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
C6H15, C2H5 has a molar mass of 29g/mol. 87 divided by 29 is 3. Then multiply each element subscript by 3
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
Answer:
5.0 38 84.0 749.7 528.0 729.0 738.9 739.0