By Dr. N. W. H. Addink (auth.)
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0·136Tbp(KJ/A = 22,400//A x S, Therefore: or S, = 165//(Tbp(KJ X 10- 3 ) (16) (17) By putting graphically values of S 1 after Eq. (14) against those of Eq. (17) we obtain a fairly good agreement between both values (see Fig. 9). Comment. In case of element concentrations of less than 100%, say b%, the righthand parts of Eq. (15) as well as of Eq. (16) have to be multiplied by b/100, but the result (Eq. (17)) remains the same. Therefore s, is independent of the concentration of the element present in the sample, provided that a fixed amount of it, in our case 5 mg, is examined.
4), this group is less sensitive to additions than the alkalis. However, at the same time, the degree of ionization dependent upon the value of V1 plays a role in the same sense. The position of Li added as Li 2 C0 3 (boiling point in between both groups) explains its use as a stabilizing element generally applied. Spark lines of the alkaline earths, originating from the core, appear to be very sensitive to a complete absence of other elements. 2), which means that in the atomion equilibrium, [M] is large and [M+] is small.
2) reads as follows: logs, = 3·06 - 0·44 Tbp