Answer:
Explanation:
Hello,
In this case, given the balanced reaction:
We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:
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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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During feasting, there is an increase in the supply of carbohydrates, fats and proteins in the body. Since these macronutrients are in excess, carbohydrates and proteins are utilized or oxidized as a source for energy production. On the otherhand, fats are accumulated as the rate of oxidation of this source does not increase during feasting.
Answer:
T₂ = 51826.1 K
Explanation:
Given data:
Initial Volume = 2.3 L
Final volume = 400 L
Initial temperature = 25 °C (25+ 273 = 298 K)
Final temperature = ?
Solution:
V₁/T₁ = V₂/T₂
T₂ = V₂ T₁/V₁
T₂ = 400 L . 298 K / 2.3 L
T₂ = 119200 K. L / 2.3 L
T₂ = 51826.1 K
