Second law of thermodynamics states entropy of the universe is always increasing. entropy always increases in a spontaneous process .
option B) second law of thermodynamics
for example if we keep a hot cup of coffee the heat is lost to surrounding and entropy( randomness increases) since it is a natural process entropy of universe increases.
Kdd was he e shxiuw r thai ed
Answer:
For the purposes of your question, we can think of speed and velocity as being the same thing. Therefore, the potential energy of an object is proportional to the square of its velocity (speed). In other words, If there is a twofold increase in speed, the potential energy will increase by a factor of four.
Answer:
To prevent the mixture from separating substances called emulsifiers can be added. These help to form and stabilise the emulsions, preventing or slowing the water and fat/oil from separating. ... The hydrophilic end of the emulsifier molecule is attracted to the water and the hydrophobic end is attracted to the fat/oil.
Answer : The entropy change of reaction for 1.62 moles of 
reacts at standard condition is 217.68 J/K
Explanation :
The given balanced reaction is,
The expression used for entropy change of reaction 
is:
![\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BBr_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28Br_2%29%7D%2Bn_%7BF_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28F_2%29%7D%5D-%5Bn_%7BBrF_3%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28BrF_3%29%7D%5D)
where,
= entropy change of reaction = ?
n = number of moles
= standard entropy of formation
= 245.463 J/mol.K
= 202.78 J/mol.K
= 292.53 J/mol.K
Now put all the given values in this expression, we get:
![\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B1mole%5Ctimes%20%28245.463J%2FK.mole%29%2B3mole%5Ctimes%20%28202.78J%2FK.mole%29%7D%5D-%5B2mole%5Ctimes%20%28292.53J%2FK.mole%29%5D)
Now we have to calculate the entropy change of reaction for 1.62 moles of reacts at standard condition.
From the reaction we conclude that,
As, 2 moles of has entropy change = 268.74 J/K
So, 1.62 moles of has entropy change = 
Therefore, the entropy change of reaction for 1.62 moles of reacts at standard condition is 217.68 J/K
