In a voltaic (galvanic) cell, oxidation occurs at the <u>anode</u> and is where <u>anions</u> in the salt bridge moves toward.
<h3>What is Galvanic Cell ?</h3>
Galvanic Cell or Voltaic Cell is an electrochemical cell that converts the energy of spontaneous redox reactions into electrical energy. In galvanic cell oxidation occurs at the anode and reduction occurs at the cathode. The anode is positive and cathode is negative, anode attracts anions from solution in an electrolytic cell.
Thus from the above conclusion we can say that In a voltaic (galvanic) cell, oxidation occurs at the <u>anode</u> and is where <u>anions</u> in the salt bridge moves toward.
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Answer:
The first two options are correct
Explanation:
The first two options are part of the benefits of a parallel connection of bulbs in a circuit. Here, the voltage of each connecting bulb is the same as the voltage of the bulb in the circuit hence all the bulbs have the same voltage running through them. Thus, when one bulb is removed/burns out, it does not affect the remaining bulbs (those ones will remain lit). Also, the addition of bulb(s) does not cause the remaining bulbs in the circuit to get dimmer (since they will all have the same voltage).
Cryo-EM is used to preserve and characterize cycled positive electrodes. Under regular cycling conditions, there isn't an intimate coating layer like CEI.A small electrical short can cause a stable conformal CEI to form in place. The conformal CEI's chemistry is revealed by EELS and cryo-(S)TEM.
It has been assumed that the intimate coating layer generated on the positive electrode, known as cathode electrolyte interphase (CEI), is crucial. However, there are still numerous questions about CEI. This results from the absence of useful instruments to evaluate the chemical and structural characteristics of these delicate interphases at the nanoscale. Here, using cryogenic electron microscopy, we establish a methodology to maintain the natural condition and directly see the interface on the positive electrode.
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Size (length+width) approx.
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
