Answer: Concentration of
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of
= 0.729 M
The given balanced equilibrium reaction is,

Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :



concentration of
in the equilibrium mixture = 
Thus concentration of
in the equilibrium mixture is 0.31 M
Answer:
Rock layer and sandwich layer is almost the same.
Why are rock layers are like sandwiches?
This geologic structure is called an anticline. Now gently bend your sandwich so that the layers are bowed downwards, thus creating a syncline. The presence of anticlines and synclines indicate strong forces that often form mountain ranges.
Answer:
10.5 g
Explanation:
Step 1: Given data
- Molar concentration of the solution (C): 0.243 M
- Volume of solution (V): 0.580 L
Step 2: Calculate the moles of solute (n)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
n = M × V
n = 0.243 mol/L × 0.580 L = 0.141 mol
Step 3: Calculate the mass corresponding to 0.141 moles of KCl
The molar mass of KCl is 74.55 g/mol.
0.141 mol × 74.55 g/mol = 10.5 g
Answer:
T =76.13 K
Explanation:
Given data:
Temperature of gas = ?
Volume of gas = 250 mL(250/1000 = 0.25 L)
Mass of helium = 0.40 g
Pressure of gas = 253.25 kpa (253.25/101 = 2.5 atm)
Solution:
Formula:
PV = nRT
First of all we will determine the number of moles of helium.
Number of moles = mass/ molar mass
Number of moles = 0.40 g/ 4 g/mol
Number of moles = 0.1 mol
Now we will put the values.
R = general gas constant = 0.0821 atm.L/ mol.K
T = PV/nR
T =2.5 atm× 0.25 L /0.1 mol ×0.0821 atm.L/ mol.K
T = 0.625 /0.00821/K
T =76.13 K
As we know, NaOH is a strong Base and completely dissociates in water as given below,
NaOH + H₂O → Na⁺ ₍aq₎ + OH⁻ ₍aq₎
So, the concentration of OH⁻ Ions will be the same as that the concentration of NaOH i.e. 0.018 M.
So,
[OH⁻] = 0.018 M