You are given pure samples of pentane, CH 3CH 2CH 2CH 2CH 3( l), and 1,3-pentadiene, CH 2=CHCH=CHCH 3( l). What prediction would you make concerning their standard molar entropies at 298 K? S°pentane < S°1, 3-pentadiene S°pentane > S°1, 3-pentadiene S°pentane = S°1, 3-pentadiene + 2 S°H2 S°pentane ≈ S°1, 3-pentadiene More information is needed to make reasonable predictions.

Complete Question

The complete question is shown on the first uploaded image  

option A is the correct option

Explanation:

Generally the more the disorder the higher the entropy of a gas  and secondly the more the atom of a compound present the more the disorder

Now from the question we can deduce that pentane has more atoms than 1,3-pentadiene so the pentane would have a higher disorder than 1,3-pentadiene which implies that the pentane will have a higher entropy than 1,3-pentadiene

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

hope i !