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1) Если концентрацию раствора выразить в долях единицы, то массу растворенного вещества в граммах находят по формуле:
m(в-во) = ω * m(p-p)
где: m(p-p) - масса раствора в граммах, ω - массовая доля растворенного вещества, выраженная в долях единицы.
41% - это 0,41 в долях единицы.
2) Обозначим концентрацию полученного раствора (массовую долю растворенного вещества) через Х. При добавлении воды масса растворенного вещества не изменяется, а масса раствора увеличивается на массу добавленной воды, на основании всего вышеизложенного составляем уравнение:
0,41 * 350 = Х * (350 + 300)
143,5 = 650Х
Х = 0,2208 (или 22,08%)
Объяснение: 1) CuSO4 + BaCl2 = BaSO4 + CuCl2
Cu^2+ + SO4^2- + Ba^2++ 2Cl^- = BaSO4 + Cu^2+ +2Cl-
SO4^2- + Ba^2+ = BaSO4
2) 2AgNO3 + BaCl2 = 2AgCl + Ba(NO3)2
2Ag^+ + 2NO3^- + Ba^2+ + 2Cl^- = 2AgCl + Ba^2+ + 2NO3^-
Ag^+ + Cl^- = AgCl
3) Na2CO3 + 2HNO3 = 2NaNO3 + H2O + CO2
2Na^+ + CO3^2- + 2H^+ + 2NO3^- = 2Na^+ + 2NO3^ + H2O + CO2
CO3^2- + 2H^+ = H2O + CO2
4) CuO + 2HNO3 = Cu(NO3)2 + H2O
CuO + 2H^+ + 2NO3^- = Cu^2+ + 2NO3^- + H2O
CuO + 2H^+ = Cu^2+ + H2O
5) CaCl2 + 2AgNO3 = Ca(NO3)2 + 2AgCl
Ca^2+ + 2Cl^- + 2Ag^+ + 2NO3^- = Ca^2+ + 2NO3^- + 2AgCl
Cl^- + Ag^+ = AgCl
6) ZnCO3 + H2SO4 = ZnSO4 + H2O + CO2
ZnCO3 + 2H^+ + SO4^2- = Zn^2+ + SO4^2- + H2O + CO2
ZnCO3 + 2H^+ = Zn^2+ + H2O + CO2