Answer:
0.239 mol Sm(NO₃)₃
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] 80.3 g Sm(NO₃)₃
[Solve] moles Sm(NO₃)₃
Step 2: Identify Conversions
[PT] Molar Mass of Sm - 150.36 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of Sm(NO₃)₃ - 150.36 + 3[14.01 + 3(16.00)] = 336.39 g/mol
Step 3: Convert
- [DA] Set up:
- [DA] Divide [Cancel out units]:
Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
0.238711 mol Sm(NO₃)₃ ≈ 0.239 mol Sm(NO₃)₃