The electron configuration that corresponds to the valence electrons of an element for which there is an especially large jump between the second and third ionization energies is ns^2.
The valence electron configuration of an atom refers to the arrangement of electrons on the outermost shell of the atom.
Recall that a large jump in ionization energy occurs when electrons are removed from inner shells of the atom.
If we study our options closely, we will discover that option A has only two electrons in the valence shell (ns^2).
This means that the third ionization energy involves removing electrons from an inner shell which leads to large jump.
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Explanation:
It is given that vapor pressure of liquid iodomethane is 40.0 mm Hg. So, if we calculate the vapor pressure according to the given values and if its value will be greater than the the given vapor pressure of iodomethane then it means that some of the vapors has converted into liquid state.
As the given values are as follows.
= 72.0 mm Hg, 
= 404 K
= ? , 
= 249 K
As volume is constant so, according to Gay-Lussac's law pressure is directly proportional to temperature.
(at constant volume)
or, 
= k
Therefore, the formula to calculate the value of is as follows.
= 
= 
= 44.37 mm Hg
As calculated vapor pressure is more than the given vapor pressure. Hence, the liquid will convert into gas.
As a result, no condensation will occur and only vapors of iodomethane will be present.
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N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol
n = 7,44mol
M = 17g/mol
m = n * M = 7,44mol * 17g/mol = 126,48g
Answer:
2NO(g) + O2(g) --> 2NO2(g)
now 400 ml of NO × 2 mol of NO2/2 mol of NO
= 400 ml of NO2
now 500 ml of O2 × 2 mol of NO2/1 mol of O2
= 1000 ml of NO2
now 400 ml of NO2 × 1 mol of O2/2 mol of NO
= 200 ml
subtract that from 500 ml of total i.e. 500-200 =300 ml
The total volume of the reaction mixture is 1000 ml -300ml = 700 ml
