How many L are required to make 3.5 M hydrochloric acid using 1.1 moles?
1 answer:
<h3>
Answer: </h3>
0.31 L
<h3>
General Formulas and Concepts: </h3>
<u>Math </u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction
Equality Properties
Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u>
</u> <u>Chemistry</u>
<u>Atomic Structure</u>
<u>Aqueous Solutions</u>
Molarity = moles of solute / liters of solution <u>Equilibrium - Acid/Base</u>
Strong Acids Weak Acids Writing Acids <h3>
Explanation: </h3>
<u>Step 1: Define</u>
[Given] 3.5 M HCl
[Given] 1.1 moles
[Solve] Liters solution
<u>Step 2: Solve</u>
Substitute in variables [Molarity]: [Solution] Cross-Multiply [Equality Property]: [Solution] Divide:
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
0.314286 L ≈ 0.31 L
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