Respuesta :

Space

Answer:

1000 g CCl₄

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
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

3.93 × 10²⁴ molecules CCl₄

Step 2: Identify Conversions

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

Step 3: Convert

  1. Set up:                              [tex]\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})[/tex]
  2. Multiply:                                                                                                             [tex]\displaystyle 1003.77 \ g \ CCl_4[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1003.77 g CCl₄ ≈ 1000 g CCl₄

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