Covalent compounds have bonds where electrons are shared between atoms. Due to the sharing of electrons, they exhibit characteristic physical properties that include lower melting points and electrical conductivity compared to ionic compounds.
Answer:
Explanation:
A is Magnesium, B is Aluminium both are adjacent element and lie in third period.
Magnesium with atomic number of 12 consist two s electrons in it's valence shell in ground state whereas, Aluminium which has atomic number of 13 consist three electrons in it's valence shell in the ground state out of which two are s electrons and only one p electron.
We need to increase the concentration of common ion first, in order to promote the common ion effect
<h3>What is the Common ion effect?</h3>
It is an effect that suppresses the dissociation of salt due to the addition of another salt having common ions.
For example, a saturated solution of silver chloride in equilibrium has Ag⁺ and Cl⁻ . Sodium Chloride is added to the solution and has a common ion Cl⁻. As a result, the equilibrium shifts to the left to form more silver chloride. Thus, solubility of AgCl decreases.
The Equilibrium law states that if a process is in equilibrium and is subjected to a change
- in temperature,
- pressure,
- the concentration of reactant or product,
then the equilibrium shifts in a particular direction, according to the condition.
Thus, an increase in the concentration of common ion promotes the common ion effect.
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If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</h3>
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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