State substituted hydrocarbon
Elaborate a substituted hydrocarbon?
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The substituted hydrocarbon is a hydrocarbon with one or more of hydrogen is substituted with other element, (every so often a halogen like as bromine or chlorine) or other group of atoms as -OH. Such as: - The simple hydrocarbon is methane (CH4). Substitute chlorine for the hydrogen to get Methyl Chloride (CH3Cl) is used for cleaning. Sub to get- Methylene Chloride (CH2Cl2) is used as a paint stripper. Sub to get- Chloroform (CHCl3) is an antique anaesthetic. Sub to get- Carbon Tetrachloride (CCl4) is used in the fire extinguishers and cleaning. Substitute single -OH group into the - CH4 to get methanol (CH3OH) or into C2H6 to get ethanol (C2H5OH) The above instances all begin with the unbranched non-cyclic hydrocarbons, but any hydrocarbon is an appropriate target. A known instance is a double replacement of chlorine at the opposite ends of the benzene ring to form paradichlorbenzene, normally found hanging in toilet bowls. C6H6 becomes C6H4Cl2
The entropy due to the rotational motion of the molecules of a gas can be calculated. Linear molecules: as was pointed out, any rotating molecule has a set of allowed rotational energies. For a linear molecule the
i) Show that the equilibrium constant Kp for the reaction CaCo3(s) ↔ CaO(s) +CO2(g)is about unity (i.e. = 1.0) at 895 °C.ii) If two grams of calcium carbonate are pl
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Give me answer of this question. A solution contains 25%H2O 25%C2H5OH , and 50% CH3 COOH by mass. The mole fraction of H2O would be: (a) 0.25 (b) 2.5 (c) 0.503 (d) 5.03.
why pentahalids are more covalent than tetrahalids
Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o
Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M
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The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop
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