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Luda [366]
3 years ago
13

The rate constant for a first-order reaction is 2.4 x 10-4 L/(mol's) at 600 K and 6.2 x 10-4L/(mol · s) at 900 K.

Chemistry
1 answer:
Alika [10]3 years ago
6 0

Answer:

(R = 8.31 J/(mol · K)) 0.88 × 104 J/mol 1.42 × 102 J/mol 1.42 × 104 J/mol at 600 K and 6.2 × 10-4 L/(mol · s) at 900 K. Calculate the activation energy.

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In the number 12.345, the 1 is in what place
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Which subshell holds the valence electrons of barium?.
olga2289 [7]

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6s

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7 0
2 years ago
Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b
Alinara [238K]

Answer:

Explanation:

The cell reaction properly written is shown below:

              Cu|Cu²⁺_{aq} || Ag⁺_{aq} | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

  Oxidation half:

                  Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

At the anode, oxidation occurs.

  Reduction half:

                  Ag⁺_{aq} + 2e⁻ ⇄ Ag_{s}

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

             Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

              Ag⁺_{aq} + e⁻ ⇄ Ag_{s}

  we multiply the second reaction by 2 to balance up:

         2Ag⁺_{aq} + 2e⁻ ⇄ 2Ag_{s}

The net reaction equation:

Cu_{s} + 2Ag⁺_{aq} + 2e⁻⇄ Cu²⁺_{aq} + 2e⁻ + 2Ag_{s}

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

  Cu_{s} + 2Ag⁺_{aq} ⇄ Cu²⁺_{aq} + 2Ag_{s}

6 0
3 years ago
Oxidation and reduction reactions (redox) involve the loss and gain of electrons. Half-reactions are a way for us to keep track
jok3333 [9.3K]

Answer:

Electrons are lost during oxidation (LEO)

4 0
3 years ago
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