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Balance each redox reaction in acid solution:CzOi 1H;Oz702 4TeO; 2 +NO: 7 TeNO;Balance each redox reaction in basic solution_Hio +CriOz 202 + CrtTeNO;TeOz?NzOaIO; +...

Question

Balance each redox reaction in acid solution:CzOi 1H;Oz702 4TeO; 2 +NO: 7 TeNO;Balance each redox reaction in basic solution_Hio +CriOz 202 + CrtTeNO;TeOz?NzOaIO; + ReReO;"IOCl

Balance each redox reaction in acid solution: CzOi 1 H;Oz 702 4 TeO; 2 +NO: 7 Te NO; Balance each redox reaction in basic solution_ Hio +CriOz 2 02 + Crt Te NO; TeOz? NzOa IO; + Re ReO;" IO Cl



Answers

Write a balanced ionic redox equation using the following pairs of redox half-reactions.
\begin{equation}
\begin{array}{l}{\text { a. } \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}} \\ {\text { } \quad \mathrm{Te}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Te}} \\ {\text { b. } \mathrm{IO}_{4}^{-}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}} \\ {\quad \text { Al } \rightarrow \mathrm{Al}^{3+}+3 \mathrm{e}^{-}(\text { in acid solution })} \\ {\text { c. } \mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}} \\\ \quad{\mathrm{N}_{2} \mathrm{O} \rightarrow \mathrm{NO}_{3}^{-}+4 \mathrm{e}^{-}(\text { in acid solution })}\end{array}
\end{equation}

And so for party. The Peers of Redox. Half reactions to be combined are Iron I run plus two plus two electrons and delirium T plus two less two electrons. Tow lorry. Um, first, we take a look at a number of electrons released in an oxidation reaction and consumed in a reduction reaction and multiply the coefficient as needed for these numbers to become equal. Since in this instance, the number of electrons in both ingredients is already equal, we don't need to change the coefficients next. We ate together both reactions here. The reaction became F e Bless T Clarion Plus two and Iron Plus two less salary Um T. Next to be eight water. I aims if it is needed to balance the equation in this instance, we don't need to do that because the previous equation is already balanced there, for the final solution is here we have a final solution. F E plus delirium. Let's to e plus two less to Lariam solution for part B. The piers off the dogs off reactions to be combined are alimony. Um, plus alimony, um, to alimony. Um, plus three bless three electrons and I hope for negative plus two electrons toe I all three negative. First, we take a look at a number of electrons released in an oxidation reaction and consumed in a reduction reaction and multiply the coefficients as needed for these numbers to become equal. In this instance, we multiply the coefficients in oxidation reaction by to destruction by two and in reduction reaction by three in this direction by three. Next, we ate together both reactions. The reaction became you great. Three. I owe for negative plus to L yeah, Thio Tree I or three negative plus toe Ill less three. Next, we ate water irons. If it is needed to balance the equation, since the given reaction is in the aesthetic solution the water species, we can use our positive and h 20 Therefore, the final solution is here The final solution three I, or for negative plus to polonium plus six Hydrogen. Yes, positive Flying to three I or three Negative plus two Alimony Um, plus three plus three is toe and a solution for Parsi. The pair's off redox off reactions to be combined are here I dean molecule plus two electrons to to are you d nine and and to all too, and not three negative plus for electrons. There's a printing mistake in the oxidation relation here. The actual character creation should be here. The correct prediction off oxidation in tow all into all to and a tree get you bless eight electrons first to be Take a look at a number of electrons released in an oxidation direction and consumed in a reduction reaction and multiply the coefficient as needed for these numbers to become equal. In this instance, we multiply the coefficients in reduction reaction by two. Next we add together both reactions. So we have a reaction here for I to plus in tow to, you know, three negative, less it Are you denying? Next we ate water. I guess if it is needed to balance the equation, since the given reaction is in the ascetic solution the water species, we can use our X positive end h 20 there for the final solution. Here is we have for final solution you four i d molecule plus nitrogen oxide, less five molecule selfish to all and two and three plus. And are you deny nine plus 10 edge Positive. I

