Answer:
1. 
was the 
value calculated by the student.
2. 
was the 
of ethylamine value calculated by the student.
Explanation:
1.
The 
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_a=\frac{[As^-][H^+]}{[HAs]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BAs%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHAs%5D%7D)
:
was the value calculated by the student.
2.
The value of ethylamine = 11.87
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BC_2H_5NH_3%5E%7B%2B%7D%5D%5BOH%5E-%5D%7D%7B%5BC_2H_5NH_2%5D%7D)
:
was the of ethylamine value calculated by the student.
Oil is less dense than water, so the difference would be its density. Water is a good solvent, which means It can dissolve other substances.
Incorrect, temperature is directly proportional to the avg. KE of a gas.
Answer: 27 is A and 28 is C.
Explanation: I’ll explain 28 but not 27 because that’s just definitions.
In CuSO4 there is a Cu, an S, and 4 O molecules. Add them up you get 6.
