Answer:

Explanation:
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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

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<h3>
Answer:</h3>
3.03 × 10²⁵ formula units KCl
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Unit 0</u>
- Reading a Periodic Table
- Writing Compounds
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
50.3 mol KCl (Potassium chloride)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
= 3.02907 × 10²⁵ formula units KCl
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.02907 × 10²⁵ formula units KCl ≈ 3.03 × 10²⁵ formula units KCl
Answer:
49.95 g of HCl
Explanation:
Let's formulate the chemical equation involved in the process:
Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O
This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2
1 mole Ca(OH)2 ---- 2 moles HCl
0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl
Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:
1 mole of HCl ---- 36.46094 g
1.37 moles ------ x = 49.95 g of HCl