Chemistry, 30.06.2019 19:20 ladawnrowles005

An ion is an atom with a net electrical charge due to

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Answer from: gingerham1

the loss of one or more electrons & the addition of one of more electrons

Answer from: shenothomas688

Explanation:

Loss of one or more electrons by a neutral atom will help the neutral atom to acquire a positive charge.

For example, atomic number of sodium is 11 and to stabilize itself it needs to lose one electron. So, by losing an electron it will change into $Na^{+}$ ion.

On the other hand, gain of one or more electrons by a neutral atom will help the neutral atom to acquire a negative charge.

For example, Cl has atomic number 17 and to fill its octet completely it need one more electron. So, by gaining an electron is will change into $Cl^{-}$ ion.

Therefore, we can conclude that an ion is an atom with a net electrical charge due to loss or gain of electrons.

Answer from: Quest

the answer is d : ) have a good da

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.

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