In the case of water (H20) we have two molecules of hydrogen and one of oxygen converted into two molecules of water. There are four possible cases to be considered depending on the direction of rotation of the several molecules. The equation of Areas becomes for these cases, Case I. Hence, if 4€, be the atomic energy lost during collision, From (A) and (B) we find, by eliminating w", and leaving the sign of in to be determined by the resulting equation, Case I. ma + 16m + 192 = 0. Case II. / The roots of this equation are imaginary, and, therefore, these cases impossible. Case III. m2 + 16m + 96 = 0. Case IV. , " The roots of this equation also are imaginary, and these cases impossible.1 It is evident, from the foregoing, that it is impossible to satisfy equations (A) and (B) so long as the value of m is the same in both. This impossibility of hydrogen and oxygen combining together of themselves to form water is entirely in accordance with chemical facts. If a mixture of two volumes of hydrogen and one volume of oxygen be made, they will not combine unless aided from without by a lighted match or an electric spark; when this aid is given, water is at once formed with an explosion, which corresponds to a "catastrophe" in a planetary system. We must now modify equations (A) and (B) so as to suit the condition of aid from without, and give a real value of m. Equation (B) which depends solely on geometrical configuration does not admit of modification; but equation (A) which represents the conservation of areas can be easily modified; let us suppose that the effect of the lighted match or electric spark is to add a quantity a to the right-hand side of the equation of areas. We now have Case I.\ „ . 8 n TT )«i"=l ±-+a, Case II./ nh Case III.\ „ 8 where o and a' are the areas necessary to be added; from which we see that a' = 1 + o. Cases I. and II. are therefore to be preferred to Cases III. and IV., because they require a less amount of external aid. We now have n>i I have already shown that in the case of hydroxyl the most prebahle value of mi is mi = - 7, from which I found mn= 12564 a, = 1-50. Now, since H20 > HO, we have «"»> 1-2564. If W--10, <o"* = + 1-3600, at - 1-3090, t, = + 0 0984. If m = -11, t0",= 1-2066, a,= 1-2412, €, =-0-0217. The last of these must be rejected, for e2 must be positive. u"3 > u'*, a, < at. Therefore select m = - 10. As this differs from ml = - 7 tlie most probable value of m in forming hydroxyl, it follows that the n' = 25829, where is the coefficient of attraction of oxygen upon hydrogen in hydroxyl, and ft." the coefficient of attraction of oxygen upon hydrogen in water. 2. Sulphuretted Hydrogen. Sulphur, selenium, and tellurium, behave towards hydrogen in the same manner as oxygen; and chemists recognize HS analogous to hydroxyl, and H2S (sulphuretted hydrogen) analogous to water. Like hydroxyl, HS is not known to exist as a separate compound, hut it may exist in combination in higher compounds. Introducing f3 = 32 in the general equation (7) bit, we find (66e, + 1) w» - 2048m - 31712 = 0. If ci = 0, we find ml = - 15, the nearest whole number to the negative root. From this we find 32 17 •'■slTs3-1,0524' 0)"= 1-1076, a, = 0-7687. It is easy to show, as in the case of oxygen, that no dyad atom can enter into combination with two hydrogen atoms without external aid; for, if y3 be the atomic weight of any dyad, we have Conservation of areas \Case II. 2m i a 3 m n /Case I. m1 + fim + - )8» = 0 C 4 xCase II. and , n 3 m „ /Case III. Both equations give imaginary values of m for real values of /?. We must therefore suppose, as in the case of water, that an addition o is made to the right-hand side of the equation of conservation of areas. This gives us, Areas, . . to" = ——- } + a. \ 2m,; Configuration, . v>m= 1 + ( —J • 'Sow, since HaS > HS, or to'" > to" |