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Masa hydratu= 208g + 18x
Masa baru w tym hydracie = 137g
56,15% * (208 + 18x) = 137
0,5615 * (208+18x) = 137
116,792 + 10,107x = 137
10,107x = 20,208
x= 2( w przyblizeniu)
Wzór hydratu:
BaCl2 * 2H2O
masa: 208g + 18x
masa baru w hydracie: 137g
0,5615(208+18x) = 137
116,792 + 10,107x = 137
10,107x = 20,208
x= 2
wzór: BaCl₂ × 2H₂O