Answer:
ΔG° = 41.248 KJ/mol (298 K); the correct answer is a) 41 KJ
Explanation:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)
⇒ Kf = 1.7 E7; T =298K
⇒ ΔG° = - RT Ln Kf.....for aqueous solutions
∴ R = 8.314 J/mol.K
⇒ ΔG° = - ( 8.314 J/mol.K ) * ( 278 K ) ln ( 1.7 E7 )
⇒ ΔG° = 41248.41 J/mol * ( KJ / 1000J )
⇒ ΔG° = 41.248 KJ/mol
#7 is D because you move the decimal 4 places
Answer:
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
Explanation:
Conversion of cyclopropane into propene follows first order kinetics.
The integrated rate of first order kinetic is given by :
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= Initial concentration of reactant
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= final concentration of reactant after time t
k = rate constant of the reaction
We have :
Rate constant of the reaction = k = 
![[A_o]=0.150 M](https://tex.z-dn.net/?f=%5BA_o%5D%3D0.150%20M)
t = 22.0 hour
[A] =?
![[A]=0.150 M\times e^{-5.4\times 10^{-2} hour^{-1}\times 22.hour}](https://tex.z-dn.net/?f=%5BA%5D%3D0.150%20M%5Ctimes%20e%5E%7B-5.4%5Ctimes%2010%5E%7B-2%7D%20hour%5E%7B-1%7D%5Ctimes%2022.hour%7D)
![[A]=0.0457 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0457%20M)
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
<span> the </span>vapor pressure<span> of the liquid at a temperature T</span>2<span> ... Now, </span>it's<span> important to realize that the </span>normal boiling point<span> of a substance is measured at an atmoshperic ... ΔHvap=−ln(</span>134mmHg760mmHg<span> )⋅8.314J mol−1K−1 (1(273.15+</span>0)−1(273.15+40))K−1 ... Give equations that can be used tocalculate<span> the .
Now try it yourself :)</span>
