Answer:
Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:
Which occurs in aqueous phase, therefore, the law of mass action is written by:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
That in terms of the change 
due to the reaction's extent we can write:
But we prefer to compute the Kb due to its exceptional weakness:
Next, the acid dissociation in the presence of the base we have:
![Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E-%5D%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D1x10%5E%7B6%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.1-x%7D)
Whose solution is 
which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
![pOH=-log([OH^-])=-log(0.0999)=1](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.0999%29%3D1)
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:
Regards.
Explanation:
atom X
it is neutral
and may not exist independently
atom X+1
it is ion which has a charge on it positive or negative
it exists independently
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
