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:
it could be 25degrees c... not sure
The standard addition equation is as followsI_(S+X) (V/V_O )=I_X+I_X/[X]_i [S]_4 (V_S/V_0 ) Here, [X]_i is the initial concentration of analyte, [S]_i is the initial concentration of standard, I_X is signal for analyte, I_(S+X) is signal for standard and analyte, V_0 is the initial volume, V_S is the added standard volume, and V is the total volume.Added volume of standard V_S is-23.3 mL. Initial volume of the sample V_0 is 10.00 mL. Initial concentration of standard ([S]_i) is 0.156 ng/mL.[X]_i= -[S]_i (V_S/V_0 )〖[X]〗_(i )= -(0.156 ng/mL)((-23.3 mL)/(10.00 mL))=0.363 ng/mL
Concentration of U(III) in ground sample is 0.363 ng/mL
Answer:15 +18
Explanation:15 fuel +18 oxygen + 33 cd