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)
Answer:
It is a combination reaction; nitrogen and hydrogen combine to form ammonia.
Explanation:
Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be
FRANCIUM, which is at the base of Group One.
Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.
To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.
The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.
Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .
Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."
To know more about electron
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A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
E) 3048 mm
