Answer:
–187.9 J/K
Explanation:
The equation that relates the three quantities is:

where
is the Gibbs free energy
is the change in enthalpy of the reaction
T is the absolute temperature
is the change in entropy
In this reaction we have:
ΔS = –187.9 J/K
ΔH = –198.4 kJ = -198,400 J
T = 297.0 K
So the Gibbs free energy is

However, here we are asked to say what is the entropy of the reaction, which is therefore
ΔS = –187.9 J/K
Its already balanced, no balancing required
Answer:

Explanation:
Hello.
In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:

In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:

Consider that the given initial pressure is also equal to Pa:

Which stands for a pressure increase when volume decreases.
Regards.
Answer:
Each orbit has a specific energy level.
Would you mind marking it the brainliest:).