nzabbie06
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Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f​

Sagot :

Cricetus

Answer:

The correct answer is "[tex]4.991\times 10^{-45} \ kg.m^2[/tex]".

Explanation:

According to the question,

[tex]R_{C-Cl} = 177 \ pm[/tex]

or,

         [tex]=1.77\times 10^{-10} \ m[/tex]

[tex]\alpha = 107^{\circ}[/tex]

[tex]m_{Cl}=34.97 \ m.u[/tex]

or,

      [tex]=34.97\times 1.66\times 10^{-27}[/tex]

      [tex]=5.807\times 10^{-26} \ kg[/tex]

The moment of inertia around the rotational axis will be:

⇒  [tex]I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha[/tex]

By putting the values, we get

       [tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)[/tex]

       [tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452[/tex]

       [tex]=4.991\times 10^{-45} \ kg.m^2[/tex]

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