Amount (mol) of mn atoms in 62.0 mg of mn

               0.0011 mol

Data Given:

                  Mass of Mn  =  62.0 mg  =  0.062 g

                  Atomic Mass of Mn  =  54.94 g.mol⁻¹

Calculate Moles of Mn as,

                               Moles  =  Mass ÷ M.Mass

Putting values,

                               Moles  =  0.062 g ÷ 54.94 g.mol⁻¹

                                Moles  =  0.0011 mol

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

          Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is

Of F or Se, the larger atom is

Of Sn or I, the larger atom is

Of Cr or W, the larger atom is

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.