Answer:
30.17 × 10²³ atoms
Explanation:
Given data:
Number of moles of lead = 5.01 mol
Number of atoms = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
In given question:
1 mole = 6.022 × 10²³ atoms
5.01 mol × 6.022 × 10²³ atoms / 1 mol
30.17 × 10²³ atoms
Answer:
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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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D:
When electrons are gained, the charge of the atom decreases.
When you are given an atom with a charge, the oxidation of that atom is the charge. So by going from a Cr^3+ (Oxidation Number = 3) to a Cr^2+ (Oxidation Number = 2), the Oxidation Number thus decreases.
