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Question

Meod:Kowoo Rc @l 43 - Honeleh f 1-3Naieefm [echon O2 fincu br culaln NuJathr * Wcheue YAAAnden at VYLlalHl) = s6)-r, then 4 03 Tennnnealne EYu} telaLiun umcron Faba The = Jomain con tartnAnc ~uty" {nMn eenrmnenantninc dnmln False Tec indrprndcnt rrubk omemimzn Telctied [04i lhz Alumenial Ihc (enclioa JutAAD Ierihad celmn Alain InumeemIEANEA Hn-Ann n GlmdinhrelemtalAattM the dorli FuaIl ralaumber In the tnlcryal o7i Attain humbcr mnlcryal 4Sae nc ekimnhin entubcr irectaLeinaaleFameenIhnba

Meod: Kowoo Rc @l 43 - Honeleh f 1-3 Naie efm [echon O2 fincu br culaln NuJathr * Wcheue YAAAnden at VYLlal Hl) = s6)-r, then 4 03 Tennnnealne EYu} telaLiun umcron Faba The = Jomain con tartnAnc ~uty" {nMn eenrmnenantninc dnmln False Tec indrprndcnt rrubk omemimzn Telctied [04i lhz Alumenial Ihc (enclioa JutAAD Ierihad celmn Alain Inumeem IEANEA Hn-Ann n Glmdinhr elemtal Aatt M the dorli FuaIl ralaumber In the tnlcryal o7i Attain humbcr mnlcryal 4Sae nc ekimnhin entubcr irecta Leinaale Fameen Ihnbancanthcidoteirt cleh Uanneia 4xun U LEc tuection f(*} Pranica Hctnm Tnetnt cunain dclncr Wanctng Ga "V-u a Hablc" ~lollnd Lic {otlonttr (D) f(1) f(-W) - 0+1) {x) Jul -



Answers

$$41-44=$ Find $y^{\prime}$ and $y^{\prime \prime}$$
$$y=e^{\alpha x} \sin \beta x$$

Hello. So today we're going to be looking at thes reduction potentials and we want to figure out this reduction potential right here. So how can we do that? Well, let's see what thes reduction potentials are telling us while we have this becoming this and then this becoming this. So how about we take a look at our oxidation states? So the oxidation state of this here is so oxygen is negative two and there are four of them. So that's negative. Eight and then we have ah, too high traditions that are positive one. So that's negative. Six. So this has to be a positive six oxidation state. And here we have two oxygen's each negative too. So we've got negative four. So this has to be positive for So we're being reduced from positive 16 positive for in oxidation state between these two. So it's gaining two electrons and now we take a look here. This is the same as above. It's positive six. But here we see that it's positive. Three. And so we consort of write it like this. We start off with this compound right here, which is the most oxidized. Then we reduce it once here and then we reduce it a second time. So this production potential right here is referring to this part. And then this reduction potential right here is referring to this and we want to figure out this part. So it seems like if we took this energy change when we subtracted this energy change, we would get this energy change. However, sell potentials. You can't add and subtract sell potentials, but you can add and subtract free energies. So let's figure out the free energy of this right here, so still touchy is equal to negative Z fair days, constant and standard self potential or reduction potential. So free energy of this compound right here, coming this compound that would be gaining three electrons SOS Z is three fair days constant and then 0.4 to 8. And then self potential of this becoming this that would be gaining two electrons. So Z is too. And then the self, the reduction potential a 0.646 So now we can just take this and subtract this from it. So 0.4 to 8 times three would be negative 1.284 Saturday subtracting. Okay, two times. 20.646 Yeah, is negative. One point tune nine to Fair day. So that would become 0.0 eight. Fair day. And now let's make that into a reduction potential. So Negative z Fair day reduction potential. And so we know that c Well, we're going from a plus four oxidation state to a plus three. So Z would just be one. And so we get that the reduction potential off this compound right here. Coming. This compound right here is negative. 0.0 eight boots. So there you go.

In this problem he had X is And AHC 03. It has molecule er wait, it has a molecular wait it's equal to 84. So according to the option, option B is correct here, option D is correct answer for this problem Here. The compound exit any HC three. It has a molecular weight. Mhm. Of 50 food. I hope you understand the answer.

The best way to answer this question is to first convert the SL or the E standard values for the half reactions into Delta G standard values. Then you can combine the two half reactions in order to get the half reaction of interest and in the process combined the Delta G values to get the Delta G for the reaction of interest. Once you have the Delta G value for the reaction of interest, you can then convert it into an e standard value. So the first reaction the first couple is something like this. It doesn't. It's not so important that we include everything else involved in the half reaction, but it is most important that we identify how many electrons are being transferred. To do that, we recognize the oxidation state of M O is four plus and over here it is six plus. So that's a difference of two electrons he is given to us. We can calculate Delta G now knowing that N is 22 electrons are involved and we get negative 1.25 times 10 to the fifth jewels For the next half reaction we have mn I'm sorry we have M O. Over here at six plus going to m O at three plus. So that's a difference of three. So three electrons are involved. So with an e standard given to us, we can calculate a Delta G standard as negative end, which is three multiplied by F. Faraday's constant multiplied by the value provided. Then, to get the reaction of interest, we're going to have to reverse this first reaction. If we reverse this first reaction, we're going to change the sign on our Delta G value. When we reverse that reaction, we do get M N 002 going to m 03 plus, which is the couple of interest where only one electron is involved. And then there are other things also, as a part of this chemical reaction which really aren't important. Delta G then is going to be negative 1.24 times 10 to the fifth plus or minus the negative 1.25 times 10 to the fifth and we get 772 times 10 to the two jewels. That's the Delta G for this half reaction to calculate the E for this half reaction, it's going to be G divided by negative end multiplied by F and we get points negative 0.8 volts

In this video, we're gonna go through the answer to question number 37 to chapter 9.3 so as to show that the Matrix X satisfies the given matrix differential equation. So, firstly, let's find the derivative off Matrix acts this book 50. So just do this element wise. So top left is gonna be two e two. The two tea till Friday is gonna be three. Hey, three day. But that's gonna remind us, too. Eat the TT and bomb, right? It's gonna be minus two times three. We just might have six e to the three T. Okay, it's, um next up, we want to find the matrix one one. It's one minus 1 to 4 times by X. So this is gonna be one minus one too. Four times. Bye into the two today. Piece of three T minus each. That ct minus two into the three day. Okay, so this is gonna be ah, each of the two team minus minus. T t t t. So that's two. He's the TT. It's all right. We're gonna have He's the three t minus minus to each of the three. Teaches us, please. The three t plus two eating three. Think so? That's three into three thing. But unless we're gonna have to e to the T T minus boys into T, that's minus two. It's the two teams on, but I might have enough to eat. Three. T minus eight. It's a three d. So that's minus two. Minus eight. Is my six e to the treaty? Uh, this matrix is the same thing. This matrix. So therefore the matrix X of tea satisfies this system off defensive equations.


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