Explanation:
The given data is as follows.
= 286 kJ = 
= 286000 J
, 
Hence, formula to calculate entropy change of the reaction is as follows.
= ![[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20S_%7BO_%7B2%7D%7D%29%20-%20%281%20%5Ctimes%20S_%7BH_%7B2%7D%7D%29%5D%20-%20%5B1%20%5Ctimes%20S_%7BH_%7B2%7DO%7D%5D)
= ![[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20205%29%20%2B%20%281%20%5Ctimes%20131%29%5D%20-%20%5B%281%20%5Ctimes%2070%29%5D)
= 163.5 J/K
Therefore, formula to calculate electric work energy required is as follows.
= 
= 237.277 kJ
Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.
The correct answer to this question is that the length of 14 is it’s half Which would be 7
D.) the original components retain their individual properties
Answer:
C. 0.4.
Explanation:
<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>
<em></em>
- no. of moles of acetic acid = 2, no. of moles of water = 3.
- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.
<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>
The answer would be D because from my research it's the only one that didn't have a catalyst
