Дано:m (AgNO3) = 170 г
Найти:m (осадка) - ?
Решение:1) Уравнение реакции:AgNO3 + NaCl => AgCl ↓ + NaNO3;2) Молярные массы:M (AgNO3) = Mr (AgNO3) = Ar (Ag) * N (Ag) + Ar (N) * N (N) + Ar (O) * N (O) = 108 * 1 + 14 * 1 + 16 * 3 = 170 г/моль;M (AgCl) = Mr (AgCl) = Ar (Ag) * N (Ag) + Ar (Cl) * N (Cl) = 108 * 1 + 35,5 * 1 = 143,5 г/моль;3) Количество вещества:n (AgNO3) = m (AgNO3) / M (AgNO3) = 170 / 170 = 1 моль;4) Количество вещества:n (AgCl) = n (AgNO3) = 1 моль;5) Масса:m (AgCl) = n (AgCl) * M (AgCl) = 1 * 143,5 = 143,5 г.
Ответ: Масса AgCl составляет 143,5 г.
Автор:
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Exercise 4: Put the words in the correct order to get the question. Write it down. Then answer the
question you received..
1. laboratory / you / do /know / analysis /What?
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4. Why / a / a / come / does / doctor / to see / patient?
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5. a nurse / What / does / do?
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