Na = 23 x 2.40 = 55.2
O = 16 x 2.40 = 38.4
H = 1 x 2.40 = 2.40
55.2 + 38.4 + 2.4 = 96
2.40 mol of NaOH = 96 amu
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.
To know more about electronic configuration, go to URL
brainly.com/question/26084288
#SPJ4
<span>A 50 kg student runs up a flight of stairs that is 6 m high. How much work is done? 3000 J. Calculate the work done when a force of 1 N moves a book 2 m. 2 J .... 200 W, 100 W. Calculate the power expended when a 500. N barbell is lifted 2.2 m in 2.0 seconds. 550 W. energy. the property of an object or system that ...</span>
The micromoles of mercury(II) iodide : 0.013 μ moles
<h3>Further explanation</h3>
Given
215.0mL of a 6.0x10⁻⁵mmol/L HgI₂
Required
micromoles of HgI₂
Solution
Molarity(M) = moles of solute per liters of solution
Can be formulated :
M = n : V
n = moles
V = volume of solution
V = 215 mL = 0.215 L
so moles of solution :
n = M x V
n = 6.10 mmol/L x 0.215 L
n = 1.312 . 10⁻⁵ mmol
mmol = 10³ micromol
so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles
+2
Barium has a positive charge of 2
