B because it shows the COOCH
Here's link to the answer:
tinyurl.com/wpazsebu
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction
.

![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Answer:
D
Explanation:
John is not a very good businessman.
:D
Answer:
A decrease in temperature would decrease kinetic energy, therefore decreasing collisions possible.
Explanation:
A gas at a fixed volume is going to have collisions automatically. If you decrease the temperature (same thing as decreasing kinetic energy) you are cooling down the molecules in the container which gives them less energy and "relaxes" them. This decrease in energy causes them to move around much slower and causing less collisions, at a much slower rate. In a perfect world, these collisions do not slow down the molecule but we know that they do, just a very very small unmeasurable amount.