What is the equilibrium expression for the reaction below? 2N₂(g) + O₂(g) 2N₂O(g)
1 answer:
Equilibrium expression =. Hence, option D is correct.
Equilibrium is generally defined as a state of rest, where there is no change.
The ratio of the concentrations of the reactants and products is known as the equilibrium constant expression.
The equilibrium expression for the reaction is:
Hence, option D is correct.
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This hypothetical process would produce actinium-230.
<h3>Explanation</h3>An alpha decay reduces the atomic number of a nucleus by two and its mass number by four.
There are two types of beta decay: beta minus β⁻ and beta plus β⁺.
The mass number of a nucleus <em>stays the same</em> in either process. In β⁻ decay, the atomic number <em>increases </em>by one. An electron e⁻ is produced. In β⁺ decay, the atomic number <em>decreases </em>by one. A positron e⁺ is produced. Positrons are antiparticles of electrons.
β⁻ are more common than β⁺ in decays involving uranium. Assuming that the "beta decay" here refers to β⁻ decay.
Gamma decays do not influence the atomic or mass number of a nucleus.
Uranium has an atomic number of 92. 238 is the mass number of this particular isotope. The hypothetical product would have an atomic number of 92 - 2 ⨯ 2 + 1 = 89. Actinium has atomic number 89. As a result, the product is an isotope of actinium. The mass number of this hypothetical isotope would be 238 - 2 ⨯ 4 = 230. Therefore, actinium-230 is produced.
The overall nuclear reaction would involve five different particles. On the reactant side, there is
- one uranium-238 atom.
On the product side, there are
- one actinium-230 atom,
- two alpha particles (a.k.a. helium-4 nuclei),
- one electron, and
- one gamma particle (a.k.a. photon).
Consider: what would be the products if the nucleus undergoes a β⁺ decay instead?