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(C) Let the mass of the $H$ atom be $m_1 = 1 \ 	ext{unit}$ and the mass of the $Cl$ atom be $m_2 = 35.5 \ 	ext{units}$.Place the $H$ atom at the ori

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$1.24$

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(C) Let the mass of the $H$ atom be $m_1 = 1 \ 	ext{unit}$ and the mass of the $Cl$ atom be $m_2 = 35.5 \ 	ext{units}$.Place the $H$ atom at the origin $(0, 0)$. Then the position of the $H$ atom is $r_1 = 0$ and the position of the $Cl$ atom is $r_2 = 1.27 \ \mathring A$.The center of mass $R_{cm}$ is given by the formula:$R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$Substituting the values:$R_{cm} = \frac{(1)(0) + (35.5)(1.27)}{1 + 35.5}$$R_{cm} = \frac{35.5 	imes 1.27}{36.5}$$R_{cm} \approx 1.24 \ \mathring A$Thus,the center of mass is at a distance of $1.24 \ \mathring A$ from the $H$ atom.

In an $HCl$ molecule,the distance between the nuclei of the two atoms is $1.27 \ \mathring A$. The $Cl$ atom is approximately $35.5$ times heavier than the $H$ atom. The center of mass of this molecule will be at a distance of approximately ....... $\mathring A$ from the center of the $H$ atom.

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