Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
A catalyst is a chemical substance that hastens the chemical reaction. This does not participates in the creating the product(s) but allows it to be formed easily. With this, it is now known that the rate of the reaction becomes relatively higher compared to the uncatalyzed reactions.
Therefore, the answer to this item is the rate of the reaction becomes faster.
A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.
<span>C4H4
The compound in question has an equal ratio of hydrogen and carbon. The atomic weight of carbon is roughly 12 and the atomic weight of hydrogen is roughly 1. The mass of the compound in question is roughly 52.
52/13=4
C4H4</span>
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>
First, the equation of the reaction:
The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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