Answer:
Explanation:
Combustion. Have fun with that.
The electron configuration filling patterns of some elements in group 6b(6) and group 1b(11) reflect the increasing stability of half-filled and completely filled sublevels.
<h2>
What is electronic configuration?</h2>
The distribution of electrons in an element's atomic orbitals is described by the element's electron configuration. Atomic subshells that contain electrons are placed in a series, and the number of electrons that each one of them holds is indicated in superscript for all atomic electron configurations. For instance, sodium's electron configuration is 1s22s22p63s1.
Almost all of the elements write their electronic configurations in the same style. When the energies of two subshells differ, an electron from the lower energy subshell occasionally goes to the higher energy subshell.
This is due to two factors:
Symmetrical distribution: As is well known, stability is a result of symmetry. Because of the symmetrical distribution of electrons, orbitals where the sub-shell is exactly half-full or totally filled are more stable.
Energy exchange: The electrons in degenerate orbitals have a parallel spin and are prone to shifting positions. The energy released during this process is simply referred to as exchange energy. The greatest number of exchanges occurs when the orbitals are half- or fully-filled. Its stability is therefore at its highest.
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Answer:
one move parallel to the direction of the movement and the other move perpendicular to towards the direction of the move of wave..transverse and longitudinal wave
Schrodinger developed a famous equation that allows the solutions for electron wave functions to be found given a specific potential. For the case of an atom, Schroginger's equation allows the determination of electron wave functions. These wave functions tell us how electrons are distributed in space around the atom.
Answer:
The new concentration will be 0.01 M.
Explanation:
To determine the new concentration we use the following formula.
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.1 M
volume (1) = 100 mL
concentration (2) = unknown
volume (2) = 100 mL + 900 mL = 1000 mL
concentration (2) = [concentration (1) × volume (1)] / volume (2)
concentration (2) = (0.1 × 100) / 1000 = 0.01 M
