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:
A. Frictional
Explanation:
Friction is the force created when two solid objects in contact try to move past each other.
Answer:
5.61983 × 10^5
Explanation:
Move the decimal forward 5 spaces, each time doing so you add 10^(# of spaces moved, in this case 5)
The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
The energy of the photon is given by;
E = hc/λ
E = energy of the photon
h = Plank's constant
c = speed of light
λ = wavelength of light
For the upper boundary range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 270 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 270 × 10^-9
E = 7.33 × 10^-19 J
For the lower range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ =300 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 300 × 10^-9
E = 6.6 × 10^-19 J
Hence, the energy range 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
Learn more: brainly.com/question/24857760
