Answer:
smallest to largest:
Electron, Neutron, Atom, Molecule, Nucleus
Explanation:
sorry if it's not right
Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry)

= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
The atomic number of Li is 3
Electron configuration of Li : 1s² 2s¹
The atomic number of Na is 11
Electron configuration of Na : 1s²2s²2p⁶3s¹
Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.