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:
Answer:
2
Explanation:
Each orbital can hold two electrons. One spin-up and one spin-down.
<span>Answer
is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643
1/s.
k</span>₂ = 0,00828
1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K =
0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
No, it doesn’t define a person, money doesn’t change people.
Answer:
619°C
Explanation:
Given data:
Initial volume of gas = 736 mL
Initial temperature = 15.0°C
Final volume of gas = 2.28 L
Final temperature = ?
Solution:
Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)
Initial temperature = 15.0°C (15+273 = 288 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = 2.28 L × 288 K / 0.736 L
T₂ = 656.6 L.K / 0.736 L
T₂ = 892.2 K
K to °C:
892.2 - 273.15 = 619°C
