Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 - 
Solving
-
=75.5 kJ/mol - 2*90.25 kJ/mol
-
=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
The order of the answers are as follows:
B
C
D
A
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Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

Here,
is mole fraction of A,
is mole fraction of B,
is partial pressure of A and
is partial pressure of B.
The mole fraction of A and B are related to each other as follows:

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.
Now, vapor pressure can be calculated as follows:

Putting the values,

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.