Oblicz masę glukozy, która uległa fermentacji, jeżeli powstało 100cm3 etanolu o gęstości d= 0,7851 g/cm3
V = 100cm³
d = 0,7851g/cm³
m = dV
m = 100cm³ x 0,7851g/cm³ = 785,1g
180u 92u
C₆H₁₂O₆ --> 2C₂H₅OH + 2CO₂
x 785,1g
z proporcji wyliczam x
x = 1563g
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Feci, quod potui, faciant meliora potentes
Pozdrawiam :)
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V = 100cm³
d = 0,7851g/cm³
m = dV
m = 100cm³ x 0,7851g/cm³ = 785,1g
180u 92u
C₆H₁₂O₆ --> 2C₂H₅OH + 2CO₂
x 785,1g
z proporcji wyliczam x
x = 1563g
----------------------------------------------------------------------------------------------------------
Feci, quod potui, faciant meliora potentes
Pozdrawiam :)