Gotta balance is Redox reaction that occurs in acidic solution. Start by breaking out the half reactions here, figuring out where the electrons are being gained or lost. And so we've gotta sign oxidation numbers to do that. And so we see that the antimony is going from plus three, two plus five. And so that's a loss of two electrons. So we're gonna put two electrons in on the product side. We're gonna take a look at the other half reaction. The permanganate I am is going to mn two plus so again assigning oxidation numbers to figure out where the electrons are going. So magnesite plus seven on the react inside and plus two on the product side. So that is a gain of five electrons. So now we're gonna balance oxygen were necessary. So that means we have to add four water molecules to the right side of that second half reaction there. And then we have to balance the hydrogen by adding a hydrogen ions to left side top Reaction also needs some oxygen, so we're gonna add four water molecules to the left side. And now to balance out the hydrogen, we've gotta add eight protons to the product side. Now that the atoms and the charge are all balanced in the individual half reactions, we have to make sure that the number of electrons lost is equal to the number of electrons gained. So those electron numbers there, we have to multiply that top reaction by five and the bottom reaction by two. And so I'm gonna write out the new totals there with the new coefficients. And that's what I'm writing out now in green. So 20 water five antimony three plus we get five of that eye on with the intimate e and oxygen In it. We get 10 electrons and 40 protons. And again for the second reaction, distributing that to all the way through 16 protons, 10 electrons to permanganate on and then to mag unease to ions and eight water molecules. I'm gonna simplify things before I add the 2/2 reactions together. In the second half reaction, we're using 16 protons and in the first half, reaction were making 40. So that's gonna change down to 24 protons as a product in that first half reaction, water also needs to be adjusted here. We're making eight in the second, but we're using 20 in the first. So that's a net use of 12. Water molecules are Electrons are canceling out. 10 lost, 10 gained. So now I can write to the combination of the 2/2 reactions here. Now, I forgot here to add the water, and I'm gonna still do that at the end. I just lost track of it. There should be 12 water molecules on the reactant side. So the products are to mag unease two plus ions, five of that Polly atomic eye on and let's see, 24 hydrogen ions or protons and then the 12 water molecules. They're on the reactant side. So that is that equation all balanced? We're gonna balance this reaction in basic solution. And we'll start by figuring out the half reactions, figuring out where electrons are being gained and where they're being lost. So we've got this one with the night die nitrogen monoxide in the night right? I am. We're gonna balance the nitrogen atoms first. Then we're gonna look at the oxidation numbers and figure out where the electrons are going. And once we do that, we'll find that four electrons are being lost in this half reaction, we'll put those on the product side. Then we'll explore the other half reaction. The hypo Cool, right? I am going to chloride ion chlorine zor balanced, so we'll go ahead and sign oxidation numbers minus one on the right and plus one on the left. So that's a gain of two electrons. So we'll put that on the react inside. And since we're in basic solution, we will balance the charge by adding hydroxide. I'll so the left side of that first half reaction doesn't have enough negative charge, so we'll add six hydroxide there. Second half reaction will have to hydroxide to the product side. And now we've got a balance oxygen atoms. So we're gonna have three water to the product side of the first half reaction and one water molecule to the left side of the second reaction. And now all that's left is to balance out the number of electrons gained and lost. So we have to multiply the second half reaction by two and then distribute that to through all the coefficients we can simplify before adding together here. So there's two water as a reactant. The second reaction and three is a product, so that simplifies down to just one as a product. Double check that are electrons lost and gained or the same those cancel out. Let's look at the hydroxide ion now six consumed in the first half, reaction for produced in the second half. Reactions. That's a net consumption of two hydroxide ions. So we'll go ahead and add up those 2/2 reactions Now that we've simplified everything on and you'll see the balanced equation here, one die nitrogen monoxide to hyper chloride ions, and two hydroxide ions are gonna produce to chloride ions. Two nitrates. Excuse me. Yes, Nitrite isles. Let me add that charge in and one water molecule.

Answer for part a year. The reaction to be balanced ease SPS antimony plus three plus aimed in or four and I am in to positive plus SB all tree all four training in chief. The changes of oxidation number dead occurred in this reaction. Are you SB plus three SB plus five plus two electrons and AM in plus seven less five electrons Men Vinnie's plus two Now be right the actual half reactions that occurs in this direction since the reaction occurs in the aesthetic solution we add as positive end is, two species were necessary for the number off items off these elements and churches to be equal on both sides. Off the half reaction The half reactions that occur in these reactions are here off reaction. We have oxidation, half reaction and reduction. Half reaction oxidation. Half reaction is SB three plus plus four each to all SB or forge three minus plus eight is positive plus tau electron and reduction have reaction and in off for negative. Plus it is positive less. Five electrons 2 a.m. in plus two plus four is toe. Since there are two electrons released in oxidation, half friction where s five electrons are consumed in the reduction half reaction for the number of electrons to be same on both sides. Off the equation, we multiply the oxidation, half reaction oxidation half reaction by five and reduction half reaction by two. And then we add together these two half reactions here, like the reaction become five SB three positive bliss to em in or phone I get you. Plus 16 is positive. Plus 10 electrons. Yes, 20 is to all five s B 04 three minus less 40. It's positive. It's positive, plus to em in to positive. Plus it is 20 plus 10 electrons. Finally, we cancel er deleted. Suspicious did occur on both sides of the equation unchanged to get the final solution. Here we get the final solution five SB three positive plus to em in or for negative plus 12 as to all 25 SB Awful. Three Negative plus two I am in plus two plus 24. It's positive. Okay. And next behalf solution for part B. The reaction to be balanced ease c l o negative plus and to all c l negative plus in or too negative. And the changes up oxidation number dead occur in this direction are yeah, in just one to nitrogen plus three plus two electrons, and second one is chlorine plus one plus two electrons to chlorine minus one. Now we write the actual half reactions that occur in this reaction. Since the reaction occurs in the basic solution we add, or its negative end store species were necessary for the number of atoms off eight elements and charges should be equal on both sides of the half reaction. The half reactions that occur in this reactions are oxidation half reaction and reduction, half reaction oxidation. Half reaction is in tow. All plus six or h to to end all toe get you plus three is too old plus four electrons. N T s c l o Negative Plus is to all plus two electrons. See a negative plus 20 h. Negative. Since there are four electrons released in oxidation half direction. Where is two electrons are consumed in a reduction? Have friction for the number of electrons, Toby. Same on both sides of the equation. We multiply the reduction of reaction by two and then the air. Together. These two half reaction the equation becomes Yeah, to see ill o negative less in tow plus six or h they get you plus through each to all plus for electrons. Uh huh. Do you know to I get you bless two c l negative less for O h negative. Okay. Less tree h 20 plus four electrons. Finally video later. Cancel the species dead fucker on both sides of the vision unchanged to get the final solution here two c l o get you bless in tow, less to or edge negative to end All too negative plus to cl negative Plus is to or

