Respuesta :

Space

Answer:

1.21 g HBr

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Compounds

  • Writing Compounds
  • Acids/Bases

Atomic Structure

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

9.03 × 10²¹ molecules HBr

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of HBr - 1.01 + 79.90 = 80.91 g/mol

Step 3: Convert

  1. Set up:                              [tex]\displaystyle 9.03 \cdot 10^{21} \ molecules \ HBr(\frac{1 \ mol \ HBr}{6.022 \cdot 10^{23} \ molecules \ HBr})(\frac{80.91 \ g \ HBr}{1 \ mol \ HBr})[/tex]
  2. Multiply/Divide:                                                                                               [tex]\displaystyle 1.21325 \ g \ HBr[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.21325 g HBr ≈ 1.21 g HBr

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