Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.
<h3>What is the enthalpy change?</h3>
Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.
The enthalpy change is given by,

and,

Given,
Mass of the liquid mercury (m) = 11.0 gm
The specific heat of mercury (c) = 0.14 J per g per degree Celsius
Temperature change = 15 degrees Celsius
Enthalpy change is calculated as:

Therefore, 0.0231 kJ is the change in enthalpy.
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Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
As the given reaction is as follows.

(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.
Thus, formation of
will increase.
- (b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.
Hence, formation of
will decrease with decrease in volume.
- When we increase the mount of
then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.
Thus, we can conclude that formation of
will increase then.
1) mass composition
N: 30.45%
O: 69.55%
-----------
100.00%
2) molar composition
Divide each element by its atomic mass
N: 30.45 / 14.00 = 2.175 mol
O: 69.55 / 16.00 = 4.346875
4) Find the smallest molar proportion
Divide both by the smaller number
N: 2.175 / 2.175 = 1
O: 4.346875 / 2.175 = 1.999 = 2
5) Empirical formula: NO2
6) mass of the empirical formula
14.00 + 2 * 16.00 = 46.00 g
7) Find the number of moles of the gas using the equation pV = nRT
=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)
=> n = 0.01769 moles
8) Find molar mass
molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol
9) Find how many times the mass of the empirical formula is contained in the molar mass
92.14 / 46.00 = 2.00
10) Multiply the subscripts of the empirical formula by the number found in the previous step
=> N2O4
Answer: N2O4