ME, W. CEOOKES ON THE ATOMIC WEIGHT OF THALLIUM. 
291 
Subtracting i from /?, 
(20) = 2(10) + 0-00303. 
By adding e , f, twice g, and thrice the last equation, we get 
99-991420 =10(10) + 0-000665 ; 
.*. (10)=9-998477 F. 
From equation i we get 
9-998477 = (6) + (3)+(l) + 0-00052 ; 
.-. (6) + (3) + (!) = 9-997957 G. 
Substituting these values in h we get 
(20) = 9-998477 + 9-997957 +0-00355 ; 
(20)=19-999984 II. 
From g we get 
(30) = 19-999984+ 9 -998477 + 0-00154 ; 
.-. (30)= 29*999991 I. 
From f we get 
(60) = 29-999991 + 19-999984 + 9-998477 — 0-00522 ; 
(60) = 59-993232 J. 
Again, adding i and j, 
(10)=2(3)+(2)+ 2(1) — 0-00050. 
Multiplying k by 2, 
2(3)= 2(2) + 2(1) + 0-00330. 
Subtracting m from /, 
(2)= 2(1)+ 0-00196. 
Then (i +/) + 21:+ 3(Z— m) gives 
(10) = 10(1)+ 0-00868 ; 
.*. 9-998477 = (10)1 + 0-00868 ; 
.-. 0-9998477 = (1) +0-000868 ; 
.-. (l) = 0-9989797 K 
From equation m we have 
0-9989797=(-6) + (-3) + (-l) — 0-00508; 
... (-6)+(-3)+(-l)= 1-0040597. . ... L. 
From l, 
(2) = 0-9989797 +1-0040597 — 0-00312 ; 
.-. (2)=l-9998394. ........ M. 
