Answer:
C) In[reactant] vs. time
Explanation:
For a first order reaction the integrated rate law equation is:

where A(0) = initial concentration of the reactant
A = concentration after time 't'
k = rate constant
Taking ln on both sides gives:
![ln[A] = ln[A]_{0}-kt](https://tex.z-dn.net/?f=ln%5BA%5D%20%3D%20ln%5BA%5D_%7B0%7D-kt)
Therefore a plot of ln[A] vs t should give a straight line with a slope = -k
Hence, ln[reactant] vs time should be plotted for a first order reaction.
Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
![Rate = k[AB]^2](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BAB%5D%5E2)
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
Where,
is the initial concentration = 1.50 mol/L
is the final concentration = 1/3 of initial concentration =
= 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-


<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Answer: C
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