5

Texiedo 3onrcrexide with 94Jgal "Orodce thminimorlacandIron Dctcrminwnchon (nc jollowint 7e70104 2 N(s) Fe(s) + AlOxslllmkcInercactantLuminumIconio idcPrtoeNet...

Question

Texiedo 3onrcrexide with 94Jgal "Orodce thminimorlacandIron Dctcrminwnchon (nc jollowint 7e70104 2 N(s) Fe(s) + AlOxslllmkcInercactantLuminumIconio idcPrtoeNet

texiedo 3onrcr exide with 94Jgal " Orodce thminimorlacandIron Dctcrminwnchon (nc jollowint 7e70104 2 N(s) Fe(s) + AlOxsl llmkcInercactant Luminum Iconio idc Prtoe Net



Answers

A voltaic cell consists of two $\mathrm{Sn} / \mathrm{Sn}^{2+}$ half-cells, A and B. The electrolyte in $\mathrm{A}$ is 0.13$M \mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2} .$ The electrolyte in $\mathrm{B}$ is 0.87$M \mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2}$ . Which half-cell houses the cathode? What is the voltage of the cell?

So next we're looking at electrons, electrolysis, electron, self potential and so on and so forth. So the first thing we're looking at is part A. So this is the 1.1 vault section so that we use the copper as the cathode to oxidize ze and xe and two plus that occurs at the an ode. So the cell potentially is 1.1. So we have the negative. An ode that is the oxidation. So that is the CNN Susie and two plus like we said. And then we have the positive CASS owed, which is C U two plus have two electrons generates. See you. So on that equation here is as follows C n I see you two plus is in equilibrium with C and two plus at C U. So you can see that we've got the electron transfer from sink to cover two plus. So I'm moving on here. R s l is not 0.337 Subtract negative. No 0.763 You sell us 1.11 volts. So in the second part, we're looking for 9.5 boats. So we have C N C N four to minus, which is are an ode, and we use this to reduce ZN two plus two c n at the cathode. So the cell potential is not 20.5. So are honored. We have CNN at four, see and miners that is an equilibrium with CNN C N for two miners on two electrons and then at the cathode we have C and two plus two electrons generates C n so on. That equation is as follows. So we have CNN two plus at four C and minus generates C n CNN for tu minus. Where are S O is not 0.763 volts. Subtract negative 1.26 volts. Eso is not 0.5. Remember? That is in vaults.

Yeah. And this problem? We have a simple circuit here that has a couple batteries, a couple of voltage supply and a couple of resistors. And we asked a few questions about it and let's see what they actually asked us to look into here a whole lot of questions about it. And I'll just I'll just kind of as I go through solving them. I'll just hopefully they'll all get answered here. Oh, they will all get answered here. So one thing we can do so we know that again. This has a potential, um, positive. Here. Negative. Here. Positive. Here. Negative here. So they're they're opposing one another. Um, so I I basically said that, you know, e two has a has a voltage of minus six folds. So this one, So we get that sign in to take that sign into account here. Um, now, if we go, you know, if we go, um, we do a loop around here, you know, to use a loop rule. Uh, the voltage drop across here is I R one. Yeah, And the voltage drop across. Um, this one is I r two, and we have, uh, have that equals to the voltage. Um, let's see, here that equals. Actually, we think we're going. I'm going this way. Sorry. So that equals to the voltage gang across here and plus the voltage gain across here. But that one is negative. So we we know everything in here except for I So we can solve for I and that comes out to be half an app. So we have half an app um, circling, uh, this way around the the circuit. So, yeah, I went I went counterclockwise in my loop here, so we got a positive amp. And that makes sense, because again, this thing is this battery has a higher voltage than this. So, you know, if we lump them together, we'd have a net six volts going this way, so we'd expect the current to be flowing counterclockwise and again we give also, given the resistance here, this one has eight arms. And this one is, um this one is forearms, and this one is eight homes. So again, we can just plug all those values in, and we get half an app of current flowing through. Now they ask us something about the power being dissipated. Um through these through the resistors. And so the power dissipated in a resistor is one way to write It is I squared R so we know what I is and we know what our is for each of the resistors. So for the first resistor here we have I squared. So that's a that's, um, ends up being one quarter and r is for so we wind up with, uh, one watt of power being dissipated in this resistor and this resistor again I squared is one quarter our is eight. So we wind up with two watts of energy being dissipated in resistor too. So we got a total of three watts of power being dissipated or three jewels per second, three tools of energy per second being dissipated. So now we can also figure out how much, Um, what what power the batteries are producing. So the energy being supplied by them and that's the voltage times current. And so in this battery we have a voltage of 12 volts, and we have a current of one half apps. So we have, um, we have Let's see, here we have, uh, six watts of energy being supplied by this one. And this one we have, um, the voltages minus six. So the voltage the current is flowing in the opposite direction of the potential of the, you know, plus minus. So it's flowing in the opposite way. So this energy, this battery, actually is absorbing energy, which is part of one of the following questions. Um, and that's absorbing three watts of energy. And so notice that the total energy being, um, the total energy output, which is six watts output by this three of its being absorbed here, and the other three is being dissipated as heat by the resistors. I realized that, and so they asked us, You know, which of these batteries are supplying absorbing energy. So energy is being supplied by this one? Because the voltage, um, the current is flowing. Um, from a lower potential to a higher potential. Um, And so then this one here we have the current flowing from a a higher potential to a lower potential. So that has. And that's one that is. Actually, we're pushing energy into here. So, Billy, what we're doing here is we're charging this battery from this battery. What? We're losing half of that half of the energy to these resistors. Mm.

