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3 years ago

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The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re

spectively. At what temperature will the diffusion coefficient have a value of 2.7 x 10-14 m2/s

1 answer:

Kitty [74]3 years ago

8 0

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

where,

D = diffusion coefficient = $2.7\times 10^{-14}m^2/s$

$D_o$ = pre-exponential constant = $1.7\times 10^{-5}m^2/s$

$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

$\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K$

Hence, the temperature of the system will be 1622 K

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En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex

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1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

1) <em>¿Cuál es la masa de agua a temperatura ambiente?</em>

Podemos determinar la masa inicial del agua ( $m_{o}$ ), medido en gramos, al conocer su densidad ( $\rho_{w}$ ), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado ( $V_{o}$ ), medido en mililitros, a partir de la siguiente expresión:

$m_{o} =\rho_{w}\cdot V_{o}$

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$m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)$

$m_{o} = 500\,g$

La masa del agua a temperatura ambiente es de 500 gramos.

2) <em>¿Cuál es la masa de agua cuando se congela?</em>

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

3) <em>¿Cuál es la masa de agua que queda después de la evaporación?</em>

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$m_{f} =\rho_{w}\cdot V_{f}$

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$m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)$

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Determinamos que la masa evaporada de agua ( $m_{v}$ ), medida en gramos, es igual a la diferencia entre las masas inicial y final, ambas medidas en gramos:

$m_{v} =m_{o}-m_{f}$

Si $m_{o} = 500\,g$ y $m_{f} = 400\,g$ , entonces tenemos que:

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$m_{v} = 100\,g$

Se ha evaporado 100 gramos de agua.

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