As per Le Chatelier principle, when a system in equilibrium is disturbed, the reaction will try to compensate the change to restore the equilibrium.
This reaction occurs in gas phase, so the volume is inversely proportional to the pressure.
Then a decrease in volume will cause an increase in pressure, so the system will tend to react in the direction that compensates this increase, this is the system will try to reduce the number of moles of particles to reduce the increase of the pressure.
As you see, there are 3 particles of products (2 of NO and 1 of Br) for every 2 particles of reactant (NOBr).
That means, that the equilibrium will displace to the left, this is the concentration of NOBr will increase while the concentration of NO and Br will decrease.
Answer:
The electronengativity values of given elements is as follows.
Fluorine - 4
Chlorine -3
Bromine - 2.9
Iodine- 2.5
Explanation:
Electronegativity =consant (I.E-E.A)
The electron affinity and ionization energy values of the given elements is as follows.
(In attachment)
First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".
<u>Fluorine:</u>
![4=constant[1678-(-327.8)]](https://tex.z-dn.net/?f=4%3Dconstant%5B1678-%28-327.8%29%5D)

By using this constant values we can find electronegatvity values of remaining elements.
<u>Chlorine:</u>
![Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1255%2B348.7%5D%3D3.1980%5Csim%203)
Therefore, electronegativity of chlorine is 3.
<u>Bromine:</u>
![Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1138%2B324.5%5D%3D2.91%5Csim%202.9)
Therefore, electronegativity of bromine is 2.9.
<u>Iodine:</u>
![Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1007%2B295.7%5D%3D2.59%5Csim%202.5)
Therefore, electronegativity of iodine is 2.5.
Answer:
613.0 g
Explanation:
2 L = 2000 mL
(30.65 g/100 mL)*2000 mL = 613.0 g
From the calculations performed, the Kb of the reaction is 9.9 * 10^-10
<h3>What is Kb?</h3>
The term Kb refers to the base dissociation constant of a solution. We have the following information from the question;
pH of the solution = 8.60
concentration of p-toluidine = 0.016-M
Recall that pOH = 14 - pH
pOH = 14 - 8.60 = 5.4
[OH] = Antilog (-5.4) = 3.98 * 10^-6 M
We have to set up the ICE table;
CH3C6H4NH2(aq) + H2O(l) ⇄ CH3C6H4NH3^+(aq) + OH^-(aq)
I 0.016 0 0
C -x +x +x
E 0.016 - x 3.98 * 10^-6 3.98 * 10^-6
Hence;
Kb = [3.98 * 10^-6]^2/[0.016 - 3.98 * 10^-6]
Kb = 9.9 * 10^-10
Learn more about Kb: brainly.com/question/12726841