Answer:
The percent ionization is 0,16%
Explanation:
The percent ionization is defined as the number of ions that exist in a substance.
![PI=\frac{[A-]}{[HA]} x100](https://tex.z-dn.net/?f=PI%3D%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%20x100)
First, we find the [A-] using the ka equation
HA ⇄ 
[H+] = [A-]
![Ka=\frac{[H+][A-]}{[HA]}\\ \\](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D%5C%5C%20%5C%5C)
since the ionization constant is very small we can assume that the final concentration of [HA] is the same
![Ka=\frac{[H+]^{2} }{[HA]} \\\\](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%2B%5D%5E%7B2%7D%20%7D%7B%5BHA%5D%7D%20%5C%5C%5C%5C)
![[H+]=\sqrt[2]{Ka.[HA]} \\\\](https://tex.z-dn.net/?f=%5BH%2B%5D%3D%5Csqrt%5B2%5D%7BKa.%5BHA%5D%7D%20%5C%5C%5C%5C)
![[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%5Csqrt%7B%282%2C610%5E%7B-7%7D%20%29%280%2C1%29%7D%20%20%3D%201%2C61210%5E%7B-4%7D)
Now we calculate the percent ionization using these values

PI=0,16%
Answer:
When there is less carbon (in the form of carbon dioxide) in abiotic matter, less carbon is available for producers making energy storage molecules. When there is more sunlight, producers can make more energy storage molecules from the carbon in carbon dioxide.
Hope this helps.
Answer : The entropy change of reaction for 1.62 moles of
reacts at standard condition is 217.68 J/K
Explanation :
The given balanced reaction is,

The expression used for entropy change of reaction
is:

![\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BBr_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28Br_2%29%7D%2Bn_%7BF_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28F_2%29%7D%5D-%5Bn_%7BBrF_3%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28BrF_3%29%7D%5D)
where,
= entropy change of reaction = ?
n = number of moles
= standard entropy of formation
= 245.463 J/mol.K
= 202.78 J/mol.K
= 292.53 J/mol.K
Now put all the given values in this expression, we get:
![\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B1mole%5Ctimes%20%28245.463J%2FK.mole%29%2B3mole%5Ctimes%20%28202.78J%2FK.mole%29%7D%5D-%5B2mole%5Ctimes%20%28292.53J%2FK.mole%29%5D)

Now we have to calculate the entropy change of reaction for 1.62 moles of
reacts at standard condition.
From the reaction we conclude that,
As, 2 moles of
has entropy change = 268.74 J/K
So, 1.62 moles of
has entropy change = 
Therefore, the entropy change of reaction for 1.62 moles of
reacts at standard condition is 217.68 J/K
I believe your answer is Favorable.
Hope this helps, and happy studying~!
~{Dunsforhands}
Answer:
The anode made of the impure copper
The cathode made of pure copper
The electrolyte of copper (II) sulfate CuSO₄ solution
The silver impurities at the anode due to their high tendency of accepting electrons and being a stronger reducing agent than either copper or zinc will remain relatively in place and relatively stable and will not actively take part in the oxidation reaction taking place at the anode
The zinc impurities will be the first element of the three metals to give up electrons and go into the solution as zinc ions due to their high tendency to loan out two electrons and become oxidized into Zn²⁺ ions
The drawing of the electrolytic cell created with Microsoft Visio is attached
Explanation: