Answer: Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. ...
Use coefficients of products and reactants to balance the number of atoms of an element on both sides of a chemical equation.
I've prepared some analysis and <span>cucumbers do have many comparable properties to potatoes, tomatoes, and lemons, all of which I know do work. So I would presume that cucumbers would also work. I would recommend trying it yourself to perceive. I'd love to hear the outcomes of your experiment. ;) </span>
The polarity of a water molecule
comes from the uneven distribution of electron density of hydrogen and oxygen
atom. The oxygen in the water molecule is more electronegative than the
hydrogen. Water has a partial positive charge near the hydrogen atom and a
partial negative charge near the oxygen atom. The result of this
electrostatic attraction results in the bond called hydrogen bond. Also,
because of this bond, it has the ability to dissolve most of the solutes due to
its polarity and bonding.
<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:check explanation
Explanation:
(a). HOW THE DISTANCE BETWEEN ELECTRON DONOR AND ACCEPTOR AFFECTS THE RATE OF ELECTRON TRANSFER IN BIOLOGICAL SYSTEM:
Distance between the acceptor and the donor can affect in two ways; short distance and long distance effect.
Short distance causes
electronic orbitals of donor and acceptor directly overlap whereas in LONG DISTANCE reactions this coupling is indirect because of
sequential overlaps of atomic orbitals of the donor, the intervening medium, and the orbitals of the acceptor.
(b). HOW REORGANIZATION ENERGY OF REDOX ACTIVE SPECIE SURROUNDING MEDIUM AFFECTS:
the reorganized energy does not depend on the pre-existing intra molecule electric field. The charge transferred inside the molecule interacts with its aqueous surroundings.
Reorganized energy can be calculated using Poisson-Boltzmann equation.
