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.
Answer:
A lot of them.
Explanation:
It would take hundreds of thousands of trees to clear all of the emmisions.
Answer:
2.5g
Explanation:
When the reaction goes into completion, they will produce 2.5g. This is complement the law of conservation of mass.
According to the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".
- The mass of reactants and products in a chemical reaction must be the same.
- There is no change in mass in moving from reactant to product
- So, if we start with 2.5g of reactants, we must end with 2.5g of products.
What's wrong with this setup is the substrate on which you have positioned
the drop is "dirty and unclean" meaning it is not being dampened by
the solution. This action can be corrected by comprehensively cleaning the
substrate where the drop will be positioned.
Answer:
A
Explanation:
lies to the left of periodic table