Compound a, with concentration ca, diffuses through a 4-cm long tube and reacts as it diffuses. the equation governing diffusion with reaction is given by d d 2ca dz2 − kca = 0 (7) at one end of the tube, there is a large source of a at a concentration of ca=0.1m. at the other end of the tube there is an adsorbent material that quickly absorbs any a, making the concentration 0 m. if d =1.5×10−6 cm2/s and k =5×10−6 s −1 , what is the concentration of a as a function of distance in the tube

Step-by-step explanation:

First you need to rearrange the original equation in order to have an equation with a general form:

We divide everything by D, so we have:

We know that k and D are constants, so we can leave them as just one constant that gives as result:

Then we proceed to change terms in the equation son it looks friendlier:

Knowing that

Then we give the general solution for the differential equation of second order:

The following is knowing the frontier values so we can calculate the constants of the equation:

With these values, we can replace in the general solution to find the final equation:

And with the other values:

Then we solve the equation system, as we have two unknowns, the constants, and two equations:

After solving both equations, we have;

The we see that we can cancel terms, so we have the final solution that is:

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

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