Chemistry, 30.09.2019 12:30 g0606997

Select the correct answer. which of the following tests could be safely used to distinguish a strong base from a weak base? a. taste b. touch c. litmus paper d. electrical conductivity

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Answer from: kbruner20

Option (D) is the correct answer.

Explanation:

Bases are the substances that release hydroxide ions when dissolved in water.

A strong base will dissociate completely in water and give more number of hydroxide ions $(OH^{-})$ . Whereas a weak base will slightly dissolve to give hydroxide ions $(OH^{-})$ in water.

So, a strong base will have high electrical conductivity and a weak base will have low electrical conductivity.

Since, both are bases (strong base and weak base) therefore, both of them will have bitter taste, slippery texture and they will turn red litmus into blue.

Hence, we can conclude that electrical conductivity tests could be safely used to distinguish a strong base from a weak base.

Answer from: Quest

Decomposition is the cycle at which bacteria breaks down any organic material into the soil

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.