The variations from these proportions occurring in subsequent experiments are noted in the table giving the summary of the results. Usually about 150 m.g. of manganese dioxide were used and the solutions in the three flasks were brought to the same volume by addition of distilled water. Experiment No. I Flask No. I Manganese dioxide 150 m.g. N/10 nitric acid 14.28 c.c. [= Mn. in KMnO₄ in No. II or No. III] Water, 27.12 c.c. [Total solution = 41.40 c.c.] Flask No. II Potassium permanganate, 20 c.c. = 112.48 m.g. KMnO₄. N/10 nitric acid 7.14 c.c. [= K in KMnO₄] Water, 14.26 c.c. [Total solution = 41.40 c.c.] Flask No. III Potassium permanganate, 20 c.c. = 112.48 m.g. KMnO₄. N/10 nitric acid 21.40 c.c. [= K + Mn in KMnO₄] Manganese dioxide 150 m.g. The nitric acid in No. I is calculated to be the same as that remaining free after the potassium of the permanganate in No. III has been neutralized. It will be clear that the quantities of free nitric acid are the same in flask No. I and in flask No. III provided the potassium of the permanganate is appropriated by the nitric acid. Moreover each flask contains the same quantity of manganese dioxide and the volume of the liquid is the same in both. On the other hand flask No. III contains permanganic acid. This arrangement was selected in order to determine how far the evolution of oxygen was influenced by the action of nitric acid on the manganese dioxide or by the spontaneous decomposition of permanganic acid. It was thought that if more oxygen should be obtained from flask No. III containing a mixture of permanganate, manganese dioxide and nitric acid, than from mixtures No. I and No. II, the larger volume must be due to the reduction of the permanganate by the manganese oxide.