Chemistry, 12.02.2020 03:34 Naysa150724

The partial negative charge in a molecule of water occurs because . a. the electrons shared between the oxygen and hydrogen atoms spend more time around the oxygen atom nucleus than around the hydrogen atom nucleus b. the oxygen atom donates an electron to each of the hydrogen atoms c. one of the hydrogen atoms donates an electron to the oxygen atom d. the oxygen atom has two pairs of electrons in its valence shell that are not neutralized by hydrogen atoms

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a. the electrons shared between the oxygen and hydrogen atoms spend more time around the oxygen atom nucleus than around the hydrogen atom nucleus.

Explanation:

Water molecule is a polar molecule , and hence , the atom can attain the respective partial positive and negative charge on them .

The difference in the electronegativity gives rise to the partial positive and negative charge .

Since , oxygen atom is more electronegative in comparison to hydrogen atom , and hence , slightly pulls the electrons towards itself , and attains partial negative charge , and correspondingly , the hydrogen atom attains partial positive charge .

Hence , from the question ,

The correct option is a.

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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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