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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 = 



Dispersion forces are the only type of intermolecular force operating between non-polar molecules, for example, dispersion forces operate between hydrogen (H2) molecules, chlorine (Cl2) molecules, carbon dioxide (CO2) molecules, nitrogen tetroxide (N2O4) molecules and methane (CH4) molecules.
www.ausetute.com.au/intermof.html
Answer:
Hi there I think the awnser is B. or A. lol but im 90% its A.
Explanation:
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Galvanizing protects from rust in a number of ways: It forms a barrier that prevents corrosive substances from reaching the underlying steel or iron. The zinc serves as a sacrificial anode so that even if the coating is scratched, the exposed steel will still be protected by the remaining zinc.