol CO----------6,023.10²³molecule CO
0,5 moli--------------z
x= 0,5 . 6,023.10²³molecule CO =
1mol CO ----------2 . 6,023.10²³atomi
0,5 moli CO------------y
y= 0,5 . 2 . 6,023.10²³ atomi=
c. 19,6 kg H2SO4
19,6 kg= 19600g H2SO4
n= 19600g : 98g/moli= 200 moli H2SO4
1mol H2SO4 -----------6,023.10²³molecule H2SO4
200 moli H2SO4------------x
x= 200 . 6,023.10²³molecule H2SO4=
1mol H2SO4 ------------7 . 6,023.10²³atomi
200 moli H2SO4-----------y
y= 200 . 7 . 6,023.10²³ atomi =
d. 80kg SO3= 80000g SO3
MSO3= 32 + 3.16= 80-------> 80g/moli
n= 80000g : 80g/moli=1000 moli SO3
1mol SO3 --------------6,023.10²³molecule SO3
1000 moli SO3-----------X
X= 1000 . 6,023.10²³molecule =
1mol SO3-------------4 . 6,023.10²³ atomi
1000 moli---------------y
y= 1000 . 4 . 6,023.10²³ atomi=
e . 2 kmoli He
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