I think it’s A but I’m not sure.
Answer:
K = 2.96x10⁻¹⁰
Explanation:
Based on the initial reaction:
N2O4 ⇄ 2NO2; K = 1.5x10³
Using Hess's law, we can multiply this reaction changing K:
3 times this reaction:
3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹
The inverse reaction has a K of:
6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;
<h3>K = 2.96x10⁻¹⁰</h3>
Answer:
o atom, smaller, anion is the right answer of the following
Answer: 72L of 30% and 128L of 80%
You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1
The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
0.625v1= 200L - 155L
v1= 45/ 0.625= 72L
v1+v2=200L
v2= 200L- 72L= 128L
We are given the molar concentration of an aqueous solution of weak acid and the pH ofthe solution, and we are asked to determine the value of Ka for the acid.
The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of benzoic acid can be written as seen in the attached image (1).
The equilibrium-constant expression is the equation number (2)
From the measured pH, we can calculate pH as seen in equation (3)
To determine the concentrations of the species involved in the equilibrium, we imagine that thesolution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acidinto H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ionare produced in solution. Because the pH measurement indicates that [H+] = 1x 10^-4 M atequilibrium, we can construct the following table as seen in the equation number (4)
To find the value of Ka, please see equation (5):
We can now insert the equilibrium concentrations into the expression for Ka as seen in equation (6)
Therefore, 2.58x10^-4 M is the concentration of benzoic acid to have a pH of 4.0