Answer:
Silver
Explanation:
I remember this because Ag can mean "Ain't Gold"
Reactions happen instantly even if some are slower than others they all happen instantly so the answer would be A.
Here's link to the answer:
bit.
ly/3a8Nt8n
Answer:
1.12M
Explanation:
Given parameters:
Volume of solution = 2.5L
Mass of Calcium phosphate = 600g
Unknown:
Concentration = ?
Solution:
Concentration is the number of moles of solute in a particular solution.
Now, we find the number of moles of the calcium phosphate from the given mass;
Formula of calcium phosphate = Ca₃PO₄
molar mass = 3(40) + 31 + 4(16) = 215g/mol
Number of moles of Ca₃PO₄ =
= 2.79moles
Now;
Concentration =
Concentration =
= 1.12M
Answer : The value of
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,
![2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)](https://tex.z-dn.net/?f=2H_2S%28g%29%2B3O_2%28g%29%5Crightarrow%202H_2O%28g%29%2B2SO_2%28g%29)
First we have to calculate the enthalpy of reaction
.
![\Delta H^o=H_f_{product}-H_f_{reactant}](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3DH_f_%7Bproduct%7D-H_f_%7Breactant%7D)
![\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)
![\Delta H^o=-1035.6kJ=-1035600J](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D-1035.6kJ%3D-1035600J)
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction
.
![\Delta S^o=S_f_{product}-S_f_{reactant}](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3DS_f_%7Bproduct%7D-S_f_%7Breactant%7D)
![\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)
![\Delta S^o=-153J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D-153J%2FK)
Now we have to calculate the Gibbs free energy of reaction
.
As we know that,
![\Delta G^o=\Delta H^o-T\Delta S^o](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5CDelta%20H%5Eo-T%5CDelta%20S%5Eo)
At room temperature, the temperature is 500 K.
![\Delta G^o=(-1035600J)-(500K\times -153J/K)](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%28-1035600J%29-%28500K%5Ctimes%20-153J%2FK%29)
![\Delta G^o=-959100J=-959.1kJ](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-959100J%3D-959.1kJ)
Therefore, the value of
for the reaction is -959.1 kJ