Dane:
I= 10A
t= 10 min= 600s
d Bi= 9,8 g/cm3
vBi=?
-----------------------------------
K (-): Bi³⁺ + 3e⁻ ---> Bi z=3
M Bi= 209 g/mol
mBi= (M Bi/(z*F))* It
mBi= (209/(3*96500)*10*600= 4,33g
dBi= mBi/vBi
vBi= mBi/dBi= 4,33g/(9,8g/cm3)= 0,442cm3= 442 mm3
Odp. Wydzielony bizmut zajmie objętość 442 mm3.
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Dane:
I= 10A
t= 10 min= 600s
d Bi= 9,8 g/cm3
vBi=?
-----------------------------------
K (-): Bi³⁺ + 3e⁻ ---> Bi z=3
M Bi= 209 g/mol
mBi= (M Bi/(z*F))* It
mBi= (209/(3*96500)*10*600= 4,33g
dBi= mBi/vBi
vBi= mBi/dBi= 4,33g/(9,8g/cm3)= 0,442cm3= 442 mm3
Odp. Wydzielony bizmut zajmie objętość 442 mm3.