What is the electron configuration for oxygen with a 2- charge (o2-)?
1 answer:
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First convert celcius to Kelvin.
20 + 273 = 293K
31 + 273 = 304K
Now we can set up an equation based on the information we have.
V1 = 5
P1 = 365
T1 = 293
V2 = 5
P1 = x
T2 = 304
The equation be:
Now just solve.
1825/293 = 5x/304
Cross multiply.
554800 = 1465x
Divide both sides by 1465
x = 378.7030717 which can then be rounded to 378.7 mmHg
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>