The value of the equilibrium constant for a reaction

Ответ:

jaquezdaking4919

31.01.2022

The answer is "33.95 $\bold{ \frac{kj}{mol}}\\$ ".

Explanation:

Formula:

$\bigtriangleup G_0= -R \times T l_n \times K_{eq}\\$

where

$K_{eq} = \text{equilibrium constant}\\$

Given value:

$T =35^{\circ} C\\$

convert temperature celsius (°C) to Kelvin (K):

$= (273+45) \ kelvin \\\\= 318 \ Kelvin \\\\= 318 \ K$

$R = 8.314 \ \ \frac{J}{ mol \cdot K}$

$\bigtriangleup G_0= -(8.314 ) \times 31.8 \times l_n (2.65\times 10^{-6})\\$

$=-(8.314 ) \times 31.8T l_n \times (2.65\times 10^{-6})\\\\$

After solving the value it will give:

= 33.95 $\bold{ \frac{kj}{mol}}\\$

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starfox5454

28.01.2021

a. 6.57 cm

b. 8.17 cm

c. 0.80

Explanation:

a. The compound moved from where it was first placed, to a spot 7.57 cm from the bottom of the paper, in other words:

7.57 cm - 1.00 cm = 6.57 cm

b. A similar method is made for the solvent:

9.17 cm - 1.00 cm = 8.17 cm

c. The Rf of the factor is equal to the distance that the compound moved divided by the distance the solvent moved:

6.57 cm / 8.17 cm = 0.80

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