Answer:
![r=25M^{-1}s^{-1}[A]^2](https://tex.z-dn.net/?f=r%3D25M%5E%7B-1%7Ds%5E%7B-1%7D%5BA%5D%5E2)
Explanation:
Hello there!
In this case, according to the given information for this chemical reaction, it is possible for us to set up the following general rate law and the ratio of the initial and the final (doubled concentration) condition:
![r=k[A]^n\\\\\frac{r_1}{r_2} =\frac{k[A]_1^n}{k[A]_2^n}](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5En%5C%5C%5C%5C%5Cfrac%7Br_1%7D%7Br_2%7D%20%3D%5Cfrac%7Bk%5BA%5D_1%5En%7D%7Bk%5BA%5D_2%5En%7D)
Next, we plug in the given concentrations of A, 0.2M and 0.4 M, the rates, 1.0 M/s and 4.0 M/s and cancel out the rate constants as they are the same, in order to obtain the following:

Which means this reaction is second-order with respect to A. Finally, we calculate the rate constant by using n, [A] and r, to obtain:
![k=\frac{r}{[A]^n} =\frac{1.0M/s}{(0.2M)^2}\\\\k=25M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Br%7D%7B%5BA%5D%5En%7D%20%3D%5Cfrac%7B1.0M%2Fs%7D%7B%280.2M%29%5E2%7D%5C%5C%5C%5Ck%3D25M%5E%7B-1%7Ds%5E%7B-1%7D)
Thus, the rate law turns out to be:
![r=25M^{-1}s^{-1}[A]^2](https://tex.z-dn.net/?f=r%3D25M%5E%7B-1%7Ds%5E%7B-1%7D%5BA%5D%5E2)
Regards!
Option 4. ratio of electrons to protons
Isotope that has atomic number of 82 is stable. An element that has an atomic number ∠82 more stable except Tc and Pm. Also there is the concept that isotopes consisting a combination of even-even,even-odd,odd-even, and odd-odd are all stable. Many isotopes with no magic numbers of nucleons are stable