Answer:
Kc = [Pb²⁺]³.[PO₄³⁻]²
Explanation:
Let's consider the following reaction at equilibrium.
Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)
The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [Pb²⁺]³.[PO₄³⁻]²
This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.
Answer:
1 .
2.
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which
behaves as cation and
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation
is greater than that of
and also the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
Explanation:
Physical property
1- Water is colorless and odorless in its pure form
A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into
, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.
An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as
* orbitals.
Therefore, molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
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Hence, the correct answer will be option (b)
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