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1.204 x 10^24 molecules of CH3COOH is how many grams?

Sagot :

Space

Answer:

120.1 g CH₃COOH

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

1.204 × 10²⁴ molecules CH₃COOH

Step 2: Identify Conversions

Avogadro's Number

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

[PT] Molar Mass of H - 1.01 g/mol

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

Molar Mass of CH₃COOH - 12.01 + 3(1.01) + 12.01 + 16.00 + 16.00 + 1.01 = 60.06 g/mol

Step 3: Convert

  1. Set up:                               [tex]\displaystyle 1.204 \cdot 10^{24} \ molecules \ CH_3COOH(\frac{1 \ mol \ CH_3COOH}{6.022 \cdot 10^{23} \ molecules \ CH_3COOH})(\frac{60.06 \ g \ CH_3COOH}{1 \ mol \ CH_3COOH})[/tex]
  2. Multiply/Divide:                                                                                                      [tex]\displaystyle 120.08 \ g \ CH_3COOH[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

120.08 g CH₃COOH ≈ 120.1 g CH₃COOH

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