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3 years ago

Will NaBr, CO2, Ca3(PO4), or KZS raise a liquid's boiling point the most when

Chemistry

1 answer:

Vladimir79 [104]3 years ago

6 0

Answer:

$K_2S$

Explanation:

According to the boiling point elevation law described by the equation $\Delta T_b = iK_bb$ , the increase in boiling point is directly proportional to the van 't Hoff factor.

The van 't Hoff factor for nonelectrolytes is 1, while for ionic substances, it is equal to the number of moles of ions produced when 1 mole of salt dissolves.

$NaBr$ would produce 2 moles of ions per 1 mole of dissolved substance, sodium and bromide ions.

$Ca_3(PO_4)_2$ is insoluble in water, so it would barely dissociate and wouldn't practically change the boiling point.

$K_2S$ would dissociate into 3 moles of ions per 1 mole of substance, two potassium cations and one sulfide anion.

$CO_2$ is a gas, it would form some amount of carbonic acid when dissolved, however, carbonic acid is molecular and would yield i value of i = 1.

Therefore, potassium sulfide would raise a liquid's boiling point the most if all concentrations are equal.

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Answer:

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3 years ago

Propiedades químicas del óxido

tangare [24]

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Óxidos neutros: Son compuestos por no metales. No reaccionan con agua, ácido o base, en razón del enlace covalente que une sus componentes; de ahí el por qué de ser llamados óxidos inertes. Ejemplos: monóxido de dinitrógeno (N2O) y monóxido de carbono (CO).

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Óxidos dobles o mixtos: La combinación de dos óxidos de un mismo elemento, da origen a este tipo de óxidos. Ejemplo: magnetita (Fe2O4), unión de los óxidos de hierro (Fe) y oxígeno (O).

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3 years ago

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liberstina [14]

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2 years ago

PLEASE.

ivolga24 [154]

Answer:

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3 years ago

Of the following solutions, which has the greatest buffering capacity?

Goshia [24]

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[Base]}{[Acid]} )$

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. $\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36$

b. $\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417$

c. $\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868$

d. $\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959$

Therefore, the d. solution has the best buffering capacity.

Regards.

4 0

3 years ago