Answer: As the atomic radii increases, the ionization energy decreases.
Explanation: Atomic Radii is the distance between the centre of the nucleus/atom and the outermost shell, while the ionization energy is the minimum energy required to remove the most loosely bound electron/valence electron from an isolated neutral gaseous atom/molecule.
As the number of shells increases, the atomic radius of an atom logically increases, and the force of attraction between the relatively positively charged heavy nucleus (positively charged because it contains protons and heavy because the protons & neutrons have great masses relative to the mass of an electron) and the electrons on the outermost shell reduces because of the increasing distance between this two entities. The reduction in the forces of attraction, makes it easier to remove that valence electron on the outermost shell as the size increases. This is why the ionization energy decreases with increase in atomic radius thereby leading to a decrease in ionization energy as we move down the group (of a periodic table).
Across the period (left to right), elements in the same period have the same number of shells, but the electrons in the outermost shell increases as we go from left to right, leading to a tighter hold on the increasing electrons by the nucleus. This is why atomic radius decreases & ionization energy consequently increase across a period.
More energy is required to remove the valence electrons as the force of attraction between nucleus & valence electrons on the outermost shell increases.
Solid (ice caps)
Liquid (oceans, rivers, lakes, etc)
Gas (clouds)
The amount of chemical energy a substance has is influenced by its structure and the bond the substance has.
Some bond is release energy when broken, other bond need some energy to break. The structure will also influence the chemical energy.
One of substance with high chemical energy is adenosine triphosphate which was used as energy source in a living organism. The phosphate bond in this molecule worth about 30.5 kJ/moles
Explanation:
Given
The enthalpy of formation of RbF (s) is –557.7kJ/mol
The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol
The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)
= -583.8 - (-557.7) kJ/mol
= -26.1 kJ/mol
The enthalpy is negative which means that the temperature will rise when RbF is dissolved.