According to the law of Charles--the volume of a given quantity of gas at constant pressure varies directly as the absolute temperature. To reduce from the centigrade to the absolute scale, we have only to add 273 degrees. Adding the observed temperature, we have 273 degrees + 27 degrees = 300 degrees. Applying the above law to O obtained at 300 degrees A, we have the proportion below. Since the volume of O at 273 degrees will be less than it will at 300 degrees, the fourth term, or answer will be less than the third, and the second term must be less than the first. 300 : 273 :: 1450 : x. This would give the result dependent upon temperature alone. By the law of Mariotte - Physics, - the volume of a given quantity of gas at a constant temperature varies inversely as the pressure. Applying this law to the O obtained at 758mm, we have the following proportion. The volume at 760mm will be less than at 758mm; or the fourth term will be less than the third; hence the second must be less than the first. 760: 758:: 1450: x. This would give the result dependent on pressure alone. Combining the two proportions in one:-- 300: 273 ):: 1450: x = 1316cc. 760: 758 ) 1316cc=1.316 liters. It remains to find the weight of this gas. A liter of H weighs 0.0896g. The vapor density of O is 16. Hence 1.316 liters of O will weigh 1.316 X 16 X 0.0896 =1.89g. (KClO3 = KCl + O3) From the equation (122.5 48) we make a proportion, ( 5 x) 122.5: 5:: 48: x = 1.95, and obtain, as the weight of O contained in 5g of KClO3, 1.95g. The weight we actually,obtained was 1.89g. This leaves an error of 0.06g, or a little over 4 per cent of error (0.06 / 1.95 = 0.03 +). The percentage of error, in performing this experiment, should fall within 10. Some of the liabilities to error are as follows:-- 1. Impure MnO2, which sometimes contains C. CO2 is soluble m H2O. 2. Solubility of O in water. 3. Escape of gas by leakage. 4. Moisture taken up by the gas. 5. Difference between the temperature of the gas and that of the air in the room. 6. Errors in weighing. 7. Want of accuracy in the weights and scales. 344. Hydrogen. Experiment 135.--Weigh 5g, or less of sheet or granulated Zn, and put it into a small flask provided with a thistle-tube and a delivery-tube. Cover the Zn with water, and introduce through the thistle-tube measured quantities of HCl, a few cubic centimeters at a time. Collect the H over water in large flasks, observing the same directions as in removing O. Weigh the water, compute the volume of the gas, reduce it to the standard, and obtain the weight, as before. Should any Zn or other solid substance be left, pour off the water or filter it, weigh the dry residue, and deduct its weight from that of the Zn originally taken. Suppose the residue to weigh 0.5g. Make and solve the proportion from the equation:- Zn + 2HCl = ZnCl2 + 2H. 65 2. 4.5 x.