Oblicz zawartość % wody w cząsteczce CaSO4 * 2H2O.
masa hydratu: 40+32+64+36=172g
masa wody: 18 x 2=36g
36/172 x 100%=21%
m.CaSO4 * 2H2O = 40+32+4 *16 +2 *18 =172g
m.2H2O =2 *18 =36g
172g -----100%
36g -------x%
------------------
x =36g/172g *100% = 20,9%
x =21%
=======
Odp.Masa wody stanowi 21% hydratu.
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masa hydratu: 40+32+64+36=172g
masa wody: 18 x 2=36g
36/172 x 100%=21%
m.CaSO4 * 2H2O = 40+32+4 *16 +2 *18 =172g
m.2H2O =2 *18 =36g
172g -----100%
36g -------x%
------------------
x =36g/172g *100% = 20,9%
x =21%
=======
Odp.Masa wody stanowi 21% hydratu.