Answer:
296 g BaBr₂
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
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Stoichiometry
Using Dimensional Analysis
Explanation:
Step 1: Define
[Given] 8.56 × 10²³ formula units BaBr₂
[Solve] grams BaBr₂
Step 2: Identify Conversions
Avogadro's Number
[PT] Molar Mass of Ba - 137.33 g/mol
[PT] Molar Mass of Br - 35.45 g/mol
Molar Mass of BaBr₂ - 137.33 + 2(35.45) = 208.23 g/mol
Step 3: Convert
[DA} Set up:
[DA] Multiply/Divide [Cancel out units]:
Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
295.99 g BaBr₂ ≈ 296 g BaBr₂