Fats , oils , waxes and sterols are collectively known as lipidsand the true fats contain only carbon , hydrogen and oxygen
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
Answer:
Explanation:
turn over number = R max / [E]t = K2
From given , R max = 249 * 10 ^ -6 mol. L^-1
T [E]t = 2.23 n mol. L^-1
= 2.23 * 10^-9 mol. L^-1
Putting values in above equation,
= 111.65 * 10^3 S^-1
Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.
1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.
The preferred method for determining or validating the structure of organic molecules or those containing protons is H NMR. When compared to other nuclei, a solution-state proton spectrum may be obtained relatively quickly, and it contains a wealth of knowledge regarding a compound's structure.
It can be calculated by simply counting the number of unique hydrogens on one side of the symmetry plane will give you the count of signals individual molecules emit in a 1H NMR spectrum.
Therefore, 1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.
To know more about 1-H NMR spectroscopy
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