INJURY AND RECOVERY 105 



tion of NaCl may be obtained by changing the constants 

 in formula (1) thus: 



0=89.i( — \( -KnTb -KoTe], -KoTe 



^ \Ko-Kn)\' -' J+90e 



(5) 



in which K^^ (the velocity constant of the reaction N — *■ 

 0) and Kq (the velocity constant of the reaction — >-P) 

 have the values 0.03 and 0.0297 respectively (see Table V, 

 page 98). 



We find by this formula that at the end of an exposure 

 of 15.9 minutes the value of + 10 is 92.57 ; hence it can 

 produce only (92.57 — 10) -r- (100 — 10) = 0.917 as much 

 of M in any given time as it could produce if it were 

 intact.^^ The amount it could produce, if intact, during 

 recovery in sea water is easily found by subtracting from 

 100 the resistance obtained by means of formula (1), when 

 jS:^ = 0.0036 and ^^=0.1080 (these are the normal val- 

 ues in sea water). Using these values we find that at the 

 end of 10.6 minutes the amount of resistance, as given by 

 formula (1), would be 98.55. Hence the loss during that 

 time would be 100 — 98.55 = 1.45, which is the amount 

 could produce in 10.6 minutes if intact. This value may 

 be called L and expressed as follows : 



[/ Ka \( -KaTr -KmTr\ „ „ 1 



''"^[k^^aJK: "" ;+9oe-^^^«+ioJ 



(6) 



in which JS:^= 0.0036 and K j^= 0.1080 (these are the nor- 

 mal values in sea water) and T^ is the time which has 

 elapsed since the tissue was replaced in sea water. 



" lu other words, if S, T and A were completely removed, O could raise 

 the level of M to 100 — 10 = 90 in the course of time. But if, for 

 example, half of is lost the remainder can raise the level of if to 

 only one-half its former value; i.e., to 45. 



L=ioo— 



