Answer:
the energy of the third excited rotational state 
Explanation:
Given that :
hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm
Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.
Thus; the reduced mass μ = 
μ = 
μ = 
∵ 1 μ = 1.66 × 10⁻²⁷ kg
μ = 
μ = 1.6139 × 10⁻²⁷ kg

The rotational level Energy can be expressed by the equation:

where ;
J = 3 ( i.e third excited state) &




We know that :
1 J = 



The substance would be referred to as a compound substance.
B A C ( Blood Alcohol Content ) of 0.10 means that there are 0.10 g of alcohol for every dl of blood.
5 L = 50 dl
50 * 0.10 g = 5 g
In his blood is circulating 5 grams of alcohol.
A, because the number of valence shell electrons (outer shell electrons) tells us how much the element or compound wants to bond or give up electrons. Most compounds and elements want to have eight valence ectrons in it's outer ring. So if an atom is far away from having eight, it will want to react more often.