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>
We can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = 303.15 x 300 / 333.15
<span>V2 = 272.99 cm³</span>
Answer:
6l
Explanation: convert temperature to kelvin by adding 273 and then input the values into the formula with the given constant
2*v=0.5*0.8206*288 then divide both sides by 2 and get the amount in litres which is 6
<span>The reactants have a slightly greater mass. In a nuclear reaction, a small amount of mass
is converted to energy according to the equation E = mc2. The difference in mass is referred to as the
mass defect.</span>
