Answer:
i think is c,my ans will not be 100%correct
Because they both have to do with and chemistry science
Answer:
Dispersion Forces are found between n-Pentane (CH₃-CH₂-CH₂-CH₂-CH₃) and n-Hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃).
Explanation:
Dispersion Forces are present and developed by those compounds which are non-polar in nature. In given statement n-Pentane and n-Hexane both are non-polar in nature as the electronegativity difference between Hydrogen atoms and Carbon atoms is less than 0.4.
When non-polar molecules approaches each other, a Dipole is induced in one of them, this step is known as Instantaneous Dipole, This generated Dipole on approaching another non-polar molecule induces dipole in it and the process propagates. Hence, creating intermolecular interactions.
The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.
First-order reactions are very commonplace. we have already encountered examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.
Rate constant K₁ = 0. 543 /s
T₁ = 25°C
Activation energy Eₐ = 75. 9 k j/mol.
T₂ = 32. 3 °C.
K₂ =?
formula;
log K₂/K₁= Eₐ /2.303 R [1/T₁ - 1/T₂]
putting the value in the equation
K₂ = 0.343 /s
Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s
The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.
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Answer:
Explanation:
For a flower to appear blue, "it needs to be able to produce a molecule that can absorb very small amounts of energy," in order to absorb the red part of the spectrum, Kupferschmidt said.