First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
In order to float the demsoty of the floating obkect must be lower then the liquid it is in. the only one that works for is Citric acid
Answer:
W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.
Explanation:
heat engine:
∴ Th = 485°C
∴ Tc = 42°C
∴ Qh = 9.75 KJ......heat from hot source
first law:
∴ ΔU = Q + W = 0 .........cyclic process
⇒ Q = - W
∴ Q = Qh + Qc = Qh - (- Qc)
∴ Qc: heat from the cold source ( - )
⇒ Qh - ( - Qc) = - W..............(1)
⇒ Qc/Qh = - Tc/Th...........(2)
from (2):
⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)
⇒ Qc = - 0.8443 KJ
replacing in (1):
⇒ - W = 9.75 KJ - ( - 0.8443 KJ)
⇒ - W = 10.5943 KJ
