Chemistry, 09.02.2021 04:00 singlegirlforlife541

!!! would anyone be willing to double check this :00 If 5.0 moles of O2 and 3.0 moles of N2 are placed in a 30.0 L tank at a temperature of 25° C, what will the pressure of the resulting mixture of gases be?
25° C = 298°K
P (O2) = 5.0atm
P (N2) = 3.0atm
Ptotal = 6.5atm

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Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.

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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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the volume of 0.220 m k2s solution required to completely react with 175 ml of 0.115m co(no3)2 calculate the moles of co(no3)2 = molarity x volume/1000175 x0.115/1000= 0.02 moleswrite the reacting equationk2s(aq) + co(no3)2 = 2 kno3(aq) + cos(s)by use of mole ratio between k2s to co(no3)2 which is 1: 1 the moles of k2s is therefore = 175 x0.115/1000= 0.02 molesvolume = moles/molarity=0.02/0.22=0.09 ml of k2s

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!!! would anyone be willing to double check this :00 If 5.0 moles of O2 and 3.0 moles of N2 are pla...

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