Answer:
Francium (Fr)
Explanation:
From the given choices, francium will have the lowest ionization energy.
Ionization energy is the energy required to remove the most loosely held electron within an atom.
The magnitude of the ionization energy depends on the characteristics of the atom in relation to its nuclear charge, atomic radius, stability etc.
- Generally on the periodic table, ionization energy increases from left to right on the table
- As you go from metals to non-metals and to gases, the value of the ionization energy increases steadily.
- Down the group, the value reduces.
- Since Francium is the most metallic of all the given choices, it has the highest ionization energy.
Answer is
C. 2.07 M
For explanation
M1V1 = M2V2
M2 = (M1V1)/V2
M2 = (1.5M x 345ml) / 250 ml
:. M2 = 2.07 M
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
