Question

1.Determinati vol. de oxigen exprimat in litri,masurat la temperatura 127 grade C si presiunea 2 atm,care se obtine stoechiometric din 170 g de apa oxigenata. .2.Calculati nr. atomilorde oxigen din 0,34g de apa oxigenata. .3. Determinati masa ,exprimata in g,a 18,066x10 la puterea 23 molecule de apa.

Answer 1

1)
ecuatia: H2O2 -> H2O + 1/2O2 sau 2H2O2 -> 2H2O + O2
34g..................0,5moli
H2O2 -> H2O + 1/2O2
170g..................x=?
x=170*0,5/34=2,5moli obtinuti in c.n
pentru conditiile date de tine, vom folosi ecuatia de stare termica: pV=nRT, T=t+273
V=nRT/p=2,5*0,082*400/2=41l O2
2)
1mol H2O2........34g.........2*6,022*10²³ atomi O
.....................0,34g..........................x atomi O
x=0,34*2*6,022*10²³/34=0,12044*10²³ atomi O
3)
1mol H2O.............18g............6,022*10²³ molecule H2O
.............................xg...........18,066*10²³ molecule H2O
x=18,066*18/6,022=3*18=54g apa