Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:

Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:



<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
<span> There are 97.2 mol of O in 10.8 mol of Fe(NO3)3</span>
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
En el caso del sodio, la valencia es 1, ya que tiene un solo electrón de valencia, si pierde un electrón se queda con el último nivel completo.
Explanation:
Grupo de la tabla periódica Electrones de valencia
Grupo 14 (IV) (Grupo del carbono) 4
Grupo 15 (V) (Grupo del nitrógeno ) 5