Atoms (Sodium and Fluorine)
Answer: K only has 1 valence electron. It will leave with only a little effort, leaving behind a positively charged K^+1 atom.
Explanation: A neutral potassium atom has 19 total electrons. But only 1 of them is in potassium's valence shell. Valence shell means the outermost s and p orbitals. Potasium's electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1. The 4s orbital is the only orbital in the 4th energy level. So it has a valency of 1. This means this electron will be the most likely to leave, since it is the lone electron in the oyutermost energy level (4). When that electron leaves, the charge on the atom go up by 1. The atom now has a full valence shell of 3s^2 3p^6, the same as argon, Ar.
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
