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Answer :

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Answer:

3.0 × 10²³ molecules AgNO₃

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
  • Writing Compounds
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

  • Using Dimensional Analysis

Explanation:

Step 1: Define

85 g AgNO₃ (silver nitrate)

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

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

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Step 3: Convert

  1. Set up:                              [tex]\displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})[/tex]
  2. Multiply/Divide:                                                                                                [tex]\displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3[/tex]

Step 4: Check

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

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

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