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Rama09 [41]
3 years ago
11

¿Qué tipo de Inter conversión existe en una celda galvánica o en una celda electrolítica? a) de energía química a energía eléctr

ica y viceversa b) de energía eléctrica a energía química c) de energía química a energía eléctrica d) de energía lumínica a energía eléctrica
Chemistry
1 answer:
insens350 [35]3 years ago
7 0

Answer:

Opción a)

Explanation:

En este caso, vamos a explicartelo descartando opciones. Para empezar el proceso que existe en una celda galvánica o electrolítica, es lo que uno llama un proceso de Electroquímica, y permite manipular y usar la energía electrica para generar una reacción.

En este caso, yo tengo por ejemplo una celda galvánica con dos componentes como hierro y cobre conectados mediante una celda. El proceso de reacción entre ellos es lo que ayudará a que se genere energia electrica y esto, encendería un bombillo de luz. También puede ocurrir lo contrario. Con electricidad, se genera una reacción química. En estos casos, se genera una reacción de tipo REDOX (Oxido reducción).

Tomando en cuenta esto, la respuesta correcta sería la opción a). Veamos por que las otras opciones no son:

b) Energía eléctrica a química

Esta opción es falsa, porque estaría supeditando que una reacción solo puede darse por medio de una manipulación de la energía electrica y en las celdas galvánicas no ocurre eso, sino al revés.

c) Energía química a eléctrica

Falsa, porque es igual que la anterior, solo está supeditado a que ocurra este tipo de reacciones y no es así.

d) energía lumínica a eléctrica

Falso porque la energía lumínica proviene tambien de la electricidad, y en el caso de una celda galvánica se genera una reacción por lo que existe otro tipo de energía.

Espero esto te ayude.

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The question is incomplete, here is the complete question:

How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.

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<u>Explanation:</u>

We are given:

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w = Weight of the gas = ?

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Putting values in above equation, we get:

65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g

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