Answer:
The answer is C
Explanation:
This is the answer because it is the only way that it is okay to split in half.
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
Answer:
it can be revised at any time.
Explanation:
Answer:
28.1
Explanation:
Divide the percentage abundances by 100:
92.2% ÷ 100 = 0.922
4.7% ÷ 100 = 0.047
3.1% ÷ 100 = 0.031
multiply them by their corresponding mass number:
0.922×28 = 25.816
0.047×29 = 1.363
0.031×30 = 0.93
add them all together to get your final answer:
25.816 + 1.363 + 0.93 = 28.109
to one decimal place:
28.1
I hope that helps!!!
The correct answer should be: the molar amount of the gas.
