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:
1335.12 mL of H2O
Explanation:
To calculate the mililiters of water that the solution needs, it is necessary to know that the volume of the solution is equal to the volume of the solute (NaOH) plus the volume of the solvent (H2O).
From the molarity formula we can first calculate the volume of the solution:
The volume of the solution as we said previously is:
Solution volume = solute volume + solvent volume
To determine the volume of the solute we first obtain the grams of NaOH through the molecular weight formula:
Now with the density of NaOH the milliliters of solute can be determined:
Having the volume of the solution and the volume of the solute, the volume of the solvent H2O can be calculated:
Solvent volume = solution volume - solute volume
Solvent volume = 1429 mL - 93.88 mL = 1335.12 mL of H2O
I believe that the balanced chemical reaction is:
C6H12O6 + 6 O2 → 6 CO2
+ 6 H2O
So the number of grams
of oxygen required is:
mass O2 required = 48
g C6H12O6 * (1 mole C6H12O6 / 180.16 g) * (6 mole O2 / 1 mole C6H12O6) * (32
grams O2 / 1 mole)
<span>mass O2 required =
51.15 grams</span>
You're able to figure this out by using the ideal gas equation PV=nRT Where R=.08 (L*atm/mol*K) Convert 37 C to Kelvin=310 K and 102 kPa to atm=roughly 1 atm Manipulate the ideal gas equation to n=PV/RT Plug in your values (1 atm * 6 L) / (.08 (L*atm/mol*K) * 310 K) All units should cancel except for mol and you should get .24 mol. Multiply the molar mass of air by your answer. .24 mol * 29 g/mol moles should cancel and you should get 6.9 g.
