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>
Answer: I believe the answer you are looking for is 0.02180
Answer:
58.94 mL
Explanation:
V1 = 48.3 mL V2 = v mL
T1 = 22 degree celsius OR 295 k T2 = 87 degree celsius OR 360 k
We will use the gas equation:
PV = nRT
Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,
we can say that
V / T = k , (where k is a constant)
Since this is the first case,
V1 / T1 = k --------------------(1)
For case 2:
Since we have the same constants, the equation will be the same
V / T = k (where k is the same constant from before)
V2 / T2 = k (Since this is the second case) ------------------(2)
From (1) and (2):
V1 / T1 = V2 / T2
Now, replacing the variables with the given values
48.3 / 295 = v / 360
v = 48.3*360 / 295
v = 58.94 mL
Therefore, the final volume of the gas is 58.94 mL
Answer:
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