Chemistry, 26.03.2021 14:00 zeldawhite76

What happens during beta decay? A photon is ejected from a nucleus.
A helium-4 nucleus is ejected from a nucleus.
A neutron changes into a proton, or a proton changes into a neutron.
A proton changes into an electron, or an electron changes into a proton.

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Answer from: gabflips7267

A neutron changes into a proton or a proton changes into a neutron

Explanation:

Here, we want to know what happens during a beta decay

In a beta decay, we have a conversion in the nucleus of an atom

Generally, there must be an excess of either the proton or a neutron in the nucleus of a radioactive atom

What we are saying here is that both do not have same values

Hence, what we have is that during a beta decay, there is a transformation of a proton to a neutron or a neutron to a proton, depending on which of the two are in excess

What this mean is that the one that changes to the other is largely dependent on which of the two is more than the other

That brings us to the principle of having a beta positive decay or a beta negative decay

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

The higher the temperature, the increased rate constant for a reaction.(positive linear association)