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Sagot :
Answer:
0.017 moles PbNO₃
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Chemistry
Atomic Structure
- Reading a Periodic Table
- Moles
Stoichiometry
- Using Dimensional Analysis
Explanation:
Step 1: Define
[Given] 4.5 g PbNO₃
[Solve] moles PbNO₃
Step 2: Identify Conversions
[PT] Molar Mass of Pb - 207.2 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of PbNO₃ - 207.2 + 14.01 + 3(16.00) = 269.21 g/mol
Step 3: Convert
- [DA] Set up: [tex]\displaystyle 4.5 \ g \ PbNO_3(\frac{1 \ mol \ PbNO_3}{269.21 \ g \ PbNO_3})[/tex]
- [DA] Multiply/Divide [Cancel out units]: [tex]\displaystyle 0.016716 \ moles \ PbNO_3[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 2 sig figs.
0.016716 moles PbNO₃ ≈ 0.017 moles PbNO₃
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