Different radioactive nuclides decay into their respective daughter nuclides at distinct rates. Some of the nuclides decay briskly, while others decay gradually. The time it consumes for half of the parent nuclide in a radioactive sample to decay to the daughter nuclides is known as the half-life of the radioactive sample.
The nuclides, which decay briskly exhibit short half-lives and are very active. The half-life can be utilized to find the rates of radioactive decay. In the given question, the half-lives of various nuclides are given. So, the order to the most active (shortest half-life or largest number of decays per second) to least reactive (largest half-life or the smallest number of decays per second) is:
Tc-99m > Y-90 > In-111 > I-131
Answer:
K = [H2] [CO] / [HCHO]
Explanation:
HCHO(g) ⇌ H2(g) + CO(g)
We can obtain the expression for the equilibrium constant for the above equation as follow:
Equilibrium constant, K for a given reaction is the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.
Thus, the equilibrium constant, K for the above equation can be written as follow:
K = [H2] [CO] / [HCHO]
Human use of land has negative impacts. Human activities contribute to the erosion and pollution of beaches. Deforestation of land can also lead to desertification and a loss of biodiversity.
Those elements with similar properties are in the same column.
The correct answer is:
a positron is emitted when proton converts to a neutron.
The reaction can be described as following:
₁¹p (proton) → ₀¹n (neutron) + ₁°e (positron or ₁⁰β)
Positron is an antiparticle of a β particle (₋₁°β), which means it has an oposite charge to it, but same mass.
