How many grams of barium chloride (BaCl2) are there in 0.187 mol?
0.187 mol
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Answer:
38.9 grams of [tex]BaCl_2[/tex]
Explanation:
0.187 mol BaCl2 x [tex]\frac{208 g}{1 mol}[/tex]
0.187 m x 208 g/m
0.187 x 208 g
38.896 g --> 38.9 g BaCl2