Chemistry, 17.07.2019 04:30 thegreentnt5025

You are experimenting on the effect of temperature on the rate of reaction between hydrochloric acid (hcl) and potassium iodide (ki). when the reaction is completed at 273k there are approximately 250,000 collisions per mole of reactant. you run the experiment again at 350k. which of the following would you expect to be the number of collisions recorded at 350k a. 2000 b. 700,000 c. 0 d. 200,000

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Answer from: oliviaprejean18

The correct answer is Option B.

Explanation:

Collision is defines as the interaction of two bodies in motion. It depends on the kinetic energy of the particles. Number of collisions increases as kinetic energy of the particles increases.

Kinetic energy is the measure of temperature of the system. It is given by the equation:

$E=\frac{3}{2}kT$

where,

E = kinetic energy of the particles

k = Boltzmann constant

T = temperature

We are given:

Number of collisions at 273 K = 250,000

We need to find the number of collisions at 350 K. As, the number of collisions increases as we increase the temperature. So, from the given options, the collisions at 350 K will be 700,000.

Hence, the correct answer is Option B.

Answer from: Quest

Dsolubility let me know if you have any other questions for chemistry

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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