Oxygen is called by right the vital chemical element. Besides, it is a part of many connections, some of Them are not less important for life, than he. The set of oxygencontaining substances finds application in the industry and agriculture. Skills of determination of mass of oxygen are necessary as that who studies chemistry, and to the staff of chemical laboratories and plants.
Instruction
1. Without oxygen the life on Earth would be impossible. This element meets in the connections which are a part of air, water, the soil, living organisms. Oxygen has interesting properties. First it represents the strong oxidizer participating in processes of burning and corrosion of metals. Secondly, it forms oxides with many elements. In addition, atoms of oxygen are present at some inorganic acids (H2SO4, HNO3, HMnO4). Oxygen is considered one of the most active elements therefore it is widely applied in metallurgy, chemical industry and also as oxidizer to rocket engines.
2. Quite often conditions of chemical tasks look as follows. Oxygen is in a free look and its volume is equal to V. It is required to determine the mass of O2 oxygen under normal conditions. The molar volume of Vm gas under normal conditions is 22.4l/mol. There is also a formula, having used which, it is possible to find amount of substance, and then and weight in the oxygen relation. This formula is given below: n(O2) = V (O2)/Vm where l Vm=const=22.4 / molteper, knowing amount of substance, it is possible to determine weight: m(O2)=n(O2) * M (O2) As a molecule of oxygen consists of two atoms, and the molecular mass of this element according to data from Mendeleyev's table is equal 16, g/mol M(O2)=32. From this it follows that m(O2) =32n(O2) = 32V/22,4, where V - the volume set in a statement of the problem.
3. In the industry pure oxygen is usually received decomposition of razilchny oxides. For example: 2HgO=2Hg+O2Vsledstvie it, tasks in which it is required to find amount of substance and weight on the set equations of reactions meet. If the amount of substance concerning oxide from which oxygen turns out is given, then such problem can be solved as follows, paying special attention to equation coefficients: 2HgO (n) =2Hg+O2 (x) 2 mol 2 mol 1molza x in the equation the unknown amount of substance of oxygen, for n - amount of HgO oxide is accepted. The equation can be transformed to a proportion: x/n=1/2 where 1 and 2 - coefficients uravneniyasootvetstvenno, n(O2) =n (HgO)/2tak as is well-known amount of substance of oxygen, it is possible to find its weight. It is equal: m(O2)=n(O2) * the M (O2) =n(HgO)/2*M(O2)