H₃O⁺ = 
×
OH = 
×
pH = 2.22
pOH = 11.78
<h3>What is pH?</h3>
The term pH, which originally stood for "potential of hydrogen" (or "power of hydrogen"), is used in chemistry to describe how acidic or basic an aqueous solution is. Lower pH values are summarized for acidic solutions (solutions with higher H+ ion concentrations) than for basic or alkaline solutions.
The pH scale is inversely indicates to the concentration of hydrogen ions in the solution and is logarithmic.
⇒pH = -log(
)
Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C (77 °F). At this temperature, solutions with a pH of 7 are neutral (e.g. pure water). The pH neutrality relies on temperature, falling below 7 if the temperature rises above 25 °C.
Learn more about pH
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Answer: The image from the question has the correct answers.
Explanation:
As summarized in the attached table.
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:
The value of the missing equilibrium constant ( of the first equation) is 1.72
Explanation:
First equation: 2A + B ↔ A2B Kc = TO BE DETERMINED
⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]
Second equation: A2B + B ↔ A2B2 Kc= 16.4
⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]
Third equation: 2A + 2B ↔ A2B2 Kc = 28.2
⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²
If we add the first to the second equation
2A + B + B ↔ A2B2 the equilibrium constant Kc will be X(16.4)
But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2) with Kc = 28.2
So this means: 28.2 = X(16.4)
or X = 28.2/16.4
X = 1.72
with X = Kc of the first equation
The value of the missing equilibrium constant ( of the first equation) is 1.72
Answer:
it is directly related to the intermolecular forces present between its molecules
