Three elements have the electron configurations 1s2 2s2 2p6 3s2 3p6, 1s2 2s2 2p6 3s2, and 1s2 2s2 2p6 3s2 3p6 4s1. The first ionization energies of the three elements (not in the same order) are 0.4189, 0.7377, and 1.505 MJ/mol. The atomic radii are 1.60, 0.94, and 2.35 .Identify the three elements and match the appropriate values of ionization energy and atomic radius to each configuration.

Explanation:

First element

1s2 2s2 2p6 3s2 3p6

Atomic Number = 18

Element = Argon

Second Element

1s2 2s2 2p6 3s2

Atomic Number = 12

Element = Magnesium

Second Element

1s2 2s2 2p6 3s2 3p6 4s1

Atomic Number = 19

Element = Potassium

In the periodic table;

The general trend is for ionization energy to increase moving from left to right across an element period. It would decrease down the group.

Both Argon and Magnesium are on the same period. Potassium is o the period below.

This leads us to;

Magnesium - 0.7377 MJ/mol

Argon -  1.505 MJ/mol

Potassium - 0.4189 MJ/mol

The atomic radius of atoms generally decreases from left to right across a period.  It would increase down the group.

This leads to;

Magnesium - 1.60

Argon -  0.94

Potassium - 2.35

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j