0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
Answer:
I) Change in solubility
II) Change in boiling point
III) Change in colour
Explanation:
A chemical change involves formation of new products and is not reversible.
So, once two liquid solutions are mixed and a chemical change takes place, the new product will have the following:
- a new solubility rate, i.e it will dissolve at a rate different from the two liquid solution
- a new boiling point i.e it takes a new point at which its molecules liberate to yield vapour
- a new colour might be detected, as the individual solution each has its own colour
Answer:
Give any two application for isotopes
Isotopes of iodine are used for radiotherapy in treatment of hyperthyroidism, cancer, etc.
Uranium, Radium, Polonium isotopes are used in atomic reactors.
Cobalt isotopes are used for irradiation of food products.
Answer:
Explanation:
Hello,
In this case, the chemical reaction is:
Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:
In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:
Regards.
Answer : The equilibrium constant for this reaction is, 
Explanation :
The given main chemical reaction is:
; 
The intermediate reactions are:
(1) 
; 
(2) 
; 
We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:
(1) 
; 
(2) 
; 
Thus, the equilibrium constant for this reaction will be:
Thus, the equilibrium constant for this reaction is,
