Answer:
The rate law for second order unimolecular irreversible reaction is
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
![v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}](https://tex.z-dn.net/?f=v%20%3D%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k.%5BA%5D%5E%7B2%7D)
rearranging the ecuation
![-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Integrating between times 0 to <em>t </em>and between the concentrations of
to <em>[A].</em>
![\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_t%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%5Cint%5Climits%5EA_%7B0%7D%20_A%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Solving the integral
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Li(s) (answer A)
Li is strongest reducing agent because of the lowest standard reduction potential. when something is oxidized, it reduces another substance, becoming a reducing.Hence Lithium is strongest reducing agent. Reducing agent is stronger when it has a more positive oxidation potential.
Answer : The density of an object is, 
Solution : Given,
Mass of an object = 60 g
Volume of an object = 
Formula used :

Now put all the given values in this formula, we get the density of an object.

Therefore, the density of an object is, 
Answer:
a. Phosphoric Acid
b. Acetic Acid
c. Hypochlorous Acid
Explanation:
A buffer works when the pH of this one is in pKa ± 1. That means, to find which buffer system works in some pH you need to find pKa:
pKa = -log Ka
<em>pKa Acetic acid:</em>
-log1.8x10⁻⁵ = 4.74
<em>pKa phosphoric acid:</em>
-log7.5x10⁻³ = 2.12
<em>pKa hypochlorous acid:</em>
-log3.5x10⁻⁸ = 7.46
a. For a pH of 2.8 the best choice is phophoric acid because its effective range is: 1.12 - 3.12 and 2.8 is between these values.
b. pH 4.5. Acetic acid. effective between pH's 3.74 - 5.74
c. pH 7.5. Hypochlorous acid that works between 6.46 and 8.46