Chemistry, 30.10.2019 10:31 rachampt76

How many moles of sodium hydroxide are needed to make 250 ml of a 3.4 m naoh solution?

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i'm pretty sure it is letter c.) car

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most historians agree that gabriel mouton, the vicar of st. paul’s church in lyons, france, is the “founding father” of the metric system. he proposed a decimal system of measurement in 1670. mouton based it on the length of one minute of arc of a great circle of the earth (now called a nautical mile, 1852 meters). he also proposed the swing-length of a pendulum with a frequency of one beat per second as the unit of length (about 25 cm). a pendulum beating with this length would have been fairly easy to produce, thus facilitating the widespread distribution of uniform standards. over the years, his work was revised, improved, and extended by a number of french scientists.

the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).

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Answers are 1,2,4 its correct on edgenuity.

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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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