Answer:
Explanation:
H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.
Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:
- H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
- HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)
Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.
After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.
You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):
By the stequiometry you can build this table:
H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)
Initial A 0 0
Change - x +2x +x
Equilibrium A - x 2x x
As explained, A - x is very low, and 2x is twice x. Thus,
The rank of the concentrations from highest to lowest is:
The normal atomic orbitals are joined mathematically during the process of hybridization to create new atomic orbitals known as hybrid orbitals. Even if hybrid orbitals are not identical to regular atomic orbitals.
<h3>What are atomic orbitals?</h3>
Atomic theory & quantum mechanics use the mathematical concept of a "atomic orbital" to describe the location and wavelike behavior of an electron within an atom. Each of those orbitals can contain a maximum of electron pairs, each with a unique spin quantum number s.
<h3>How are atomic orbitals calculated?</h3>
Within every of an atom's shells, various orbital combinations can be found. The n=1 shell has just s orbitals; the n=2 shell contains s and p orbitals; the n=3 shell contains s, p, and d orbitals; and the n=4 up shells include all four types of orbitals.
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answer is B
because both sides have equal numbers of atomos
