Respuesta :

Space

Answer:

14900 g 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
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

  • Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 5.29 × 10²⁵ molecules AgNO₃

[Solve] grams AgNO₃

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. [DA] Set up:                                                                                                        [tex]\displaystyle 5.29 \cdot 10^{25} \ molecules \ AgNO_3(\frac{1 \ mol \ AgNO_3}{6.022 \cdot 10^{23} \ molecules \ AgNO_3})(\frac{169.88 \ g \ AgNO_3}{1 \ mol \ AgNO_3})[/tex]
  2. [DA] Multiply/Divide [Cancel out units]:                                                           [tex]\displaystyle 14923 \ g \ AgNO_3[/tex]

Step 4: Check

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

14923 g AgNO₃ ≈ 14900 g AgNO₃