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.
Answer:
https://www.chegg.com/homework-help/questions-and-answers/1-10-points-solution-15-percent-mass-kcl-benzene-new-boiling-point--901-c-b-921-c-c-821-c--q63751186
Explanation: Thats your answer
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
What items are you separating? Please be more specific with your answer!
