Answer:
2%
Explanation:
.98 is 98% of one and therefore they are missing 2%
<em>Same group element have same</em><em><u> Valence electron</u></em><em> and behave similarly in </em><em><u>Chemistry.</u></em>
<u>Explanation:</u>
For example. First group elements Alkali metals:- H, Li, K, Rb, Cs, Fr
Valance electron will take part in forming a bond with other elements and compound will form. All the above-given elements (H-Fr) have valence electron 1 in outer most 'S' shell. All have electronic configuration S1
Behavior: Since valence electrons are the same so the behavior of all the elements in this group is the same. All are metal (from Li-Fr, except Hydrogen), all are very reactive, does not found in native state in the environment, and all react with water.
This problem is describing the state two gases have when separated and together as shown on the attached picture. First of all, diagram 1 shows how they are separated in two containers with apparently equal volumes, whereas diagram 2 shows the removal of the barrier so that they get mixed together.
In this case, we can analyze that each gas has its own pressure and due to the removal of the barrier, both pressure and volume undergo a change. Thus, we can infer that the final volume is doubled with respected to the initial one for each gas, causing the pressure of each gas to be halved and the total pressure the half of the added ones, in agreement to the Boyle's law (inversely proportional relationship between pressure and temperature).
Therefore, the correct choice is:
C. The partial pressure of each gas in the mixture is half its initial pressure; the final total pressure is half the sum of the initial pressures of the two gases.
Learn more:
<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