The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.
The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.
If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).
In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.
As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.
In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.
Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.
Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² / 0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.
Answer: The energy of activation for the chirping process is 283.911 kJ/mol
Explanation:
According to the Arrhenius equation,
The expression used with catalyst and without catalyst is,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate of reaction at 
= 194/min
= rate of reaction at 
= 47.6 /min
= activation energy
R = gas constant = 8.314 J/Kmol
tex]T_1[/tex] = initial temperature = 
tex]T_1[/tex] = final temperature = 
Now put all the given values in this formula, we get
![\frac{194}{47.6}=\frac{E_a}{2.303\times 8.314}[\frac{1}{278}-\frac{1}{301}]](https://tex.z-dn.net/?f=%5Cfrac%7B194%7D%7B47.6%7D%3D%5Cfrac%7BE_a%7D%7B2.303%5Ctimes%208.314%7D%5B%5Cfrac%7B1%7D%7B278%7D-%5Cfrac%7B1%7D%7B301%7D%5D)
Thus the energy of activation for the chirping process is 283.911 kJ/mol
Answer:
The formula used to calculate heat of fusion:
q = m·ΔH f
Explanation:
The formula used to calculate heat of fusion:
q = m·ΔH f
