<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>
Answer:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Explanation:
Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
I'm assuming you need to know the percentage yield of the reaction
To calculate the percentage yield = (actual yield x 100%) / predicted yield
actual yield is 56,9 g
predicted yield is 36,6g ( is the amount that's expected if nothing had got lost)
(56,9 x100)/36,6=
= 155%