i think it is 6 valence electrons
First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT
when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K
12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L
The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.
The alkali earth metal beryllium (
) belongs to the second group of the periodic table. The ground state electronic configuration of
is:
From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (
).
After losing all valence electrons that is 2 electrons from
orbital. The electronic configuration will be:

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as
.
Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.
States that the properties of elements are periodic or recurring and are correlated to their atomic number.