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: Mass Of CFC that needs to evaporate for the freezing of water = 328.24 g
Explanation: Heat gained by the CFC = Heat lost by water
Heat lost by water = Heat required to take water's temperature to 0°c + Heat required to freeze water at 0°c
Heat required to take water's temperature from 33°c to 0°c = mCΔT
m = 201g, C = 4.18 J/(gK), ΔT = 33
mCΔT = 201 × 4.18 × 33 = 27725.94 J
Heat required to freeze water at 0°c = mL
m = 201g, L = 334 J/g
mL = 201 × 334 = 67134 J
Heat gained by CFC to vaporize = mH = 27725.94 + 67134 = 94859.94 J
H = 289 J/g, m = ?
m × 289 = 94859.9
m = 328.24 g
QED!!
False. Because Science is really a primarily concerned with understanding how the natural world works.
Answer: D.Solar power produces no carbon dioxide
Explanation:
