They are all stable and have eight valence electrons
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.
The answer is TRUE.
If the Energy is on the left, then the problem is true. If it is on the right then it would be negative, false, and considered as exothermic.
Endothermic reaction = the products are higher in energy than the reactants.
Exothermic reaction = a chemical reaction that releases energy by light or heat.
So we know that the equation for density is:
where D is the density, m is the mass in grams, and V is the volume in mL.
So since we know two of the variables, mass and density, we can solve for the volume:
Therefore, the volume of this urine sample is 144.12mL.
