222 REPORT— 1870. 



46-69=31-12; therefore the amount of hydrogen required to combine with 

 the oxygen of the nitrous oxide is 146-23— 31-12=115*11 ; hence the 

 volume of the oxj'gen contained in the nitrous dxide is equal to ii|-!-=57"5;j, 

 differing by 1"9G per cent, from the calculated volume of oxvgen, -which is 

 iy:3?=58-()9. 



Hence in 100 volumes of nitrous oxide we find : — 



By experinieiit. Calculalod. 



Nitrogen 100-52 100 



Oxygen 4iJ-02 50 



(2) Volume of nitrous oxide iised 116-93 



Volume after the admission of hydrogen 266-20 



Volume after explosion 151-69 



Volume after the admission of oxygen 207-19 



Volume after explosion 1 55-71 



Hence in 100 volumes of nitrons oxide — 



By expei-Iiiif-nt. Caleiiliited. 



Nitrogen 100-38 100 



Oxygen 49-14 50 



(3) Volume of nitrous oxide usel 126-42 



Volume after the admission of hydrogcu 284-38 



Volume after explosioii 160-55 



Volume after the adm!Fsio:i of oxygon 217-53 



Volume after explosion 107-25 



Hence in ] 00 volumes of the gas — • 



By experiment. 'Calculated. 



Nitrogen ." 100-49 100 



Oxygen 49-22 50 



(4) Volume of nitrous oxide used 149-39 



Volume after the admission of liydrogen 345-92 



Volume after cxplo.sion ] 99-65 



Volume after the admission of oxygen 326-34 



Volume after explosion 252-43 



Hence in 100 volumes of the gas — 



By oTpei-iment. Calculated. 



Nitrogen lOO-Gti 100 



Oxygen '. . . 49-28 50 



IT. Nitroits oxide obtained by lieating nitrate of ammonium. 



(5) Volume of nitrous oxide used 128-20 



Volume after the admission of hydrogen 297*05 



Volume after explosion 171-29 



Volume after the admission oi" oxj-gen 229-80 



Volume after explosion 166-58 



Hence in 100 volumes of the gas — 



By exporiiiu^nt. Calcidatixi- 



Nitrogen Imi-T:'. 100 



Oxvjrou i-'-4i? 50 



