Non-stoichimoetric cuprous oxide. `Cu_2O` can be prepared in laboratory. In this oxide, copper to oxygen ration is slightly less than 2:1 can you account for the face that this substance is a p-type semiconductor?

Non-stoichiometric cuprous oxide (Cu_2O) can be prepared in the laboratory. In this oxide, copper to oxygen ratio is sloghtly less than 2 : 1 can you account for the fact that this substance is p-type semiconductor

The discs of moments of inertia I_1 and I_2 about their respective axis (normal to the disc and passing through the centre), and rotating with agnular speeds omega_1 and omega_2 are brought into contact face to face with their axis of rotation coincident. Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take omega_1 ne omega_2 .

Knowing the electron gain enthalpy values for O rarr O^- and O rarr O^(2-) as -141 and 702 kJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O^(2-) species and not O^-

Deep sea divers have been using the compressed air containing N_2 in addition to O_2 for breathing. When the sea divers breathes in compressed air at a depth, more N_2 dissolves in the blood and other body fluids than would dissolved at the surface because the pressure at the depth is far greater than surface atmospherice p ressure. when the diver comes towards the surface, the pressure decreases and N_2 comes out of the body quickly forming bubbles in the stream which restrict blood flow and affect the transmission of nerve impuses. this results into a condition called 'the bends' which is dangerous and painful. To avoid this condition, professionals now use air diluted with helium. As a student of chemistry, can you analyse as to why helium is used?

The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) " solution " =2xx " molarity of" H_(2)O_(2) solution What is the molarity of "11.2 V" H_(2)O_(2) ?

STATEMENT-1: If SOP of substance is less than -1.23 V and over voltage=0V, then its oxidation in its aqueous solution is not possible at 298 K. , STATEMENT-2: Standard reduction potential (SRP) of water is +1.23V.