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Answer:
Enthalpy change is 6007 J
Entropy change is ![20.15\,JK^{-1}](https://tex.z-dn.net/?f=20.15%5C%2CJK%5E%7B-1%7D)
Gibb's free energy change is 0 J.
Explanation:
From the given,
The enthalpy change for the melting of ice = ![60007\,J\,mol^{-1}](https://tex.z-dn.net/?f=60007%5C%2CJ%5C%2Cmol%5E%7B-1%7D)
Temperature = ![25^{0}C](https://tex.z-dn.net/?f=25%5E%7B0%7DC)
Let's convert the temperature centigrade into Kelvin.
![=25^{0}C \,+273.15\,=298.15K](https://tex.z-dn.net/?f=%3D25%5E%7B0%7DC%20%5C%2C%2B273.15%5C%2C%3D298.15K)
Number of moles of ice = 1.00 mol
Enthalpy change of 1.00 mol
temperature
![\bigtriangleup H=(1\,mol)(\frac{6007\,J}{mol})](https://tex.z-dn.net/?f=%5Cbigtriangleup%20H%3D%281%5C%2Cmol%29%28%5Cfrac%7B6007%5C%2CJ%7D%7Bmol%7D%29)
Therefore, Enthalpy change (
) for the melting of 1.00 mole of ice at
temperature is 6007 J.
Entropy change for the melting of 1.00 mole of ice at
:
![\Delta S_{sys}\,=\frac{\Delta H}{T}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsys%7D%5C%2C%3D%5Cfrac%7B%5CDelta%20H%7D%7BT%7D)
![=\,\frac{6007\,J}{298.15\,K}\,=20.15\,JK^{-1}](https://tex.z-dn.net/?f=%3D%5C%2C%5Cfrac%7B6007%5C%2CJ%7D%7B298.15%5C%2CK%7D%5C%2C%3D20.15%5C%2CJK%5E%7B-1%7D)
Therefore, Entropy change(
) for the melting of 1.00 mole of ice at
temperature is
.
The Gibb's free energy change is expressed by the following formula.
![\Delta G^{o}\,=\Delta H^{o}-T\bigtriangleup S](https://tex.z-dn.net/?f=%5CDelta%20G%5E%7Bo%7D%5C%2C%3D%5CDelta%20H%5E%7Bo%7D-T%5Cbigtriangleup%20S)
![=\,6007\,J-(298.15\,K)(20.15\,J\,K^{-1})](https://tex.z-dn.net/?f=%3D%5C%2C6007%5C%2CJ-%28298.15%5C%2CK%29%2820.15%5C%2CJ%5C%2CK%5E%7B-1%7D%29)
![=6007\,J-6007\,J=\,0J](https://tex.z-dn.net/?f=%3D6007%5C%2CJ-6007%5C%2CJ%3D%5C%2C0J)
Therefore, Gibb's free enrgy change (
) for the melting of 1.00 mole of ice at
temperature is ![0J](https://tex.z-dn.net/?f=0J)
Answer:
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Explanation:
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