Anti-aromatic and the non-aromatic compounds
What is main difference among anti-aromatic and the non-aromatic compounds?
Expert
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.
The Van't Hoff factor of the compound K3Fe(CN)6 is: (a) 1 (b) 2 (c) 3 (d) 4 Answer: (d) K3[Fe(CN)6] → 3K+
give atleast two application of following colligative properties
what is molarity of a solution of hcl which contains 49% by weight of solute and whose specific gravity is 1.41
While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than
Normal butane (C4H10) is stored as a compressed liquid at 90°C and 1400 kPa. In order to use the butane in a low-pressure gas-phase process, it is throttled to 150 kPa and passed through a vaporizer. The butane emerges from the vaporizer as a
Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20
The pressure-temperature relation for solid-vapor or liquid vapor equilibrium is expressed by the Clausis-Clapeyron equation.We now obtain an expression for the pressure-temperature dependence of the state of equilibrium between two phases. To be specific,
Select the right answer of the question. A solution of CaCl2 is 0.5 mol litre , then the moles of chloride ion in 500ml will be : (a) 0.25 (b) 0.50 (c) 0.75 (d)1.00
These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters Q : Dipole attractions-London dispersion Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?
Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?
18,76,764
1954427 Asked
3,689
Active Tutors
1449064
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!