Chemistry, 14.02.2021 21:00 dangelovasquez32

HELP I NEED TO SEND THEM TO MY TEACHER ITS REACTIONS AND SOME OF THEM MIGHT CANT HAPPEN

HELPPPP I NEED TO SEND THEM TO MY TEACHER ITS REACTIONS AND SOME OF THEM MIGHT CANT HAPPEN

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Answer from: genyjoannerubiera

Explanation:

follow the rule of "a weak acid can't make a stronger acid and weak base can't make a stronger base(it's pretty accurate for acid/salt reactions/replacemnts)" the last reactions won't work, but the second to last one will be kbr+pb(no3)2-> kno3 + PbBr2 (this is unballanced.) the first reaction will work. the second one i'm not too sure tbh, but i think it won't work since barium hydroxide's a weak base and if it were to react it would make koh, a super strong base

Answer from: skaterwolf1317

3Ba(OH)2 + 2K3PO4 > 6KOH + Ba3(PO4)2

2KBr + Pb(NO3)2 > 2KNO3 + PbBr2

H3PO4 + 3KCl > K3PO4 + 3HCl

Explanation:

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.

Answer from: Quest

The answer is neonhope it

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