So now we'll work on problem 36 from chapter 20. In this problem, you were asked about a voltaic cell that uses the reaction. Palladium, Uh, chloride minus was cadmium makes palladium for chlorine ions and a cadmium. So the question here is asking us The first part of this question is to write the 2/2 reactions so we can do this by splitting the palladium and cadmium reaction separately. So first we can write the reduction reaction. We have palladium C L for minus to minus plus two electrons. Well, give us a palladium and for chlorine sze and the 2nd 1 is simply just the cadmium medal will react to produce the cadmium two plus cat eye on and two electrons. So, in part B of this problem, we're asked to use it dated from Appendix E and determined e production for the reaction involving palladium. Now we're told in the problem that the e uh, the energy or the potential of the cell is equal to 1.3 votes now from the appendix e. We don't have this reaction. This reaction here we do not see an appendix C, but we do have this reaction, so we know that the reduction will go first. So you have the reduction thie cathode, and then we'll subtract the value of this reaction, which is negative 0.40 And so then we can solve for the value of the cathode or the Palladian. Reaction the potential, and we get 1.3 minus is it Q B plus 0.40 which is equal to one point for three bolt. So that's the potential for this. So the we actually are not solving for the solution. Here we're solving for the cathode. And so when we're solving for the cathode, that means we have the positive and positive here and the negative negative here will make the positive. We'll make the positive here. But since we're going to subtract it from 1.3 negative is correct. And so the value for this reaction is 0.63 bolts. The last part of this problem asks us to draw the SL so we can go ahead and put this in here. We have two beakers. We have a solution. In each. We have a salt bridge. We have our electrodes here our old meter. So over here we can put our cathode. This will be our palladium cathode! And on the left we can put our cadmium, which is an anode. And we have cadmium two plus in here and hi play and chloride here. And the last thing that it asked us to do is indicate the flow of electrons. So the flow of electrons will be from anode cathode. So we have electron is going this way.

In this problem. On the topic of circuits we have two ideal batteries with the mfc epsilon one, let alone two. There are 12 and six fold respectively. And we want to find the current and the dissipation rate in resistor one which has a resistance of forearms. Then in resisted to which has a resistance of eight homes. And then the energy transfer rate in battery one and battery to. And lastly we want to know if the energy is being supplied or absorbed by each of these two batteries. Now these consists of two batteries into resistors and we can apply cache of slough cruel to solve for the current. Well, let I be the current in the circuit and taking positive if it is to the left in our one. And so the loop rule gives us absalon one minus I are too minus I are one minus epsilon to equal to zero. Now we have to solve for the current I And we get this current I two b. Absalon 1- Absalon too Divided by R. one blessed are too which is 12 olds minus six folds over. Okay for rooms plus eight arms. So we get the current I to be 0.5 and peers and a positive value means the current is counter clockwise around the circuit. Now for part B For resistor R one, the dissipation participation rate P one is I squared times resistance are one Which is zero five mps squared Times the resistance of the resistor one which is four owns. And so this is a dissipation rate in resistor one of one. What For our to the dissipation rate? P two is again the current I squared times this time. The resistance are too, Which is 0.5 amperes squared times the resistance, eight arms. And so this gives a dissipation rate of two watts. Now, if I is the current in a battery with the MF epsilon, then the battery supplies energy at the rate of items epsilon provided the current and E M F I in the same direction. On the other hand, the battery absorbs energy at the rate of I. Absalon if the current and E M F I in opposite directions. So for ε one we have P one equal to I. Absalon won. Which is 0.5 m. s. Times 12 vaults, which gives a dissipation rate of six watts. And for part E we have yeah, P two equal to I. Absalon too, Which is zero five mps times the E M F. Absalon two, which is six faults, which gives Power P 2 to be three watts. Now, for part F. In battery one, the current is in the same direction as the E M F. Therefore this battery supplies energy to the circuit. And so we can see that this battery is discharging and for part G, the current and battery to his opposite the direction of the E M F. So the this battery absorbs energy from the circuit and his hands charging


Similar Solved Questions

5 answers
13AaBbCcDdEe AaBbCcDdEe AaBbcbuppeds ONJuipeahNormalAnormally distributed set of population scores has & mean of 88 and a standard deviation of 12.2. What i8 the sample mcan; for samples of size 64 equals3.For the population and sample size given in question the standard crror of the mean equals
13 AaBbCcDdEe AaBbCcDdEe AaBbc buppeds ON Juipeah Normal Anormally distributed set of population scores has & mean of 88 and a standard deviation of 12.2. What i8 the sample mcan; for samples of size 64 equals 3.For the population and sample size given in question the standard crror of the mean...
5 answers
ULI(C) {Enf,1/n, definition that the following sequences converge. 2. Prove using the € ~ no 2+5n 16 8+ 1ln 80 (1+2n)2 (6) (g) VSn 5+3n + 3n2 (1+4n)2 1-3 O} (h) n ~ 3n 6n2
ULI (C) {Enf, 1/n, definition that the following sequences converge. 2. Prove using the € ~ no 2+5n 16 8+ 1ln 80 (1+2n)2 (6) (g) VSn 5+3n + 3n2 (1+4n)2 1-3 O} (h) n ~ 3n 6n2...
5 answers
Consider binomial experiment with n = 11 and P = 0.3 Compute f(0) (to decimals)_ f(o) Compute f(6) (to decimals) f(6) Compute P(z < 4) (to decimals) P(c < 4) = Compute P(z > 3) (to decimals) . P(z > 3) = Compute E(z) (to decimal) _ E(z) Compute Var(r) adVar()decimals)decimals}
Consider binomial experiment with n = 11 and P = 0.3 Compute f(0) (to decimals)_ f(o) Compute f(6) (to decimals) f(6) Compute P(z < 4) (to decimals) P(c < 4) = Compute P(z > 3) (to decimals) . P(z > 3) = Compute E(z) (to decimal) _ E(z) Compute Var(r) ad Var() decimals) decimals}...
5 answers
A marble is selected at random from jar containing 2 red marbles yellow marbles, and green marbles What is the probability that the marble is red?What is the probability that the marble is red?
A marble is selected at random from jar containing 2 red marbles yellow marbles, and green marbles What is the probability that the marble is red? What is the probability that the marble is red?...
5 answers
Find the arc length of the curve given by:r(t) = sinht i - (+2) j + exp(-t?) k[0, 2]:
Find the arc length of the curve given by: r(t) = sinht i - (+2) j + exp(-t?) k [0, 2]:...
5 answers
14. A home freezer operates between 18*C and +25*C. Assuming it operates atits theoretically maximum possible coefficient of performance (COP_ (that is as a Carnot engine working in- reverse), how much electrical energy would it take to freeze 300 g water initially at 0*C into ice at 0*C? (Li = 333 kJlkg; COP = QuW)a) 20.2 kJ b) 22.5 kJ c) 28.1 kJd) 33.7 kJe) 16.8 kJ
14. A home freezer operates between 18*C and +25*C. Assuming it operates atits theoretically maximum possible coefficient of performance (COP_ (that is as a Carnot engine working in- reverse), how much electrical energy would it take to freeze 300 g water initially at 0*C into ice at 0*C? (Li = 333 ...
1 answers
The magnetic field in a hospital's cyclotron is $0.50 \mathrm{T}$. Find the magnitude of the magnetic force on a proton with speed $1.0 \times 10^{7} \mathrm{m} / \mathrm{s}$ moving in a plane perpendicular to the field.
The magnetic field in a hospital's cyclotron is $0.50 \mathrm{T}$. Find the magnitude of the magnetic force on a proton with speed $1.0 \times 10^{7} \mathrm{m} / \mathrm{s}$ moving in a plane perpendicular to the field....
5 answers
Question 33 (1 point) One stcp of RNA processing is5"-capping~polyadenylationhaupin loopingligation~methylation
Question 33 (1 point) One stcp of RNA processing is 5"-capping ~polyadenylation haupin looping ligation ~methylation...
5 answers
Part 2. Now write the appropriate equilibria and associate the correction Kn values- Remember; we will want to calculate the concentrations of all species in 530 M Na_SO; (sodium sultite) solution The ionization constants for sulfurous acid are Kul 144*10 and K_ 63x10Enter the appropriate Kb values (25 *C)S0}-(aq)- H,0 = HSO; (aq) OH (aq)HSO; (1q) H,o = HSO,(aq) OH-(aq)
Part 2. Now write the appropriate equilibria and associate the correction Kn values- Remember; we will want to calculate the concentrations of all species in 530 M Na_SO; (sodium sultite) solution The ionization constants for sulfurous acid are Kul 144*10 and K_ 63x10 Enter the appropriate Kb values...
5 answers
The stock price per share of a certain commodity can go up or down An investor believes that the chance of it going up is 90%. If stock price goes up the market is good 75% of the time, fair 20 % of the time and bad 5% of the time. When stock price goes down; those numbers are 60%, 30% and 10% respectively. Given that the market is fair, find the investors updated belief of the price going up. [4]
The stock price per share of a certain commodity can go up or down An investor believes that the chance of it going up is 90%. If stock price goes up the market is good 75% of the time, fair 20 % of the time and bad 5% of the time. When stock price goes down; those numbers are 60%, 30% and 10% respe...
5 answers
Simplify each expression.$$-4(-3 x+3)-(6 x-4)-2 x+1$$
Simplify each expression. $$ -4(-3 x+3)-(6 x-4)-2 x+1 $$...
3 answers
Graph G with vertex set {v1,02, U3, U4, U5, V6, U7} has adjacency matrix To 14 8 13 13 where ELh # 1 13 13 13 13 28 Sketch the graph of G. b) How many V1 U4 walks of length exist? Name them_ How many V2 U4 walks of length exist? Name them. How many U7 U4 walks of length 4 exist? How many distinct cycles of length begin and end at Uz?
graph G with vertex set {v1,02, U3, U4, U5, V6, U7} has adjacency matrix To 14 8 13 13 where ELh # 1 13 13 13 13 28 Sketch the graph of G. b) How many V1 U4 walks of length exist? Name them_ How many V2 U4 walks of length exist? Name them. How many U7 U4 walks of length 4 exist? How many distinct cy...
1 answers
For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{aligned}-4 x-3 y-8 z &=-7 \\ 2 x-9 y+5 z &=0.5 \\ 5 x-6 y-5 z &=-2 \end{aligned} $$
For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{aligned}-4 x-3 y-8 z &=-7 \\ 2 x-9 y+5 z &=0.5 \\ 5 x-6 y-5 z &=-2 \end{aligned} $$...
5 answers
Problem 2: (a) Find the electric field at a point located 50 cm from 9z along the +X axis if 9, 50x10-6 € and 9z 50x10-6 C? (3 points) (b) What is force on a particle with a charge of Qat that point? point)9125 cm50 cm92V+X
Problem 2: (a) Find the electric field at a point located 50 cm from 9z along the +X axis if 9, 50x10-6 € and 9z 50x10-6 C? (3 points) (b) What is force on a particle with a charge of Qat that point? point) 91 25 cm 50 cm 92 V+X...
5 answers
Propose mechanism for the following general Claisen condensation: NaOEt; HOEt dil HO_
Propose mechanism for the following general Claisen condensation: NaOEt; HOEt dil HO_...
5 answers
Determine the oxidation number 0t The suitur atom:HzsH2SO4 d. 5?"Hs"f SOz 9. SOaIndicate the oxidation number of phosphorus in each of the following compounds: HPOa HaPoab. HiPOzHAPzOrHiPOaf. HsPaO1o
Determine the oxidation number 0t The suitur atom: Hzs H2SO4 d. 5?" Hs" f SOz 9. SOa Indicate the oxidation number of phosphorus in each of the following compounds: HPOa HaPoa b. HiPOz HAPzOr HiPOa f. HsPaO1o...
5 answers
Use Taylor's formula for f(z,y) at (0,0) to find the quadratic approximations of near tne origin whenf(z,y) 4-Ij 4y + 2Iyf(z,y)
Use Taylor's formula for f(z,y) at (0,0) to find the quadratic approximations of near tne origin when f(z,y) 4-Ij 4y + 2Iy f(z,y)...
5 answers
Part 3 0f 4If the systolic pressures two patients differ by 14 millimeters_ by how Much would vou predict differ? Round the answer to three their diastolic pressures to decimal places:Their diastollc pressures would differ bymilllmeters.
Part 3 0f 4 If the systolic pressures two patients differ by 14 millimeters_ by how Much would vou predict differ? Round the answer to three their diastolic pressures to decimal places: Their diastollc pressures would differ by milllmeters....
5 answers
At a certain temperature; 0.600 mol SO3 is placed in a 4.50 L container.2SO,(g) = 2SOz(g) + 0,(g)At equilibrium; 0.110 mol 0z is present. Calculate Kc.Kc0.09Incorrect
At a certain temperature; 0.600 mol SO3 is placed in a 4.50 L container. 2SO,(g) = 2SOz(g) + 0,(g) At equilibrium; 0.110 mol 0z is present. Calculate Kc. Kc 0.09 Incorrect...

-- 0.022025--