C
This is because 10+5=15
15/45=0.3
Carbon has a tendency to form covalent bonds.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Principle quantum number describes the energy of an electron and most probable distance of the electron from the nucleus.
<h3>What is the significance of principle quantum numbers and azimuthal quantum numbers?</h3>
A principal quantum number signifies size and energy of the orbital.Azimuthal quantum number signifies three dimensional shape of the orbital.
Magnetic quantum numbers signifies spatial orientation of the orbital.
Principal quantum numbers is the quantum numbers denoted by n which indirectly describes the size of the electron orbitals. It is always assigned an integer value but its value never be 0. The feature of a principal quantum numbers is the energy of an electron and most probable distance of the electron from the nucles.
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