It would help if I could see the specific answer choices, but an environment that would have lots of oxygen, water, decaying organisms and sunlight would be the answer (ex. the forest or riverbed)
Answer:- 
Explanations:- The solution we have is a buffer solution and we know that a buffer solution resists a change in its pH if a strong acid or base is added to it.
Here, the buffer solution we have is of a weak base and it's conjugate acid. So, a strong acid(nitric acid) is added to this buffer then it reacts with the base present in the buffer so that the acid could be neutralized. This is called buffer action.
The net ionic equation is written as:

Note that
is a strong acid and nitrate ion is the spectator ion so it is not included in the net ionic equation.
Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Best regards!