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True [87]
3 years ago
8

Use the titration results to calculate the moles of h2so4

Chemistry
1 answer:
Aloiza [94]3 years ago
4 0
Missing in your question: the molarity of H2SO4 M = 0.1 M
and the equivalence point (mL) = 19.31 mL = 0.019 L
So by using the molarity formula:

molarity of H2SO4 = no of moles of H2SO4 / Volume L

So by substitution:
0.1 M = no of moles of H2SO4  /  0.019 L
∴ no of moles of H2SO4 = 0.1 M * 0.019 L
                                          = 0.0019 mol
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How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution
dalvyx [7]
<h3>Answer:</h3>

0.024 kg CaO

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Aqueous Solutions</u>

  • Molarity = moles of solute / liters of solution

<u>Atomic Structure</u>

  • Reading a Periodic Tables
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

<u>Step 2: Identify Conversions</u>

1000 g = 1 kg

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Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})
  2. Multiply:                              \displaystyle 0.024114 \ kg \ CaO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>

0.024114 kg CaO ≈ 0.024 kg CaO

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