The enthalpy of the creation of Methanol is negative. This means heat is released when the reaction proceeds. You would want to use a low temperature.
Answer:
e. adiabatic process
Explanation:
Adiabatic process -
In the thermodynamic system , an adiabatic process is the one which involves no transfer of mass or heat of the substance , is referred to adiabatic process.
In this process , the temperature need not be constant ,
But only the heat is transferred into or out of the system .
Hence, from the given information of the question,.
The correct option is e. adiabatic process .
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
Answer:
Sodium bicarbonate
Explanation:
Sodium bicarbonate ( NaHCO₃ ) -
Sodium bicarbonate , according to the IUPAC nomenclature , its name is sodium hydrogen carbonate ,and in common terms also refereed to as baking soda .
It is a white crystalline solid , it is basic in nature .
<u>The cation and anion of this salt are the sodium ion ( Na⁺) and the anion bicarbonate anion (HCO³⁻) .</u>
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Answer: D
Explanation:
the atom of the metal loses one electron which becomes delocalised and is attragted by the positive nucleus leading to formation of metallic bond.
