Chemistry, 09.10.2019 22:30 yennie123

Calcium has atomic number of 20 how many electrons of a calcium atom are involved in chemical bonding

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Answer from: brendahndz8676

The answer is 2 electrons.

The electron configuration of calcium is 2:8:8:2

Calcium has two electrons in its outermost shell. These are its valence electrons and are the ones used in bonding with other elements. Valence electrons of an atom are those electrons that are in its outer energy shell or that are available for bonding.

Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known as an ionic compound.

For example, when calcium metal reacts with chlorine gas, calcium gives up its two valence electrons and Chlorine accepts them resulting in a new substance called calcium chloride in which the two elements have ended up forming ionic bonds.

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

Answerδh > 0explanationc₆h₆ (s) → c₆h₆ (l)if you are changing it from a solid to a liquid, you are pretty much melting the solid. this requires heat energy. therefore, the reaction is endothermic (takes in energy); the system enthalpy increases as the reaction proceeds: δh > 0.