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
The temperature , <u>165 degree F</u> should soup that contains cooked beef be reheated for hot-holding.
Food should reach a temperature serving 135°F before it can be opened in the restaurant and warmed for hot holding. Within two hours, quickly reheat the food. If food is being served right away, it can be served anywhere at temperature as long that it as it was cooked and chilled appropriately.
TCS food must be totally chilled to 41°F or even below in 6 hours after being thoroughly cooled between 135°F to 70°F in 2 hours. Within four hours, TCS food made with ingredients at room temperature needs to be chilled to 41°F or lower.
Therefore, the temperature , <u>165 degree F</u> should soup that contains cooked beef be reheated for hot-holding.
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Answer:
38.3958 °C
Explanation:
As,
1 gram of carbohydrates on burning gives 4 kilocalories of energy
1 gram of protein on burning gives 4 kilocalories of energy
1 gram of fat on burning gives 9 kilocalories of energy
Thus,
27 g of fat on burning gives 9*27 = 243 kilocalories of energy
20 g of protein on burning gives 4*20 = 80 kilocalories of energy
48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy
Total energy = 515 kilocalories
Using,

Given: Volume of water = 23 L = 23×10⁻³ m³
Density of water= 1000 kg/m³
So, mass of the water:
Mass of water = 23 kg
Initial temperature = 16°C
Specific heat of water = 0.9998 kcal/kg°C

Solving for final temperature as:
<u>Final temperature = 38.3958 °C </u>
The crust of the Earth is thickest beneath the continents.