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Wittaler [7]
3 years ago
8

Explique por qué se necesita una temperatura alta para derretir el yoduro de potasio.​

Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
7 0

Answer:

El yoduro de potasio es una sal cristalina de fórmula KI, usada en fotografía y tratamiento por radiación. Al ser menos higroscópica que el yoduro de sodio, es más utilizada como fuente de ion yoduro.

Explanation:Se porta como una sal simple. El ion yoduro, al ser un reductor débil, es fácilmente oxidado por otros elementos como el cloro para transformarse en yodo:

2 KI(ac) + Cl2(ac) → 2 KCl + I2(ac)

Tiene un pH neutro (pH = 7) ya que el catión potasio y el anión yoduro son iones espectadores, por lo que no reaccionan con el agua, manteniéndose inalterado el pH. El yoduro se oxida aún más fácilmente al formar ácido yodhídrico (HI), el cual es un reductor más fuerte que el KI. El yoduro de potasio forma el anión triyoduro(I3−) al combinarse con yodo elemental.

A diferencia del yodo, los triyoduros son altamente solubles en agua, por lo que el yoduro de potasio aumenta considerablemente la solubilidad del yodo elemental en agua, que por sí solo se disuelve en muy bajas cantidades.

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The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the
Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and NH_3.

\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles

\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl

From the balanced reaction we conclude that

As, 1 mole of NaCl react with 1 mole of NH_3

So, 0.243 mole of NaCl react with 0.243 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and NaCl is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaHCO_3

From the reaction, we conclude that

As, 1 mole of NaCl react to give 1 mole of NaHCO_3

So, 0.243 moles of NaCl react to give 0.243 moles of NaHCO_3

Now we have to calculate the mass of NaHCO_3

\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3

Molar mass of sodium bicarbonate = 84 g/mol

\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0
3 years ago
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