Question

Calculate the kinetic energy of 5 moles of nitrogen at 27 cen

Answer 1

vhjn NBC gbhehd UFC ehsheesgevekegev ed.

Answer 2

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\purple{R}}}}$

• ❤..$\red{\tt{Kinetic}}$$\orange{\tt{Energy}}$=$\frac{3}{2}$
• ❤.Where n=5 moles;R =8.314J${mol^-1k^-1}$ T= ${27\degreeC}$+273=300K
• KINETIC ENERGY
• ${Ek}$=$\frac{3}{2}$×5mol×8.314J${mol^-1k^-1}$=300k=18706.50J.