Answer:
The correct answer is option E.
Explanation:
Structures for the reactants and products are given in an aimage ;
Number of double bonds in oxygen gas molecule = 1
Number of double bonds in nitro dioxide gas molecule = 1
Number of single bond in in nitro dioxide gas molecule = 1
Number of triple bonds in nitrogen gas molecule = 1

![\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CNO_2%7D%5D-%5B1%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CN_2%7D-2%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CO_2%7D%5D)

 (pure element)
 (pure element )

The enthalpy of the given reaction is 15.86 kcal.
 
        
             
        
        
        
Conservation efforts most effective if <span>Conservation efforts are most effective when multiple groups cooperate. The answer is letter A. The rest of the choices do not answer the question above</span>
        
                    
             
        
        
        
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression : 
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added . 
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
 i = Von't Hoff factor 
 kf =molal freezing point depression constant of solvent.
 m = molality of solute (m or 
 )
Given : kf = 1.86 
 m = 0.907 
 )
 Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 
 * 0.907
 )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C