Transformation in chemistry is scientifically used to explain the process of changing one compound to another in a chemical reaction.
<h3>What is transformation?</h3>
The word "transformation" has a very special significance in chemistry. We know that in English, to transform would simply imply to change from one form to another. This is not quite far from its meaning in the parlance of chemistry.
The word transformation is normally applied in the area of chemical reactions especially as it has to do with reaction with in organic chemistry. It has to do with the change from one molecule to another and this is of great importance in the discussion of synthetic chemistry.
As such, the word transformation in chemistry is scientifically used to explain the process of changing one compound to another in a chemical reaction.
Learn more about chemical transformation:brainly.com/question/8210521
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Answer: A new model of the atom that described electrons as being in a cloud
Explanation:
Why does the chemical reaction seen here obey the law of conservation of matter?
<u><em>Answer:</em></u>
- Because there are the same number of atoms of each element shown on both sides
<u><em>Explanation</em></u>
- As in chemical reactions, atoms bonds are break and new bonds are formed. As new substance are formed but overall they have same elements, no new elements come from outside or go to outside. In other words , rearrangement of atoms take place but number of atoms remained same.
NaOH + HCl -----> NaCl + H2O
- As in above reaction there are the same number of atoms of each element shown on both sides .
?? Is that the whole question?
<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
