Answer : The value of 
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction 
.
![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)
Now we have to calculate the Gibbs free energy of reaction 
.
As we know that,
At room temperature, the temperature is 500 K.
Therefore, the value of for the reaction is -959.1 kJ
Answer:
Individual solute particles are broken apart from the solid by the;
c. Solvent
Explanation:
A solution is the homogeneous mixture that is made up of two or more substances formed by dissolving a substance which can be a solid, liquid or gas in another substance known as the solvent which normally the larger part of the fraction of the solution than the solute and can also be a solid, liquid or a gas
In a solution the solvent particles serves to brake of and disperser parts of a solid solute to form a more or less homogeneous mixture
Therefore, the solute particles are broken by the <u>solvent</u> particles in a solution
Answer:
Let me give it a try.
H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O
Balancing this reaction
2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.
Moles= Molarity x Volume
Volume = 38.5ml = 0.0385L
Moles of Ca hydroxide = 0.150m/L x 0.0385L
(Notice the units canceling out...leaving moles).
=0.005775moles of Ca(OH)2.
From balanced reaction...
3moles of Ca(OH)2 completely reacts with 2moles of H3PO4
0.005775moles of Ca(OH)2 would completely react with....
= 0.005775 x 2/(3)
=0.00385moles of H3PO4.
Now we're looking for its Concentration in Mol/L
Molarity=Moles of solute/Volume of solution(in L)
Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L
Molarity = 0.00385/0.0535
=0.072Mol/L.
If this is wrong
then Simply Try The formula for Mixing of solutions
C1V1 = C2V2
0.15 x 38.5 = C2 x (15+38.5)
C2 = 0.11M/L.
