Relative formula mass C₅H₁₁ = 71
Now divide the molar mass by the RFM = 142.32 / 71 = 2
Now C₍₅ₓ₂₎H₍₁₁ₓ₂) = C₁₀H₂₂
Hope that helps
Answer:
1.33 × 10²⁴ molecules CO₂
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Reading a Periodic Table
- Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
97.3 g CO₂
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Convert</u>
= 1.33138 × 10²⁴ molecules CO₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules.</em>
1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂
Answer:
1.The substance(s) to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction.The combustion of methane or octane is exothermic; it releases energy. CH4 + 2 O2 → CO2 + 2 H2O + energy The energies of the products are lower than the energiies of the reactants.
Explanation: