The reactions based on the absorption and release of the energy are called endothermic and exothermic reactions. The reaction is exothermic.
<h3>What is an exothermic reaction?</h3>
Exothermic reactions are the reaction in which the reactant produces products that release energy from the system to the surroundings. In the reaction bond energy of the reactant is less than the product.
Energy from the system is released in the form of heat, sound, light and electricity. The weak bonds of the compounds are replaced with stronger ones and the standard enthalpy of the reaction is negative.
Therefore, option c. reaction is exothermic is correct.
Learn more about the exothermic reactions here:
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Answer:
C) Both
Explanation:
Whenever we mix any pure form of a compound with some other form of a compound which is not in the other standard pure state, this results in the melting point of mixture to get dispersed and it becomes broad form.
Thus, when a known compound of 3-Nitroaniline mixes with both 3-Nitroaniline and 4-Nitrophenol, the melting point of the compound becomes depressed and board.
Thus the correct option is (C).
Explanation:
Kb: The base ionization constant
2) Ammonia (formula = NH3) is the most common weak base example used by instructors. ... This constant, Kb, is called the base ionization constant. It can be determined by experiment and each base has its own unique value. For example, ammonia's value is 1.77 x 10¯5.
Explanation:
According to Clausius-Claperyon equation,
![ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%20%3D%20%5Cfrac%7B-%5Ctext%7Bheat%20of%20vaporization%7D%7D%7BR%7D%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
The given data is as follows.
= (63.5 + 273) K
= 336.6 K
= (78 + 273) K
= 351 K
= 1 atm, 
= ?
Putting the given values into the above equation as follows.
![ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]](https://tex.z-dn.net/?f=ln%20%28%5Cfrac%7B1.75%20atm%7D%7B1%20atm%7D%29%20%3D%20%5Cfrac%7B-%5Ctext%7Bheat%20of%20vaporization%7D%7D%7B8.314%20J%2Fmol%20K%7D%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7B351%20K%7D%20-%20%5Cfrac%7B1%7D%7B336.6%20K%7D%5D)
= 
= 
= 3813.1 J/mol
Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.
