The rate constant for the decomposition of n2o5 is 7.78 × 10−7 at 273 k and 3.46 × 10−5 at t2. if the activation energy is 1027 kj/mol, what is the final temperature

Answer is: the temperature of the reaction is 275,48 K.
k₁ = 7,78·10⁻⁷ 1/s.

T₁ = 273 K.

Ea = 1027 kJ/mol = 1027000 J/mol.

k₂ = 3,46·10⁻⁵ 1/s.

R = 8,314 J/K·mol.
T₂ = ?
Natural logarithm of Arrhenius' equation:
lnk₁ = lnA - Ea/RT₁.
lnk₂ = lnA - Ea/RT₂.
ln(k₂/k₁) = (Ea/R) · (1/T₁ - 1/T₂).
ln(3,46·10⁻⁵ 1/s / 7,78·10⁻⁷ 1/s.) = (1027000 J/mol ÷ 8,314 J/K·mol) · ·(1/273K - 1/T₂).
3,79 = 123526,58 K · (0,00366 1/K - 1/T₂).
3,79 = 452,47 - 123526,58 · (1/T₂).
1/T₂ = 0,00363.
T₂ = 275,48 K.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72