Answer:
Both are highly reactive.
Explanation:
A has 1 valence electron D has 3
A is sodium D is aluminum
How am I supposed to help you if there's not picture of what the problem is
The amount of the solute present in the given solution is called the concentration. The best way to represent the concentration of the solution is ![\rm [K_{2}CrO_{4}].](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D.)
<h3>What is molar concentration?</h3>
Molar concentration is the molarity of the solution that is the measure of the concentration of the solute dissolved in the solution.
The formula for calculating molar concentration is given as,

The concentration of any substance is represented in the square bracket like
or ![\rm [K_{2}CrO_{4}].](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D.)
Therefore, option B.
is the representation of the concentration.
Learn more about the molarity here:
brainly.com/question/1532164
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.