C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>
Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
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From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:

The substance having highest positive
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:


Hence, the standard electrode potential of the cell is 4.53 V.
The best and most correct answer among the choices provided by your question is the second choice or letter B.
The iron cam is larger than the aluminum cam even if with the same size.
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<span>I would say only if one of your data points is the origin. But your experiment could have started with a non-zero velocity, for instance, which would rule out the origin as one of your data points. Even so, a "best fit" is not meant to be perfect, it is only meant to be the best that you can do with your particular data set.</span>
Answer:
4244.48 g to the nearest hundredth.
Explanation:
The molar mass of Glucose = 6*12.011 + 12*1.008+ 6*15.999
= 180.156.
So 23.56 moles = 180.156 * 23.56 = 4244.48 g