T_(2)=273+350=623K D_(2)=(7.8times 10^-11m^2/s)exp[(-41,500Jmml)/(8.314Jmolcdot K)((1)/(623K)-(1)/(573K))]

The is:<br /><br />$D_{2}=(7.8\times 10^{-11}m^{2}/s)exp[\frac {-41,500Jmml}{8.314Jmol\cdot K}(\frac {1}{623K}-\frac {1}{573K})]$<br /><br />This equation represents the diffusion coefficient ($D_{2}$) for a substance at a temperature of 623 K. The equation is based on the Arrhenius equation for diffusion, which relates the diffusion coefficient to the temperature and the activation energy for diffusion.<br /><br />The first part of the equation, $T_{2}=273+350=623K$, calculates the temperature in Kelvin. The temperature is an important factor in determining the diffusion coefficient, as it affects the kinetic energy of the particles and their ability to move and diffuse.<br /><br />The second part of the equation calculates the diffusion coefficient using the Arrhenius equation. The activation energy for diffusion is given as $41,500 J/mol$, and the gas constant is $8.314 J/mol \cdot K$. The equation takes into account the difference in the reciprocal temperatures between 623 K and 573 K.<br /><br />Therefore, the correct answer is the second equation, which calculates the diffusion coefficient at a temperature of 623 K.