Answer:
east to west is the answer
Answer:
The process or result of oxidizing or being oxidized.(Rust)
Explanation:
Pluto
Answer:
The correct option is: Carbonate ion < Carbon dioxide < Carbon monoxide
Explanation:
Bond energy is defined as the average energy needed to break a chemical covalent bond and signifies the strength of chemical covalent bond.
The bond strength of a covalent bond depends upon the <u>bond length and the bond order.</u>
Carbon monoxide molecule (CO) has two covalent bond and one dative bond. Bond order 2.6
Carbon dioxide (CO₂) has two carbon-oxygen (C-O) double bonds of equal length. Bond order 2.0
Carbonate ion (CO₃²⁻) has three C-O partial double bonds. Bond order 1.5
Also, the bond length is <u>inversely proportional to the bond order and bond strength.</u>
Therefore, <u>order of C-O bond length:</u> Carbon monoxide<Carbon dioxide<Carbonate ion
<u>Order of C-O bond order</u>: Carbonate ion<Carbon dioxide<Carbon monoxide
<u>Order of C-O bond strength or energy</u><u>: Carbonate ion<Carbon dioxide<Carbon monoxide</u>
The azimuthal quantum number (l) determines its orbital angular momentum and describes the shape of the orbital.
s-orbitals (for example 1s, 2s) are spherically symmetric around the nucleus of the atom.
p-orbitals are dumb-bell shaped. l = 0,1...n-1, when l = 1, that is p subshell.
d-orbitals are butterfly shaped.
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!