I thank that your answer is C.
Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol)
Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L
Dissociation equation:
Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq)
3 s 2 s
Ksp = [Sr²⁺]³ [AsO₄³⁻]²
= (3s)³ (2s)²
= 108 s⁵
Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹
<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
When drawing lewis dot diagram structure we consider the number of valence electrons of the atom.Se has six valence electrons since it is in group six thus it Lewis dot diagram is as follows
..
: Se:
Hydrogen is in group one hence has one valence electron.The lewis dot diagram for 2H is therefore
H:H
Now ,
C + O2 → CO2
According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.
No of moles = mass of the substance/molecular mass of the substance.
In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.
No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.
No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles
Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.
Also number of moles of O2 = mass of O2÷ molar mass of O2
Substituting number of moles of O2 as 0.1 we get
mass of O2(x) = Number of moles of O2 × Molar mass of O2
Mass of O2 (x) = 0.1 × 32= 3.2 g
Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.
