How many moles of MgSiO3 are in 237g of the compound?
1 answer:
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<h3>Answer:</h3>
1.1 × 10²² atoms Au
<h3>General Formulas and Concepts:</h3><u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Step 1: Define</u>
3.7 g Au
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
<u /> = 1.13121 × 10²² atoms Au
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules and round.</em>
1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au