Answer:
Van't Hoff factor for AlCl₃ = 3 (Approx)
Explanation:
Given:
Number of observed particular = 1.79 M
Number of theoretical particular = 0.56 M
Find:
Van't Hoff factor for AlCl₃
Computation:
Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular
Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M
Van't Hoff factor for AlCl₃ = 3.19
Van't Hoff factor for AlCl₃ = 3 (Approx)
Halogen are the most reactive due to their electronic configuration
AC2
A will lose 2
C will gain 1
Need two Cs to match A
Answer:
Element Lithium
Explanation:
The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals
It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.
Its atomic number is 3. The electronic configuration is 1s2 2S1
