The explanation I got for this was not correct. Please help with this problem. Units must be SI units.

Thank you in advance.

A sphere of naphthalene of 6cm diameter is sublimated in stagnant air. Derive the equation AND calculate the half-life time of this sphere (the time it takes for the sphere to have a diameter of 3cm ). Additional information: Air is at 113F and 1atm. The molecular weight and density of naphthalene are 128g/mol and 71.1lb/ft3, respectively. The vapor pressure of naphthalene at these conditions is 0.8mmHg. Diffusion of naphthalene in air at 45C is 6.92106m2/s. You will derive the equation starting from the two equations derived in class: dr=NAMWdt(1)NAr2=r0DABCln(1yAsat1)(2) You have to replace (2) in (1) and solve the integral with limits t=0 to t=t1/2 and r=r0 to r= ro/2 This is the final solution: t21=(7ro2)(24DABCMWln(1yAsat1))