Ile gramów KOH otrzymasz po odparowaniu do sucha 0,5dm³ 20% roztworu tego wodoro tlenku o gęstości na d=1,18g/cm³
v=0,5dm3=500cm3, d=1,18g/cm3, m=?
m=v*d
m=500*1,18
m = 590g
mr=590g
Cp=20%
ms=?
Cp=ms*100%/mr
ms = Cp*mr/100%
ms = 20*590/100
ms = 118g KOH
1dm^3 to 1000cm^3
20%*500cm^3*1,18=100cm^*1,18=118g
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v=0,5dm3=500cm3, d=1,18g/cm3, m=?
m=v*d
m=500*1,18
m = 590g
mr=590g
Cp=20%
ms=?
Cp=ms*100%/mr
ms = Cp*mr/100%
ms = 20*590/100
ms = 118g KOH
1dm^3 to 1000cm^3
20%*500cm^3*1,18=100cm^*1,18=118g