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3 years ago

1. Que factores provocan la oxidación?

Chemistry

1 answer:

Katyanochek1 [597]3 years ago

4 0

Answer:

Mira la explicación.

Explanation:

Hola,

Las reacciones de oxidación se definen como aquellos procesos químicos en las cuales uno de los elementos involucrados sufre un aumento en su estado de oxidación, causado por el efecto de un agente oxidante, el cual es capaz de aceptar los electrones perdidos por dicho elemento oxidante. Dentro de los factores promotores de la oxidación se pueden encontrar:

1. Presencia de oxígeno gaseoso: al ser un agente oxidante fuerte, es capaz de recibir electrones de casi cualquier sustancia.

2. Presencia de humedad: a la larga, cuando una sustancia reacciona con agua tiende oxidarse ya que el agua posee oxígeno y como se vio anteriormente, es el principal promotor de oxidación.

3. La presencia de algunas sales iónicas como cloruro de sodio también promueven la oxidación, por ello es que no debemos introducir joyería al mar, debido a que esta salinidad induce reacciones electroquímicas que son capaces de oxidar metales.

4. Reacciones con ácidos oxidantes: ácidos como el nítrico, son oxidantes fuertes que tienen la capacidad de disolver metales (oxidarlos) y formar nitratos, por ejemplo el cobre en presencia de ácido nítrico, sufre una oxidación bastante violenta.

Saludos!

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A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f

Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

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Answer: Option (b) is the correct answer.

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If particles of a soil are flat and even if they are sorted then there will rarely be any space available for fluid or water to pass through it.

If soil particles are round in shape and are sorted then there will be space available for the fluid to pass through it. Hence, this type of soil will be most porous.

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Answer:

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Explanation:

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