N_H2 = %Al/100 (0.0566) + (100-%Al/100) (0.0153) In your lab notebook on a separ
ID: 976493 • Letter: N
Question
N_H2 = %Al/100 (0.0566) + (100-%Al/100) (0.0153) In your lab notebook on a separate page, construct a graph of N_H2, vs. % Al (N_H2 in the y-axis and % Al in the x axis). To do this, refer to Equation 11 and the discussion preceding it. Note that a plot of N_H2 vs. % Al should be a straight line (why?). The most obvious way to do this is to find N_H2, when % Al = 0 and when % Al = 0 and when % Al = 100. If you wish you may calculate some intermediate points (for example, N_H2, when % Al = 50, or 20, or 70); all these points should lie on the same straight line.Explanation / Answer
nH2 = %Al/100(0.0566) + (100 - %Al)/100(0.0153)
= 1/100{0.0566.%Al + 1.53 – 0.0153.%Al} =1/100(1.53 + 0.0413%Al)
or, nH2 = 0.0413.%Al/100 + 1.53/100
This equation is of the form y=mx + c where y =nH2, x = %Al; m = 0.0413/100 and c =1.53/100. The plot of nH2 vs %Al will be a straight line since nH2 varies linearly with %Al, i.e, as x is doubled, nH2 is also doubled and so on.
We construct the plot of nH2 vs %Al by calculating the following values of nH2 for given values of %Al
%Al
NH2 = 0.0413.%Al/100 + 1.53/100
0
0.0153
25
0.0256
50
0.0359
75
0.0463
100
0.0566
%Al
NH2 = 0.0413.%Al/100 + 1.53/100
0
0.0153
25
0.0256
50
0.0359
75
0.0463
100
0.0566
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