You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
0.340 mol Br2 ( </span>6.022 x 10^23 molecules / mol ) = 2.05 x 10^23 molecules
Answer:
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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.