As you move down through the family of alkali metals, the first ionization energy because a) increases; the atomic radii increase. b) increases; the atomic radii decrease. c) decreases; the atomic radii increase. d) remains constant; all the elements lose one electron.

As you move down through the family of alkali metals, the first ionization energy decreases because the atomic radii increases.

C

Explanation

The first ionization energy is the energy required to remove the electron from the outermost shell or valence shell of any atom.

Generally, the ionization energy increases along the periods and decreases along the group.

The decrease of ionization energy along the group is because of the increase in atomic radii of atoms along the group.

As the atomic radii increases , the influence of the nuclei towards the outermost shell electrons will be decreasing because of insertion of new inner shells leading to electron shielding effect.

Thus the valence electrons will have less binding energy with the nuclei with the increase of atomic radii.

As the binding energy of the valence electrons with the nuclei decreases ,the valence electrons can be knocked out of the atom with less amount of energy.

This leads to decreasing ionization energy. Thus for the present problem, as you move down through the family of alkali metals means moving down the alkali metal group will lead to increase in the atomic radii along the group. This in-turn will lead to decrease in the 1st ionization energy.

Thus, the first ionization energy will decrease on moving downward in the family of alkali metals because of the increase in atomic radii along the downward direction of elements in alkali metals.

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole