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51How man} mautee #ould be needed to plaze # eraphite electrode Wuth !5.0 mg Af zinG Mbkng 25 miA cunail? [IO potats]...

Question

51How man} mautee #ould be needed to plaze # eraphite electrode Wuth !5.0 mg Af zinG Mbkng 25 miA cunail? [IO potats]

51How man} mautee #ould be needed to plaze # eraphite electrode Wuth !5.0 mg Af zinG Mbkng 25 miA cunail? [IO potats]



Answers

The half-reaction at an electrode is $$\mathrm{Mg}^{2+}(\text { molten })+2 e^{-} \longrightarrow \mathrm{Mg}(s)$$ Calculate the number of grams of magnesium that can be produced by supplying $1.00 \mathrm{F}$ to the electrode.

We need to determine the mass of solid magnesium and grams that we can produce if we apply one Faraday of charge to an electrode where this reaction is taking police. So if we stirred with 1.0 Faraday's, we know that one Faraday is equal to 96,485 cool ums per mole of electrons. So for one fair a day that corresponds to one mole of electrons. So one Faraday is equal to one mole of electrons. And based on this reaction, we know that we require two moles of electrons to produce one mole of solid magnesium. So we include that ratio into our dimensional analysis. Now, two moles of electrons are required to produce one mole of solid magnesium, and now we just convert this amount of magnesium and two grams through its more mass. You see from the periodic table that one mole of magnesium corresponds to about 24.31 grams of magnesium, and now we see when we cancel off units, Fair Day's most of electrons and moles of magnesium. We're left with the total mass in grams of magnesium that we can produce from one Faraday of charge, and that comes out to about 12.2 grams of magnesium

I am in the given problem, the platinum ball is having a capacitance of 10 Nana far it. Or we can say the system into tended bar minus nine carriage which will come out to be attended for minus state carriage. Then the diameter of platinum ball is given us 20 micrometer, so X radius will be given us. A is equal to half of this. 20 micrometer means 10 my group meter. Then the resistance of this electrode is given to be, which is also known as the excess resistance. And that is given us up is equal to roll by 10 8. There This role is the resistive itty of fluid and that has given us 100 on in two centimeter. And this is Dan. And for these radios, you know its value is 10 micrometer or we can say the sustained into 10 dish par minus 6 m in America. In place of this 100 forms centimeter, we can write it one on and to meet and divided by Spanish bar minus four. So finally this resistance will come out to be, then race to the part four home. Now we will find it's the capacity react tense, which is gonna ask Excess C is going to run by Omega Psi. Or we can say this is one by two by FC so plugging in all known values it becomes one by two by for frequency. This is given in the problem as 5000 bye bye And for capacitance is having a value of 10. We should bottom line estate parent. So solving this we get to attend this part whole home the capacity re attends hence impedance of the circuit uses said sequel to Exercises Square plus are square. Or we can say this is the square of 10 dish par four once again square of damage part four. Or we can say this is two times of squared off. And this bar four which comes out to be equal to 10 this part four route to home. Or we can write it like route to into 10 days. Part four home. So here we can say of an option, See? Is correct. Here. Thank you.

The reason for this seat? The reason for this. Mhm. The reason for this age at T. G. Each more in our then PT actually eat more in it than PT. So option B. So option B. H, correct answer. Well, this problem, so correct answer for this problem actually is more inert than pity. And this is the reason for this question.

Given that the resistance is much smaller than the inductive or the capacitive reactor since, and the system is operating at a frequency much higher than the resident's frequency. So we can say the frequency is much higher than the resonance frequency. We can say that the resistance is much smaller then the inductive reactant and the resistance. It's much smaller than the capacitive reactant ce. We can also therefore say that except AL minus except C is still going to be much, much larger than the resistance of your sister. And so, therefore, for the phase angle Phi, this would be equaling to arc 10 of the inductive react. It's minus the capacitive reactant divided by the resistance are. And because our is so small, this actually approaches infinity. And so we're saying that the phase angle would be equaling. Two are tend of infinity or approaching infinity essentially, and this would be approaching 90 degrees. And so we can say that then and we could say, approaching. So the power factor of the R. O. C. Circuit, then co sign of they equaling the current times. The resistance are divided by I, the current multiplied by the impotence R squared. Plus, they're inductive reactant spine. With the capacitive reactor, it's quantity squared. This is going to be small because the inductive reactant ce and the inductive capacitance as much greater than our. And so the phase angle is close to 90 degrees. However, the power factor is gonna be very small because as X as the capacitive reactor it's and the inductive reactant increase. This approaches zero. So we can say that in this case, option C would be the best choice. That is the end of the solution. Thank you for watching.


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