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How many milliliters of 12.0 M hydrochloric acid contain 3.646 g of HCl (36.46 g/mol)?
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Answer:

8.3ml of 12M HCl contains 3.646 grams HCl

Explanation:

moles HCl = Molarity X Volume => Volume (Liters) = moles HCl/Molarity

Vol(L) = (3.646g/36.46g/mole)/12M = 0.0083Liter x 1000ml/L = 8.3ml

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