Oblicz i porównaj zawartość procentową (procent masowy) wody w CuSO4·5H2O i CuCl2·5H2O
Z góry dziekuję
m.cz. CuSO4 * 5H2O = 63,5u + 32u + 4*16u = 159,5u + 90u = 249,5u
5 m.cz. H2O = 5 * (2*1u + 16u) = 5*18u = 90u
%H2O = (90u/249,5u)*100% = 36,07%
m.cz. CuCl2 * 5H2O = 63,5u + 2*35,5u = 134,5u + 90u = 224,5u
%H2O = (90u/224,5u)*100% = 40,08%
pozdrawiam :)
249,5u - 100%
90u - x
_____________
224,5u - 100%
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m.cz. CuSO4 * 5H2O = 63,5u + 32u + 4*16u = 159,5u + 90u = 249,5u
5 m.cz. H2O = 5 * (2*1u + 16u) = 5*18u = 90u
%H2O = (90u/249,5u)*100% = 36,07%
m.cz. CuCl2 * 5H2O = 63,5u + 2*35,5u = 134,5u + 90u = 224,5u
5 m.cz. H2O = 5 * (2*1u + 16u) = 5*18u = 90u
%H2O = (90u/224,5u)*100% = 40,08%
pozdrawiam :)
249,5u - 100%
90u - x
_____________
224,5u - 100%
90u - x