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' = ?
= 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
I can't explain the like dissolves like part but the water and oil I can
when oil is poured into water the oil will stay at the top of the water meaning that water is more dense than oil
Answer:
Personal Question
Explanation:
I wish I could help more, but it's based on your own prior knowledge, desires and experiences. Only you can answer those questions. For the first one, if you know examples of genetic engineering you could write that, or you could define it. For the second one, you could talk about things that interest you. The last one you could talk about the concepts you learned in the lesson and the things you didn't know before the lesson and then learned afterward. You could talk about the methods of genetic manipulation, or just what you learn in the class.
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
![Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change
, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

Thus, we obtain a quadratic equation whose solution is:

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.