292 ME. W. CEOOKES ON THE ATOMIC WEIGHT OF THALLIUM. 
From Jc, 
(3)=T9998394 + 0-9989797 + 0-00165 ; 
.-. (3)=3-0004691 N. 
From j, 
(6) = 3*0004G91+1*9998394+0'9989797-0-00102; 
.-. (6) = 5*9982682 O. 
Again, adding m and n, 
(l)=2(-3) + (-2)+2(-l)-0*00768. 
Multiplying o by 2 we get 
2(-3)=2(-2)+2(-l)+0-00450. 
Subtracting q from p, 
(•2)=2(-l)+0-00702. 
Then (m+w)+2o+3Q}— q) gives 
(1)=10 (T) + 0 , 01788 ; 
.-. 0-9989797 = 10(T) + 0-01788 ; 
.-.0-09989797= (-1) + 0-001788 ; 
(-1) = 0-09810997 P. 
From q we get 
0-09810997 = (-06) + (-03) + (-01) — 0-00802 ; 
(-06) + (-03) + (-01)=0-10612997. . . . Q. 
From q), 
(-2)=0-09810997 + 0-10G12997-0-00100; 
.-. (-2) = 0-20323994 R. 
From o, 
(-3) = 0*20323994 + 0-09810997 +0-00225 ; 
.-. (-3)=0-30359991 . S. 
From n we get 
(•6) = 0-30359991 + 0-20323994 + 0-09810997 — 0-002G0 ; 
.-. (-G)=0-60234982 T. 
Again, adding q and r, 
(-1) = 2(-03) + (-02) + 2(-01)— 0-01409. 
Multiplying s by 2 we get 
2(-03)=2(-02) + 2(-01)- 0-01284. 
