Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
The group that magnesium is in has all the elements that are alike to it.
Magnesium is in group 2 which is the alkaline earth metals. So elements similar to it would be calcium, beryllium!
----If you need anymore help in chemistry.....just ask me....im very good at chemistry!
Barium has a 2+ charge as it is in group 2 in the periodic table and so it has two electrons in its outer shell and chloride has a -1 charge on its chloride ion. So we will need two of the chloride ions as we have a 2+ charge to match the amount of charge on one barium ion- forming barium ion
BaCI2
Answer:
The required volume is 1.6 x 10³mL.
Explanation:
When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:
C₁ . V₁ = C₂ . V₂
where,
C₁ and V₁ are the concentration and volume of the concentrated solution
C₂ and V₂ are the concentration and volume of the dilute solution
In this case, we want to find out V₁:
C₁ . V₁ = C₂ . V₂
Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>
Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :
a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol
