Answer:
The change in entropy of the surrounding is -146.11 J/K.
Explanation:
Enthalpy of formation of iodine gas = 
Enthalpy of formation of chlorine gas = 
Enthalpy of formation of ICl gas = 
The equation used to calculate enthalpy change is of a reaction is:
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28ICl%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28I_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Cl_2%29%7D%29%5D)
![=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol](https://tex.z-dn.net/?f=%3D%5B2%5Ctimes%2017.78%20kJ%2Fmol%5D-%5B1%5Ctimes%200%20kJ%2Fmol%2B1%5Ctimes%2062.436%20kJ%2Fmol%5D%3D-26.878%20kJ%2Fmol)
Enthaply change when 1.62 moles of iodine gas recast:

Entropy of the surrounding = 

1 kJ = 1000 J
The change in entropy of the surrounding is -146.11 J/K.
Answer:
I had the same question and I put a total lunar eclipse.
Explanation:
A total lunar eclipse would be less widely visible because, in a partial lunar eclipse the moon only has to partly be in the Earth's shadow.
I don't know if this is helpful or not, but this is what I put if you still needed it.
Still stuck? Get 1-on-1 help from an expert tutor now.
<u>Answer</u>: Light
<em>Computer is an example of light energy which is the third option out of the given four choices.
</em>
<u>Explanation:</u>
We know how a computer works it takes in <em>the electrical energy</em> and does a <em>lot of mathematical mechanical work</em> and for giving answers. It uses a screen on which light blinks in pattern such that it represents letters or mathematical numbers or expressions.
Hence by using this statement we can say <em>computer converts electrical energy into light energy.
</em>
About 5 millimeters which weighs about 5 grams. Hope this helps! ^-^
Answer: The yield of dibromide product will be approximately one‑half of the expected yield.
Explanation: