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Temka [501]
3 years ago
12

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>Math</u>

<u>Pre-Algebra</u>

  1. Order of Operations: BPEMDAS
  2. Brackets
  3. Parenthesis
  4. Exponents
  5. Multiplication
  6. Division
  7. Addition
  8. 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>

  1. Substitute in variables [Molarity]:                                                                  \displaystyle 3.5 \ M = \frac{1.1 \ moles}{x \ L}
  2. [Solution] Cross-Multiply [Equality Property]:                                                \displaystyle x \ L = \frac{1.1 \ mol}{3.5 \ M}
  3. [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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There is some info missing. I think this is the original question.

<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>

<em />

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