You can't usually just use a single spectrum line to confirm the identity of an element because there are cases that the emission line id not clearly defined. When the emission line is very weak compared to surrounding noise, in which case the more datapoints you have to build up confidence for the existence of a particular emission spectra, the better.
The aim is to use less space while demonstrating the distribution of electrons in shells
If you want to depict how an atom's electrons are scattered across its subshells, an orbital notation is more suited.
This is due to the fact that some atoms have unique electronic configurations that are not readily apparent from textual configurations.
<h3>How does electron configuration work?</h3>
The placement of electrons in orbitals surrounding an atomic nucleus is known as electronic configuration, also known as electronic structure or electron configuration.
<h3>What sort of electron arrangement would that look like?</h3>
- For instance: You can see that oxygen contains 8 electrons on the periodic table.
- These 8 electrons would fill in the following order: 1s, 2s, and finally 2p, according to the aforementioned fill order. O 1s22s22p4 would be oxygen's electron configuration.
learn more about electronic configuration here
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Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5
Q1)
We have been given the OH⁻ concentration, therefore we first need to find the pOH value and then the pH value.
pOH = -log [OH⁻]
pOH = -log (0.225 M)
pOH = 0.65
pH + pOH = 14
pH = 14 - 0.65 = 13.35
Q2)
pOH = -log[OH⁻]
pOH = -log (0.0015 M)
pOH = 2.82
pH + pOH = 14
pH = 14 - 2.82
pH = 11.18
