202 MEMOIR OF DB. DALTON, AND 



No. 2. 

 Sulphuric Acid a = 526, 5 = 90 



Alumina., a =526 = 526 



A Magnesia. 0+ b =526+ 90 = 616 



A Lime . . , a-\- Sb =526+ 3.90 = 796 



* a+ 5b — S =526+ 5.90 — 3 =973 

 Alumina..(a+ eb —13) =526+ (6.90 — 13) =1053 



* a+ 75 — (3+ 1)=526+ 7.90 — (3+ 1)=1152 



* a+ 95 __ (3+ 3) =526+ 9.90 — (3+ 3) =1330 



* o+lU_(3+ 5) =526+11.90 — (3+ 5) =1508 



* a+135— (3+ 6) =526+13.90 — (3+ 6) =1687 



* a+156 — (3+ 7) =526+15.90 — (3+ 7) =1866 



* a+175 — (3+ 8) =526+17.90 — (3+ 8) =2045 

 Baryta., a+195 — (3+ 9) =526+19.90 — (3+ 9) =2224 



* a+215 — (3+11) =526+21.90 — (3+11) =2402 

 ^ a+235 — (3+13) =526+23.90 — (3+13) =2580 



&c. &c. 



" C. If we convert the differences into simple consecutive 

 odd numbers, the quantity of the determining element and 

 all the terms of the series have only to be divided by b, and 

 we receive : 



No. 1. 



Muriatic Acid, ^ ^^a i 2*9 Muriatic Acid, ^+8 

 4 «=734, b=H^ ^_ ^S^^g 



Alumina.. -? = _»i_ := 5 + ||| 



A Magnesia, -f- + 1 = ^»f = 6 + If^ 



A Lime f + 3 = 1^- 8 + H ? 



13564 1 n _J_ ?A? 



124i ^^1" 249 



1605i 1 O _1_ ?^3 



1244 1-4 -f- 249 



18544 1 4. _i. 2U 



1244 ^^1 249 



2 1£34 Ifi _!_ ^1^ 



T244 AU -f- 2 49 



