Answer:
7.3 × 10⁻⁷ g Ni
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
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
Step 1: Define
7.5 × 10¹⁵ atoms Ni
Step 2: Identify Conversions
Avogadro's Number
Molar Mass of Ni - 58.69 g/mol
Step 3: Convert
- Set up: [tex]\displaystyle 7.5 \cdot 10^{15} \ atoms \ Ni(\frac{1 \ mol \ Ni}{6.022 \cdot 10^{23} \ atoms \ Ni})(\frac{58.69 \ g \ Ni}{1 \ mol \ Ni})[/tex]
- Multiply: [tex]\displaystyle 7.30945 \cdot 10^{-7} \ g \ Ni[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 2 sig figs.
7.30945 × 10⁻⁷ g Ni ≈ 7.3 × 10⁻⁷ g Ni
