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romashka310
@romashka310
September 2021
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определите массавую долю воды в веществе.
пожалуйста срочно))
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Alexei78
M(FeSO4*7H2O)=56+32+16*4+7*(1*2+16)=152+126=278 g/mol
W(H2O)=Ar(H2O)*n / M(FeSO4*7H2O)*100% = 18*7/278*100%=45.32%
ответ 45.32 %
M(CaCO3*2H2O)=40+12+16*3+2*(1*2+16)=100+36=136 g/mol
W(H2O)=Ar(H2O)*n / M(CaCO3*2H2O)*100%= 18*2 / 136*100%=26.47%
ответ 26.47%
M(Na2CO3*10H2O)=23*2+12+16*3+10*(1*2+16)=106+180=286g/mol
W(H2O)=Ar(H2O)*n / M(Na2CO3*10H2O)*100%
W(H2O)=18*10 / 286*100%=62.94%
ответ 62.94%
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Answers & Comments
W(H2O)=Ar(H2O)*n / M(FeSO4*7H2O)*100% = 18*7/278*100%=45.32%
ответ 45.32 %
M(CaCO3*2H2O)=40+12+16*3+2*(1*2+16)=100+36=136 g/mol
W(H2O)=Ar(H2O)*n / M(CaCO3*2H2O)*100%= 18*2 / 136*100%=26.47%
ответ 26.47%
M(Na2CO3*10H2O)=23*2+12+16*3+10*(1*2+16)=106+180=286g/mol
W(H2O)=Ar(H2O)*n / M(Na2CO3*10H2O)*100%
W(H2O)=18*10 / 286*100%=62.94%
ответ 62.94%