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lvicyna24
@lvicyna24
July 2022
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К 500 г 10%-ного раствора сульфата меди (II) добавили 100 г медного купороса. Массовая концентрация соли в полученном растворе равна _ %.
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SashaB11
Verified answer
Дано:
m1(p-pa) = 500г
w₁(CuSO₄) = 10%
+m₂(CuSO₄) = 100г
Найти:
w₂(CuSO₄)
Решение:
w₂(CuSO₄) = m₃(CuSO₄):m₂(p-pa)
m₁(CuSO₄) = 500*0,10 = 50г
m₃(CuSO₄) = 50+100 = 150г
m₂(p-pa) = 500+100 = 600г
w₂(CuSO₄) = 150:600 = 0,25*100% = 25%
Ответ: w₂(CuSO₄) = 25%
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Answers & Comments
Verified answer
Дано:m1(p-pa) = 500г
w₁(CuSO₄) = 10%
+m₂(CuSO₄) = 100г
Найти:
w₂(CuSO₄)
Решение:
w₂(CuSO₄) = m₃(CuSO₄):m₂(p-pa)
m₁(CuSO₄) = 500*0,10 = 50г
m₃(CuSO₄) = 50+100 = 150г
m₂(p-pa) = 500+100 = 600г
w₂(CuSO₄) = 150:600 = 0,25*100% = 25%
Ответ: w₂(CuSO₄) = 25%