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Question

InlbetulnIndilIcion bcunriotn Groblaro Ita * Jxis betteen K = Jnnee 64 410 Oakna [o VEdpi Lctha Dgqroltable_Clice s ebvlen 64a 2ou_iha aleokabl Clkd tgre lovler 64oe J otha OxkhotloYem Poca 5 Iho rtegrolybin Clcaheta yigw D aolir "UrulalbiyClct ntogu Uibt Iha glvon renion | hlnin LRa (Typa ansuur_/

Inlbetuln Indil Icion bcunrio tn Grobl aro Ita * Jxis betteen K = Jnnee 64 410 Oakna [o VEdpi Lctha Dgqroltable_Clice s ebvlen 64a 2ou_iha aleokabl Clkd tgre lovler 64oe J otha OxkhotloYem Poca 5 Iho rtegrolybin Clcaheta yigw D aolir "UrulalbiyClct ntogu Uibt Iha glvon renion | hlnin LRa (Typa ansuur_/



Answers

Determine $E_{\mathrm{MoO}_{2} / \mathrm{Mo}^{3+}}^{\circ},$ given that $E_{\mathrm{H}_{2} \mathrm{MoO}_{4} / \mathrm{MoO}_{2}}^{\circ}=$ $0.646 \mathrm{V}$ and $E_{\mathrm{H}_{2} \mathrm{MoO}_{4} / \mathrm{Mo}}^{3+}=0.428 \mathrm{V} .$ (See page 875).

Hello students in this question we have an eyes which is initially at a temperature T1 equals to 0° integrate. Which was first melted. Okay melted and converted into the water. And then it is heated to a temperature T two equals 200 degrees integrate and then it is evaporated. Okay evaporation has been taken place so after evaporation this has been converted into the weapons. Okay so we have to calculate the increase in entropy. S. So here delta. S. Win in this process data as to and here delta. So delta S. Will be equal to the during the melting process that it will be equal to m replaceable latent heat of fusion to ever be temperature. This is steven. Okay this zero C is 2 73 Calvin and this 100 C is 3 73 Calvin. Okay Plus Delta H. two which is during the temperature change. So this will be given by M particularly specific heat of water and natural log of temperature T. To develop it temperature T. V. Okay Plus Delta S. three during the weaponization. So member player BQ weaponization development temperature T. Two. So we can now substitute the values so mass M. This can be taken as one. Okay because parquet G. We can calculate so this M. Can be taken as one. So cute Fusion. It is 3:33 multiplied by 10 to the power three june per kilogram never be temperature T. Which is 2 73 plus one particularly specific heat 4.18 molecular beaten to the power three Ellen of 3 73. There will be 273 plus one. More popular. Be latent heat of a tradition is double to +50, manipulated by 10 to the power three D. Barbee the verb 3 73. So after solving the inclement in entropy is 8.56 molecular weight. And to the power three. Kill a jewel jewel perp mhm Calvin. Okay. Or it can be return is delta as it is equal to 8.56 kg jewel per Calvin. So this is the answer for the problem. Okay, thank you.

UNESCO's were given the function y echoed you the to the power and for X times side bet the next and the first thing you need to do would be the first derivative. Hey, we need your bride up front row here you fee And then we know you attempts v be riveted, which would include you know, you Prem fee plus UV prem. And here we should get the each about in front. Thanks. Do you really to No time, Sam. That bit the X plus each other and for X times side. But the x the river to Yeah this one the river doing any co Judah and far each other. And for x times Yes, I bet that thanks bless each other. And for X on Now this one the revolution get equal to go sign after but X and by the General Winnetou Times about the in the front Here And then this will be the wiper Him then wind up a Prem were planned again. The power the product rule here. So the father first come here. It was getting an unfair square Egypt and four x times Say'i'm duh hair will be much is it. We, Dr Drew factor the each other and for X and signed in South, Under and for sight. But X plus, But go side of the bed that X And now the wider book Graham again, we used the broader rule here. And then we get now Egypt and for next, dearie with there was yet each other and for X and for in the front times and, uh, sigh Bet that X blessed better course, I bet the axe and now we bless. The second time will be each of the answer. Thanks On then, we have inside will be the real deal decide equal Judah go size and Phi Betta Because I bet the X the review because I go to the minus eyes or you have a minus But that square site X and that will be the answer here.

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, I'm completing the given reaction. So just look at it carefully, See us a four bypass tool, then It did at 100° integrate. Then it will break into See us a four S 20 this is compound a and when component A react AD 230°C. Then it will form CUS 04. And this compound At this compound at 800 degrees integrate will form will form, this is compound, we will form See You oh plus S 03 Therefore according to the option option we eat correct.


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