Chemistry, 15.10.2019 01:30 Justinoreilly71

The reason iodine, i2, exists as a solid at room temperature, while bromine, br2, exists as a liquid at the same temperature, is because:

iodine exhibits hydrogen bonding, while bromine exhibits only london dispersion forces.

iodine molecules create stronger london dispersion forces than do bromine molecules.

bromine exists only as a gas at all temperatures.

bromine forms fewer dipole-dipole interactions than does iodine.

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Answer from: Anshuman2002

Option (b) is the correct answer.

Explanation:

London dispersion forces are defined as weak intermolecular forces due to temporary opposite charges on the combining atoms.

London dispersion forces depend on the size of an atom. More is the size of atoms more will be the London dispersion forces present within a molecule. Hence, substance will exist as a solid.

Since, bromine is smaller in size than iodine so, bromine molecules will have weak London dispersion forces between them.

On the other hand, iodine being larger in size so, its molecules will have strong London dispersion forces.

Therefore, we can conclude that the reason iodine, I_2 , exists as a solid at room temperature, while bromine, Br_2 , exists as a liquid at the same temperature, is because: Iodine molecules create stronger London dispersion forces than do bromine molecules.

Answer from: Quest

i know insulin is primary but i dont know the others im sorry

explanation:

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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The reason iodine, i2, exists as a solid at room temperature, while bromine, br2, exists as a liquid...

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