Ile gramów CaCl2 pozostanie po całkowitym odparowaniu 500 cm3 roztworu o stężeniu 4mol/dm3?
Cm = 4mol/dm3
V = 500cm3 = 0,5 dm3
n = 0,5 * 4 = 2 mole CaCl2
MCaCl2 = 111g/m
m = 2 * 111 = 222 g - CaCl2
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Cm = 4mol/dm3
V = 500cm3 = 0,5 dm3
n = 0,5 * 4 = 2 mole CaCl2
MCaCl2 = 111g/m
m = 2 * 111 = 222 g - CaCl2