Chemistry, 29.01.2021 16:40 cjsboardinghouse

Complete the following sentences to identify the process that ice, water, or water vapor may undergo if either the temperature or the pressure is increased. Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
FREEZE 1. If ice is heated at a constant pressure of 0.00512 atm, it will.
CONDENSE 2. If ice is heated at a constant pressure of 1 atm, it will.
MELT 3. If the pressure of water vapor is increased at a constant pressure of 100 degrees Celsius, it will.
SUBLIME 4. If the pressure of water vapor is increased at a constant pressure of -50 degrees Celsius, it will__.
VAPORIZE
DEPOSIT

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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

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Answer from: Quest

The answer would be true

Answer from: Quest

answer : the correct answer is 2) hno₃ and naoh .

arrhenius concept of acid and base :

acid :

according to arrhenius acid are those which release h⁺ or hydronium ion (h₃o⁺) in aqueous solution .

example : hcl +h₂o ↔ h₃o⁺ + cl⁻

hcl is giving h₃o⁺ ion in water , hence it is an arrhenius acid .

base :

according to arrhenius , bases are those which releases oh⁻ ions in aqueous solution .

example : koh (aq) ↔ k⁺ + oh⁻

since koh is producing oh⁻ ions hence , it is an arrhenius base .

checking the options :

1) ch₃oh and naoh :

ch₃oh is known as methanol , an organic compound , which is neutral compound . the dissociation of ch₃oh is negligible . it is very very weak acid (ka = 3.2 * 10 ⁻¹² ) .

naoh being ionic gives na⁺ and oh⁻ ions in aqueous solution as :

naoh ↔ na⁺ + oh⁻ . hence it is an arrhenius base

it can be concluded that ch₃oh and naoh is not arrhenius acid and base.

2) hno₃ and naoh :

hno₃ is an ionic compound which gives ions h⁺ and no₃⁻ ion in aqueous solution as:

hno₃ + h₂o ↔ h₃o⁺ + no₃⁻

since it is giving h⁺ ion . hno₃ is arrhenius acid .

naoh being ionic gives na⁺ and oh⁻ ions in aqueous solution as :

naoh ↔ na⁺ + oh⁻

. hence it is an arrhenius base .

it can be said that hno₃ and naoh are arrhenius acid and base.

3) ch₃oh and hcl