Answer:
3.0 × 10²³ molecules AgNO₃
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Chemistry
Atomic Structure
- Reading a Periodic Table
- Writing Compounds
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Stoichiometry
- Using Dimensional Analysis
Step-by-step 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
- Set up:
![\displaystyle 85 \ g \ AgNO_3((1 \ mol \ AgNO_3)/(169.88 \ g \ AgNO_3))((6.022 \cdot 10^(23) \ molecules \ AgNO_3)/(1 \ mol \ AgNO_3))](https://img.qammunity.org/2022/formulas/chemistry/college/8ocl64ciy0ve4tldkzjqo194encb94kos9.png)
- Multiply/Divide:
![\displaystyle 3.01313 \cdot 10^(23) \ molecules \ AgNO_3](https://img.qammunity.org/2022/formulas/chemistry/college/dh893eq0olka9xybcva6ni9i7yqzc44xar.png)
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