Anti-aromatic and the non-aromatic compounds
What is main difference among anti-aromatic and the non-aromatic compounds?
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The anti aromatic compounds follows 4npi electron rule and are planar, cyclic compounds, while non aromatic are cyclic or acyclic and non-planar or planar.
Unit of measurement- These are also some systems for units: (1) C.G.S.
The ionic radii of Rb+ and I- respectively are 1.46 Å and 2.16Å. The very most probable type of structure exhibited by it is: (a) CsCl type (b) ZnS type (c) Nacl type (d) CaF2 type Q : Calculating value of molar solution Choose the right answer from following. An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is: (a)14 (b) 3.2 (c) 4 (d) 2
Choose the right answer from following. An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is: (a)14 (b) 3.2 (c) 4 (d) 2
Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal
Elaborate a substituted hydrocarbon?
Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998
Alkyl halides can be prepared from alkanes through substitution and from alkenes through addition of halogen acids or through allylic substitution. From alkanesWhen alkanes are treated with halogens, chlo
Ionic dissociation depends on the dielectric constant of the solvent.The Arrhenius that ions are in aqueous solutions in equilibrium with parent molecular species allows many of the properties of ionic solutions to be understood. But difficulties began to
Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N (b) 19.8 N (c) 39.6 N (d) 98
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
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