Answer:
2900 g Au
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
- Moles
Stoichiometry
- Using Dimensional Analysis
Step-by-step explanation:
Step 1: Define
[Given] 14.7 mol Au
[Solve] g Au
Step 2: Identify Conversion
[PT] Molar Mass of Au - 196.97 g/mol
Step 3: Convert
- [DA] Set up:

- [DA] Multiply [Cancel out units]:

Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
2895.46 g Au ≈ 2900 g Au