Chemistry, 28.07.2020 18:01 michelemosbahiow6yya

A. Determination of initial concentrations 32.94 Volume of 0.25 M NaOH added (mL) B. Determination of the Blank 5.91 Volume of 0.25 M NaOH added (mL) C. Determination of Final Concentrations Volume of 0.25 M NaOH added (ml) 14.994 (10pts) Calculations and Analysis Use the data collected above and the given volumes of acetic acid and ethyl alcohol to determine the initial and equilibrium concentrations of acetic acid. A. Determination of Initial Concentrations Volume of 0.25 M NaOH added 32.94 mL (2pts) Initial concentration of acetic acid (M)
B. Determination of the Blank 03 MacBook Air 20 999 $ A 2 # 3 4 % 5 6 & 7 8 9 0 W E R Т. Y U 0
B. Determination of the Blank 5.91 ml Volume of 0.25 M NaOH added
C. Determination of Final Concentrations Volume of 0.25 M NaOH added 14.99 mL (4pts) Volume of 0.25 M required to neutralize acetic acid (Apts) Final concentration of acetic acid 8pts)
D. Calculation of the Equilibrium Constant Enter the equilibrium concentrations for all species and calculate the Key for the reaction. (1 pts) Equilibrium concentration of acetic acid (from above) (1pts) Equilibrium concentration of ethyl alcohol ?

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answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

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Okay well gravitational energy is matter forced by gravity making energy though the kinetic energy then the electric energy produces electricity from the kinetic energy and the kinetic energy gets its energy from the gravitational energy. was that the answer you wanted or something else?

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A. it generates energy from carbohydrates.b. it requires activation by oxygen.c. it is used by cells to dispose of excess electrons

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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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