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spayn [35]
3 years ago
9

How many grams are in 3.93 x 10^24 molecules of CCl4?

Chemistry
1 answer:
Lorico [155]3 years ago
8 0
<h3>Answer:</h3>

1000 g CCl₄

<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>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.93 × 10²⁴ molecules CCl₄

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

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.93 \cdot 10^{24} \ molecules \ CCl_4(\frac{1 \ mol CCl_4}{6.022 \cdot 10^{23} \ molecules \ CCl_4})(\frac{153.81 \ g \ CCl_4}{1 \ mol \ CCl_4})
  2. Multiply:                                                                                                             \displaystyle 1003.77 \ g \ CCl_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1003.77 g CCl₄ ≈ 1000 g CCl₄

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Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

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By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?

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Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

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