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lesya [120]
2 years ago
10

The electron configuration filling patterns of some elements in group 6b(6) and group 1b(11) reflect the ______ of half-filled a

nd completely filled sublevels.
Chemistry
1 answer:
Fudgin [204]2 years ago
8 0

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

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22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp​
IceJOKER [234]

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

<em>2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.</em>

<em />

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

<em>Moles NH3 -Molar mass: -17.01g/mol-</em>

Using PV = nRT

PV/RT = n

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = 22.4L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, <em>the limiting reactant is NH3</em>

<em />

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

4 0
3 years ago
Determine physiological temperature, 98.6 F in degree C
Salsk061 [2.6K]

Answer:

37

Explanation:

( 98.6 - 32 ) × 5(100c) ÷ 9(180f) = 37

4 0
2 years ago
If element R has 2 valence electron and element Q has 6 valence electrons, what is the likely compound that will form:
kvv77 [185]

Explanation:

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4 0
2 years ago
What characteristic frequencies in the infrared spectrum of your sodium borohydride reduction product will you look for to deter
kkurt [141]

Answer:

A)The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

B)Characteristic frequency  present in the infrared spectrum will be at a peak of  3300-3400 cm-1 which will be due to O-H stretch.

C)If the product is wet with water there will be no change in the infrared spectrum

Explanation:

The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

Characteristic frequency  present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.

If the product is wet with water there will be no change in the infrared spectrum

5 0
3 years ago
You are given 100 g of a compound. The compound is composed of 37% hydrogen and 63% oxygen. How many grams oxygen are present in
USPshnik [31]

mass_{oxygen} = \%mass_{oxygen} \times mass_{compound} \\ m = 63\%(100) \\ m = 0.63(100) \\ m = 63 \: grams

<h2>Option d</h2>
7 0
1 year ago
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