A hydrocarbon 11, of molecular weight 54 reacts with an excess of Br2 in CCl4 to give compound 12, whose molecular weight is 593% more than that of 11. However, on catalytic hydrogenation with excess of H2, 11 forms 13, whose molecular weight is only 7.4% more than that of 11. 11 reacts with CH3CH2Br in the presence of NaNH2 to g...
Read More..To find the mass of 36% of 0.1M Na2CO3·10H2O in 750ml of solution, we need to consider the following steps:
Read More..Explain in detail by giving the type of hybridization, geometry, molecular shape, number of bonds and angles formed by the following compounds; (a) Ammonia (NH3) compound; (b) Ethyne (C2H2) (c) Ethanal (CH2O) (d) Sulphur trioxide (SO3)
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