Chemistry, 15.04.2021 15:20, joselynflores574

6. What is the number of moles of KF in a 29 gram sample? a. 59.0 mols
C. 2.0 mols
b. 470 mols
d. 1684.9 mols

Answers: 3

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answered: Guest

pv=nrt

explanation:

answered: Guest

The net force is 100

answered: Guest

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.

answered: Guest

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