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
It would be 335 J and of energy is released
Answer:
positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).