Answer:
Enthalpy change for the reaction is -67716 J/mol.
Explanation:
Number of moles of 
in 50.0 mL of 0.100 M of
= Number of moles of HCl in 50.0 mL of 0.100 M of HCl
= 
moles
= 0.00500 moles
According to balanced equation, 1 mol of reacts with 1 mol of HCl to form 1 mol of AgCl.
So, 0.00500 moles of react with 0.00500 moles of HCl to form 0.00500 moles of AgCl
Total volume of solution = (50.0+50.0) mL = 100.0 mL
So, mass of solution = (
) g = 100 g
Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)
= 
= ![\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}](https://tex.z-dn.net/?f=%5Cfrac%7B-100g%5Ctimes%204.18%5Cfrac%7BJ%7D%7Bg.%5E%7B0%7D%5Ctextrm%7BC%7D%7D%5Ctimes%20%5B24.21-23.40%5D%5E%7B0%7D%5Ctextrm%7BC%7D%7D%7B0.00500mol%7D)
= -67716 J/mol
[m = mass, c = specific heat capacity, 
= change in temperature and negative sign is included as it is an exothermic reaction]
Distance and period of time is the correct answer
Hope this helps!
Answer:
118.22 atm
Explanation:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
KP = 0.13 = 
Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.
- With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
- With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.
The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:
The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.
Rewriting KP and solving for PT:
