<span>After alcohol is metabolized in the liver, it is burned as fuel for the cells. </span>
After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
<u>Answer:</u> Group 1 ions are known as cations and Group 17 ions are known as anions.
<u>Explanation:</u>
Ions are formed when an atom looses or gains electrons.
If an atom gains electrons, it leads to the formation of negative ions known as anions. <u>For Example:</u> Fluorine is a Group 17 element which gains 1 electron to form
ions.
If an atom looses electrons, it leads to the formation of positive ions known as cations. <u>For Example:</u> Sodium is a Group 1 element which looses 1 electron to form
ions.
Hence, group 1 ions are known as cations and Group 17 ions are known as anions.
<u>Answer:</u> The standard free energy change of formation of
is 92.094 kJ/mol
<u>Explanation:</u>
We are given:

Relation between standard Gibbs free energy and equilibrium constant follows:

where,
= standard Gibbs free energy = ?
R = Gas constant = 
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
K = equilibrium constant or solubility product = 
Putting values in above equation, we get:

For the given chemical equation:

The equation used to calculate Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the Gibbs free energy change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag%5E%2B%28aq.%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag_2S%28s%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol](https://tex.z-dn.net/?f=285.794%3D%5B%282%5Ctimes%2077.1%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%28-39.5%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%3D92.094J%2Fmol)
Hence, the standard free energy change of formation of
is 92.094 kJ/mol
Answer : The specific heat of tin is, 0.213 J/g.K
Explanation :
Formula used :

where,
q = amount of heat lost = -399.4 J
c = specific heat capacity of tin = ?
m = mass of tin = 25.0 g
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:


Therefore, the specific heat of tin is, 0.213 J/g.K