<span>If a reaction is reversible, then it will attain the phase of Equilibrium and at that phase, the Amount of Reactants and Products would be: Equal
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Another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.
<h3>What is a microspectrophotometer?</h3>
Microspectrophotometry is a biological technique used to measure the absorption or transmission spectrum of a solid or liquid material in either transmitted or reflected light.
Microspectrophotometry can also measure the emission of light by a sample, which is usually small as the micro implies.
One advantage of microspectrophotometry is that the sample does not get damaged. However,
However, another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.
Learn more about microspectrophotometry at: brainly.com/question/5832827
H20. 2 of hydrogen and oxygen
Answer: 
Explanation:
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature = 
Now put all the given values in this formula, we get
![\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)
Therefore, the value of the rate constant at 775.0 K is
