Answer:
Here:
Explanation:
To familiarize students with experimental apparatus, the scientific method, and methods of data analysis so that they will have some idea of the inductive process by which the ideas were originated. To teach how to make careful experimental observations and how to think about and draw conclusions from such data.
Answer:
34.23 g.
M = (no. of moles of solute)/(V of the solution (L)).
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.
Learn more about activation energy here:- brainly.com/question/26724488
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Answer: The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl– has an oxidation state of -1. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2.
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