Step one write the equation for dissociation of AgNO3 and NaCl
that is AgNO3-------> Ag+ + NO3-
NaCl--------> Na+ + Cl-
then find the number of moles of each compound
that is for AgNO3 = ( 1.4 x10^-3 ) x 25/1000= 3.5 x10^-5 moles
Nacl= (7.5 x10^-4)x 60/1000= 4.5 x10^-5 moles
from mole ratio the moles of Ag+= 3.5 x10^-5 moles and that of Cl-= 4.5 x10^-4 moles
then find the total volume of the mixture
that is 25ml + 60 Ml =85ml = 0.085 liters
The Ksp of Agcl = (Ag+) (cl-), let the concentration of Ag+ be represented by x and also the concentration be represented by x
ksp (1.8 x10^-10) is therefore= x^2
find the square root x=1.342 x10^-5
Ag+ in final mixture is = moles of Ag+/total volume - x
that is {(3.5 x10^-5)/0.085} - 1.342 x10^-5=3.98x10^-4
Cl- in the final mixture is =(4.5 x10^-5 /0.085) - 1.342 x10^-5= 5.16 x10^-4
The difference between the calculated voltage (based on standard potentials) and the actual voltage required to cause electrolysis is called overvoltage.
<h3>Electrolysis </h3>
Electrolysis is the technique of causing a chemical change in a substance by passing an electric current through it. The substance either loses or gets an electron during the chemical transition (oxidation or reduction). The procedure is carried out in an electrolytic cell, a device made up of positive and negative electrodes that are kept apart and submerged in a solution with ions that are both positively and negatively charged. The chemical that needs to be converted might either be dissolved in the solution or could form the electrode. The negatively charged electrode (cathode) receives electrical current (i.e., electrons), which travels there and combines with the components of the solution to convert them (reduced).
Learn more about electrolysis here:
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Because they cannot survive long outside a living host cell.
Answer:
[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹
Explanation:
The balanced equation is
H₂O + Cl₂O ⇌ 2HOCl
Data:
Kc = 0.0900
[H₂O] = 0.004 32 mol·L⁻¹
[Cl₂O] = 0.004 32 mol
1. Set up an ICE table.
2. Calculate the equilibrium concentrations
![K_{\text{c}} = \dfrac{\text{[HOCl]$^{2}$}}{\text{[H$_{2}$O][Cl$_2$O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHOCl%5D%24%5E%7B2%7D%24%7D%7D%7B%5Ctext%7B%5BH%24_%7B2%7D%24O%5D%5BCl%24_2%24O%5D%7D%7D%20%3D%20%5Cdfrac%7B%282x%29%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%3D%200.0900%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B4x%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%26%3D%26%200.0900%5C%5C%20%5Cdfrac%7B2x%20%7D%7B0.00432%20-%20x%7D%20%26%20%3D%20%26%200.300%5C%5C2x%20%26%20%3D%20%26%200.300%280.00432%20-%20x%29%5C%5C2x%20%26%20%3D%20%26%200.001296%20-%200.300x%5C%5C2.300x%20%26%20%3D%20%26%200.001296%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B5.63%5Ctimes%2010%5E%7B-4%7D%7D%5C%5C%5Cend%7Barray%7D)
[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹
[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹
Check:
OK.
To know the electrostatic force between two charges or between two ions, you can use the Coulomb's Law. The equation is F = k*q1*q1/r^2, where F is the electrostatic force, q1 and q2 are the charger for Na and Cl, and r is the distance between the centers of both atoms. In literature, the distance is 0.5 nm or 0.5 x 10^-9 meters. The charge for Na+ and Cl- is the same magnitude but different in sign. Since Na+ is a cation, its charge is +1.603x10^-19 C (the charge of an electron). For Cl- being an anion, its charge is -1.603x10^-19 C. The constant k is an empirical value equal to 9x10^9. Using the formula:
F = (9x10^9)(+1.603x10^-19)(-1.603x10^-19)/(0.5 x 10^-9)^2
F = -9.25 x 10^-10 Newtons
The negative denotes that the net force is more towards the Cl- ion.
