Chemistry, 21.10.2020 16:01 kalieghcook

Emma wants to go swimming the first day the pool opens in May, but she is worried it will be too cold since summer only just started. She only likes to swim in the water if the pool is at least 82°F. The thermometer in the pool reads 24°C. Is the pool warm enough for Emma to swim in? Show all work to support your answer.

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Answer from: Abrahamolve

The water is not warm enough for her to swim in.

She loves to swim in water at temperature = 82°F.Thermometer in water reads = 24°C.

We have to convert Emma's preferred temperature to °C

°C= 5/9 x (F - 32)

°C=5/9 (82-32)

°C = 27.8

Therefore the temperature is 27.8°C

This means that the temperature of the water is not warm enough for her

to swim in.

Read more about Temperature here link

Answer from: jholland18

the water is not warm enough for her

Explanation:

Given that

Her preferred temperature = 82°F

Thermometer in water reads = 24°C

Convert the preferred temperature to °C

°C=5/9 (82-32)

°C = 27.8 °C

Hence the water is not warm enough for her

Answer from: Quest

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence, the electronegativity (or the ability to attract electrons) is lessened."

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.