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:
the answer I think is 76.5L
If the half life is 5800 years, and there was only 1/4 of the Carbon 14 left in the wood remains, then 1/2 of the 1/2 life or 1/4 is 2900 years so the total age of the wood should be 5800 + 2900 = 8700 years old. We found some ancient highly compressed and water logged wood branches in an old glacial lake in the Highland Valley of BC., Canada and the lake deposits were up to 40,000 yrs old or just past the limit of Carbon 14 dating.
B. sounds like the most reasonable answer.
With an increase of people studying space exploration and sharing their discoveries, then it is likely for it to inspire other people in study science as well.
Answer:
20.446cm3
Explanation:
The following were obtained from the question given:
V1 (initial volume) = 24.568 cm3
T1 (initial temperature) = 25°C = 25 + 273 = 298K
T2 (final temperature) = - 25°C = - 25 + 273 = 248K
V2 (final volume) = ?
Since the pressure is constant, the gas is obeying Charles' law. With the Charles' law equation V1/T1 = V2/T2, we can easily calculate the final volume as follow:
V1/T1 = V2/T2
24.568/298 = V2/248
Cross multiply to express in linear form as shown below:
298 x V2 = 24.568 x 248
Divide both side by 298
V2 = (24.568 x 248) /298
V2 = 20.446cm3
From the calculations made above, the volume of the gas at -25°C is 20.446cm3