Answer:
11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.
Explanation:
Concentration of sodium stearate acid : c
Moles of sodium stearate = 
Volume of the solution = 10.0 mL = 0.010 L

![[C_{17}H_{35}COO^-]=c=1.294 M](https://tex.z-dn.net/?f=%5BC_%7B17%7DH_%7B35%7DCOO%5E-%5D%3Dc%3D1.294%20M)

initially c
c 0 0
At equilibrium
(c-x) x x
Dissociation constant of an acid = 
Expression of a dissociation constant of an acid is given by:

Solving for x;
x = 0.0041 M
![[OH^-]=0.0041 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.0041%20M)
The pOH of the solution:
![pOH=-\log[OH^-]=-\log[0.0041 M]=2.39](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B0.0041%20M%5D%3D2.39)
pH = 14 -pOH
pH = 14 - 2.39 = 11.61
11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.
Answer:
In your case, you have 8 protons and 10 electrons. This means that the atom will carry a total 2 xx (1-) = 2- net charge. Therefore, you are dealing with an anion that carries a 2- net charge, i.e. it has two more electrons than it does protons.
Answer:
The new volume of gas would be 30 L.
Explanation:
This is an example of a Combined Gas Laws problem.