Respuesta :

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Answer:

3.2 g CaCO₃

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

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

Chemistry

Atomic Structure

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

  • Using Dimensional Analysis

Explanation:

Step 1: Define

1.9 × 10²² molecules CaCO₃

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ca - 40.08 g/mol

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CaCO₃ - 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

Step 3: Convert

  1. Set up:                              [tex]\displaystyle 1.9 \cdot 10^{22} \ molecules \ CaCO_3(\frac{1 \ mol \ CaCO_3}{6.022 \cdot 10^{23} \ molecules \ CaCO_3})(\frac{100.09 \ g \ CaCO_3}{1 \ mol \ CaCO_3})[/tex]
  2. Multiply/Divide:                                                                                                    [tex]\displaystyle 3.15794 \ g \ CaCO_3[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

3.15794 g CaCO₃ ≈ 3.2 g CaCO₃

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