This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
Learn more:
Copper has 29 protons, when dealing with Cu^2+ all that means is it lost two electrons. so now the element has 29 protons and 27 electrons. Protons are positive and electrons are negative and neutrons are neutral. So say you had an element X^2- then you have gained two more electrons so the element has an overall negative charge. hope that helps
Answer:
um thats tricky man
Explanation:
i dont even know what nuclei is
The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
