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.
Phosphorus + Sulfur ------> Phosphorus sulfide
2P + 3S ------> P2S3
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Answer:
400 mL
Explanation:
Given data:
Mass of barium = 2.17 g
Pressure = 748 mmHg (748/760 = 0.98 atm)
Temperature = 21 °C ( 273+ 21 = 294k)
Milliliters of H₂ evolved = ?
Solution:
chemical equation:
Ba + 2H₂O → Ba(OH)₂ + H₂
Number of moles of barium:
Number of moles = mass/ molar mass
Number of moles = 2.17 g / 137.327 g/mol
Number of moles = 0.016 mol
Now we will compare the moles of barium with H₂.
Ba : H₂
1 : 1
0.016 : 0.016
Milliliters of H₂:
PV = nRT
V = nRT/P
V = 0.016 mol × 0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm
V = 0.39 atm. L/0.98 atm
V = 0.4 L
L to mL
0.4 × 1000 = 400 mL
Atomic mass of Sulfur = 32g
32g of Sulfur is one mole.
1g of Sulfur is 
96.21g of Sulfur is 
