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<h3>Answer:</h3>
0.239 mol Sm(NO₃)₃
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[Given] 80.3 g Sm(NO₃)₃
[Solve] moles Sm(NO₃)₃
<u>Step 2: Identify Conversions</u>
[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
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.238711 mol Sm(NO₃)₃ ≈ 0.239 mol Sm(NO₃)₃