Question

Calculate the bond length of HCL molecule of its moment of inertia 2×10^-40 gm cm^2 and reduced mass 1.0 g mol^-1

Answer 1

Answer:

We know, μ=q×l

=1.6×10  

−19

C×1.275(10  

−10

)m

=2.04×10  

−29

Cm

Actual μ of HCl  = 1.03D = 1.03×3.336×10  

−30

Cm

=0.34×10  

−29

Cm

So percentage of ionic character =  

Calculatedμ

Actualμ

​  

×100

=  

2.04×10  

−29

 

0.34×10  

−29

 

​  

×100

=16.667% (approx)

Explanation:

Answer 2

Answer:

helloooo

Explanation:

same as above mate...