Chemistry, 02.02.2020 23:43, Isabelcastillox1085

Which best describes accuracy?
the smallness of the graduations on a measuring tool
the reproducibility of a measured value
the dependence of a measured value on a controlled value
the agreement between a measured value and an accepted value

Answers: 2

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answered: jesser9

I would go with b: the reproducible of a measured value. but im not 100%

answered: jaydenjohnson6310

the agreement between a measured value and an accepted value

Explanation:

Accuracy measures how close a measured value is to a standard value. For example, if the weight of a bag of rice is 1 kg and the measured value is 0.90 kg rather than 0.75 kg, then the former measure value is more accurate than the later measured value.

Precision on the other hand is how close are the measured values to one another. For example, if that same 1 kg rice bag was measured 3 times and the readings were 0.75 kg, 0.73 kg and 0.74 kg then the measured values are precise but inaccurate.

answered: Guest

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.

answered: Guest

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.