Ile gramów wodoru tlenku sodu potrzeba do przygotowania 2 dm3, 10%-owego roztworu o gęstości 1,115 gramy przez cm3
mr=2dm³*1,115g/cm³=2000cm³*1,115g/cm³=2230g
Cp=10%
ms=?
Cp=ms/mr * 100% /*mr
Cp*mr=ms*100% /:100%
ms=Cp*mr/100%
ms=10%*2230g/100%
ms=223g
dane:
Vr(roztworu) =2dm3 =2000cm3
dr= =1,115g/cm3
Cp =10%
szukane:
mr =?
ms =?
d =m/V I*V
mr =d * V
mr =1,115g/cm3 *2000cm3
mr = 2230g
--------------
Z def. stężenia procentowego:
Cp =ms/mr *100% I*mmr
ms *100% =Cp *mr I:100%
ms =Cp *mr/100%
ms =10% *2230g/100%
ms =223g NaOH
==============
Odp.Potrzeba 223g NaOH.
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mr=2dm³*1,115g/cm³=2000cm³*1,115g/cm³=2230g
Cp=10%
ms=?
Cp=ms/mr * 100% /*mr
Cp*mr=ms*100% /:100%
ms=Cp*mr/100%
ms=10%*2230g/100%
ms=223g
dane:
Vr(roztworu) =2dm3 =2000cm3
dr= =1,115g/cm3
Cp =10%
szukane:
mr =?
ms =?
d =m/V I*V
mr =d * V
mr =1,115g/cm3 *2000cm3
mr = 2230g
--------------
Z def. stężenia procentowego:
Cp =ms/mr *100% I*mmr
ms *100% =Cp *mr I:100%
ms =Cp *mr/100%
ms =10% *2230g/100%
ms =223g NaOH
==============
Odp.Potrzeba 223g NaOH.