Answer: False. Hope this helps!!
Answer : The process is not spontaneous.
Explanation :
As, we know that:
Change in entropy = Change in entropy of system + Change in entropy of surrounding
As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.
For the given reaction to be spontaneous, the total change in entropy should be positive.
Given :
Entropy change of system = +125J/K
Entropy change of surroundings = -125J/K
Total change in entropy = Entropy change of system + Entropy change of surroundings
Total change in entropy = 125 J/K + (-125 J/K)
Total change in entropy = 0
The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.
The mean is simply the arithmetic average of all your raw data. This can be solved methodically by summing up all of the raw data points that you have. Take note how many raw data points you used, because this will be used to divide the sum. You will obtain the mean.
Answer: (3) 15
Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.
Answer:
Equilibrium constant of the given reaction is 
Explanation:
....
....
The given reaction can be written as summation of the following reaction-


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Equilibrium constant of this reaction is given as-
![\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%5BBr_%7B2%7D%5D%7D)
![=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B%5BNOBr%5D%7D%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%29%5E%7B2%7D%28%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%29)

