Answer is: <span>the exact ratio of oxygen to octane for is 12.5 : 1.
</span>Balanced chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O or multiply by 2:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.
There same number of atoms on both side of balanced chemical reaction: eight carbon atoms, eighteen hydrogen atoms and twenty five oxygen atoms.
Answer:
The equilbrium constant is 179.6
Explanation:
To solve this question we can use the equation:
ΔG = -RTlnK
<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>
<em>R is gas constant = 8.314x10⁻³kJ/molK</em>
<em>T is absolute temperature = 298K</em>
<em>And K is equilibrium constant.</em>
Replacing:
12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK
5.19 = lnK
e^5.19 = K
179.6 = K
<h3>The equilbrium constant is 179.6</h3>
Random errors will shift each measurement from its true value by a random amount and in a random direction. These will affect reliability (since they're random) but may not affect the overall accuracy of a result.
Jill and Susan violated safety procedures by not properly listening and/or reading over the instructions to know all the materials, steps, and equipment they need for the lab. Hope this helps!
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
