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3 years ago

How many L are required to make 3.5 M hydrochloric acid using 1.1 moles?

Chemistry

1 answer:

ch4aika [34]3 years ago

4 0

<h3>Answer:</h3>

0.31 L

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

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>

Moles

<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]: $\displaystyle 3.5 \ M = \frac{1.1 \ moles}{x \ L}$
[Solution] Cross-Multiply [Equality Property]: $\displaystyle x \ L = \frac{1.1 \ mol}{3.5 \ M}$
[Solution] Divide: $\displaystyle x = 0.314286 \ L$

<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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