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:
Book on Table
Car at the Hilltop
Falling Objects
Skydiver
Hammering a Nail
Dam Water
Roller Coaster
Stretched Rubber Band
Simple Pendulum
Compressed Spring
Battery
Flashlight
Exothermic Chemical Reaction
Burning of Oil, Gas and Coal
Wind Turbine
Explanation:
have a nice day!
Chromium , silver, zinc...
Answer:1 mol of Mg(NO3)2 contains 6.022*10^23 molecules
3 mol Mg(NO3)2 contains 3*6.022*10^23 = 1.81*10^24 molecules
Each Mg(NO3)2 molecule contains 2 N atoms
Number of N atoms = 2*1.81*10^24 = 3.62*10^24 N atoms.
Answer:
See explanation
Explanation:
In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.
As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.
However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.
This accounted for their scattering through large angles throughout the foil in all directions.