Zawartosc procentowa w hydracie CuSO4*xH2O wynosi 36 %. Ustal wzór sumaryczny tego hydratu( 64u, 32u, 16u, 1u). pilne!
mCuSO4=160u mH2O=18u mCuSO4*xH2O=160u+x*18u 160+18x -------- 100% 18x -------------- 36% 18x*100 = 36(160+18x) 1800x = 5760+648x 1152x = 5760 x = 5
Wzór hydratu: CuSO4*5H2O
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mCuSO4=160u
mH2O=18u
mCuSO4*xH2O=160u+x*18u
160+18x -------- 100%
18x -------------- 36%
18x*100 = 36(160+18x)
1800x = 5760+648x
1152x = 5760
x = 5
Wzór hydratu: CuSO4*5H2O