Answer:
1.1 × 10²² atoms Au
General Formulas and Concepts:
Chemistry
Atomic Structure
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Explanation:
Step 1: Define
3.7 g Au
Step 2: Identify Conversions
Avogadro's Number
Molar Mass of Au - 196.97 g/mol
Step 3: Convert
[tex]3.7 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au} )(\frac{6.022 \cdot 10^{23} \ atoms \ Au}{1 \ mol \ Au} )[/tex] = 1.13121 × 10²² atoms Au
Step 4: Check
We are given 2 sig figs. Follow sig fig rules and round.
1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au
