Answer: The differences is Air masses cover over thousands and hundreds and millions of square kilo. A front is a boundary which two air masses but different temperature and moisture content meet.
Explanation:
Answer:
The independent variable is what you change in the experiment.
Explanation:
The independent determines the dependent variable.
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Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
