Answer : The rate constant at 785.0 K is, 
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/mole = 262000 J/mole
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{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)
Therefore, the rate constant at 785.0 K is,
The hydrogen bonding in H₂O is stronger than that of HF
Explanation:
Hydrogen bonds are special dipole-dipole attraction in which electrostatic attraction is established between hydrogen atom of one molecule and the electronegative atom of a neighboring molecule.
- The strength of hydrogen bonds depends on the how electronegative an atom is.
- Electronegativity refers to the tendency of an atom to gain electrons.
- The higher the value, the higher the tendency.
- This why oxygen with a higher electronegativity will form a stronger hydrogen bond with hydrogen compared to fluorine.
Learn more:
hydrogen bond brainly.com/question/12408823
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Answer:
When the concentration of all the reactants increases, more molecules or ions interact to form new compounds, and the rate of reaction increases. When the concentration of a reactant decreases, there are fewer of that molecule or ion present, and the rate of reaction decreases.
Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
The hydronium ion concentration at pH = 5.
![5=-\log [H_3O^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-5%7DM)
..............(1)
The hydronium ion concentration at pH = 3.
![3=-\log [H_3O^+]](https://tex.z-dn.net/?f=3%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-3%7DM)
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
![\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_3O%5E%2B%5D_%7Boriginal%7D%7D%7B%5BH_3O%5E%2B%5D_%7Bfinal%7D%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-5%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cfrac%7B1%7D%7B100%7D)
![100\times [H_3O^+]_{original}=[H_3O^+]_{final}](https://tex.z-dn.net/?f=100%5Ctimes%20%5BH_3O%5E%2B%5D_%7Boriginal%7D%3D%5BH_3O%5E%2B%5D_%7Bfinal%7D)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.
According to the illustration, the vanadium (V) oxide would be a catalyst.
<h3>What are catalysts?</h3>
Catalysts are substances that are utilized in reactions that are not themselves consumed in reactions but only speed up the rate of the reactions.
Catalysts speed up the rate of reactions by lowering the activation energy of the reactants.
Sulfur dioxide reacts with oxygen to produce sulfur trioxide. The vanadium (v) oxide is not consumed in the reaction. Thus it only serves as a catalyst.
More on catalysts can be found here: brainly.com/question/12260131
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