Question

How many molecules of CaCi2 are equivalent to 75.9 grams

Answer 1

M CaCl₂: 40+(35,5×2) = 111 g/mol


6,02·10²³ molecules ---------- 111g
X molecules --------------------- 75,9g
X = (75,9×6,02·10²³)/111
X = 4,116·10²³ molecules of CaCl₂

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