<span>Sm
For an element to have a partially filled f orbital, it will have to have an f orbital in the first place, this cancels barium, as it is the lightest of the elements listed:
Barium does not have an f orbital:
[Xe]6s^2
or
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2
Sm:
[Xe] 4f6 6s2
Does have an f orbital AND they are partially filled (the F subshell has the potential to hold 14 electrons, but Sm only holds 6 electrons on its F subshell, therefore the electrons, by the rule of maximum multiplicity, in which the electrons will try to occupy orbitals by themselves first (the F subshell has 7 orbitals because 14/2 = 7), it leaves the f subshell with partially filled orbitals.
Os:
Xe 4f14 5d6 6s2
all occupied f orbitals
Bi:
Xe 4f14 5d10 6s2 6p3
Has full F orbitals </span>
According to the principle, electrons fill orbitals starting at the lowest available energy states before filling higher states (e.g., 1s before 2s). The Madelung energy ordering rule: Order in which orbitals are arranged by increasing energy according to the Madelung Rule.
hi
I hope u understand
Answer:
it's subduction
Explanation:
i know this because I just do lol
The balanced equation for the neutralisation reaction is as follows
Ca(OH)₂ + H₂SO₄ ---> CaSO₄ + 2H₂O
stoichiometry of Ca(OH)₂ to H₂SO₄ is 1:1
equivalent number of acid reacts with base
number of H₂SO₄ mol reacting - 2 mol
according to molar ratio of 1:1
number of Ca(OH)₂ mol = number of H₂SO₄ moles
therefore number of Ca(OH)₂ moles required - 2 mol
Answer:
Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.
Step 2: Determine the number of moles of each type of atom present. ...
Step 3: Divide the number of moles of each element by the smallest number of moles. ...
Step 4: Convert numbers to whole numbers.
Explanation:
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