Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would s
2 answers:
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<h3>Answer:</h3>
0.0113 mol Ba(ClO₃)₂
<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>Chemistry</u>
<u>Atomic Structures</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
3.45 g Ba(ClO₃)₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Ba - 137.33 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Ba(ClO₃)₂ - 137.33 + 2(35.45) + 6(16.00) = 304.33 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.011336 mol Ba(ClO₃)₂ ≈ 0.0113 mol Ba(ClO₃)₂