So we're going to be balancing some Redox reactions, and all that means is that we're going to balance them. And then we're also going to see just which have some reduction is occurring and which has are where oxidation is referring. So just to give a quick background, Redox is just short reduction and oxidation, just a reaction where both are occurring. Go ahead and write that down. Simply put, that is an oxidation is just the loss of electrons and a good rule of thumb when we're trying to identify which halves of our equations are where oxidation is occurring. Is that what we need to balance things out in ad electrons to the right side of the equation, then that generally means that that is where oxidation is occurring. Now, in contrast, reduction is the gaining electrons. And again, a good rule of thumb is that we're trying to balance things out, and we end up having to add electrons to be left side of an equation. That's that's probably for that's where production is is happening in that house. And then another thing to note is that, um, for all of our given, you know, equations they're all happening in acidic conditions. That just means that we're balancing things out. We can either add hydrogen ions or water molecules in order to balance things out. Now, if if they were to be occurring in basic conditions thes, you know, reactions, then we would be adding hydroxide irons or water molecules. But we don't have to worry about that, since all our givens are currently, um occurring, you know, an acidic conditions. So we'll go ahead and get started with the first given down. And we can tell right away that there are there seem to be two main parts to this. The equation that we're going to, you know, um, divide into two, you know, half equations or half reactions using the half reaction method in order to kind of simplify and balance three overall equation in the end. So those two, all right, this apartment here with 10 and then the second half is right here. Hydrogen. So let's go ahead and balance the tin first. Mm already. Then let's just start balancing things out. Balance. Both, um, the elements or the atoms that we have. Whatever we have on each side and the chart So we have 1 10 on the left side and as your on that charge, girls want in on the right and a two plus charge, and to balance that charge out and bring it back to zero, you're going to add two electrons to the right side in order to bring that charge to zero. And since we added these two electrons to the right side, that just means, um, according to our little rule of thumb that this is behalf of the overall equation where oxidation is occurring and likewise for the second half of the equation, go ahead and write down our givens. So we have one hydrogen on the left and a plus one charge and two hydrogen is on the right and zero net charge. So the easiest way to even out the number of hydrogen we're gonna even them out first before worrying about the charge is just to add to so in. In other words, we're going to multiply this hydrogen on the left by two to bring the total number of hydrogen institute. That also increases our charge 22 plus. So how we're gonna balance that out is we're going to add two electrons to the left side, and that will bring our total charge back to zero now, since we added it on the left side again. Our rule of thumb is that when we add electrons to the left side, Um, this means that this is where reduction is occurring already before we add them both together, it's it's good to note that we don't want to see the electrons as much as possible. We don't want to see electrons in the final overall equation, so thankfully, they're both equal on, and we don't have to find a least common multi between these two house. So what we can do is we can just add them street, these two house, go ahead and do that. And this is what pitch in already and then for the right side already. And we already see here that the electrons cancel out. And so we're left with overall equation of this. There we go already, so let's go ahead and get started on the second given equation that we have already. So here's our 2nd 1 that we can go ahead and work on. So again we already have. It's, it seems, pretty obvious that we have to kind of separate parts or has of this equation one that involves chromium and one that involves iron. Copy that down. All right. So go ahead and mark of the different house, the equations. There's one. And then there's another Well, let's get started on kind of chromium have first. All right, so here we see that, um, we can go ahead and list out what we have on both sides before we start trying to balance anything again. Always good force of habit. Know what you have first. So we have two chromium, seven oxygen's and a total net charge of minus two. Then on the right side, we have one chromium, no oxygen's and a total charge of three. So here is where it gets slightly creaking. So notice how we're going to have to balance out the oxygen's like here and to do so. Another good thing to remember is that again we are in acidic conditions. When we're trying to balance out hydrogen, we usually can just add hydrogen ions, but when we're trying to bounce off, oxygen's in reactions happening in acidic conditions. Um, it's a good um, I guess Pointer ruled them to try to add some water molecules for So let's try that right here. You can go ahead and add some, uh, H 20 molecules, um, to the right side and will add seven. So that way we can balance out the number of oxygen's on both sides. But in doing so, we've now added 14 h two times seven about 14 hydrogen. And so that means that we're also going to have to add hydrogen to the north side. And so that's where our hydrogen ions come in. Let's try that 14 gin. I owns cat ions, um, to the left side. So now that seems to balance out. And then finally, we can go ahead and balance out our chromium on both sides. So let's multiply. Be chromium cat eye on it on the right side by two metal finally even out both of our chromium on both sides. That should be. And then finally, the last thing we worry about is charged. Now, since we've added a whole bunch of things to both sides and done, um, kind of we're things to both sides. That's definitely changed our total charge on both sides. Let's go ahead and first figure out what the charges on the left side on the left side first. So our charges no longer to minus. Um, it is two minus plus, uh, 14 a positive 14 charge. And that comes from the plus over here. That is, associate it with the hydrogen ions. So 14 pocket of 14 plus negative, too. That's Ah, plus 12 charge on the left side. So far. And right here we have two times the three plus charge. Here we have, uh, six plus charge on our right side. So in order to balance things out, we'll just go ahead and add. Let's move this down. Go ahead and add six electrons to the left side, and then that will bring our total charge back to six. Plus. All right. And so there we have it. Okay. And then the second half of the equation, Woods, go ahead and figure that out. It's the half involving the iron. But before we do that, one important thing to know before we forget is since we added the electrons to the left side again, a rule of thumb states. And when we add electrons to the left side. That is where reduction is occurring. So let's go ahead. And just like this, half down as the reduction before we get our rig and then moving back to, um are iron, you know, kind of half of the equation. All right, let's go ahead and write that down, and then we'll try to balance that out as well before adding the two together. So again, we have one iron on this side and a total charge of plus two. And we have what, one iron on this side and a total charge of plus three. All right, so the only thing that we need to do is balance out the charges. Since it seems, um, the irons are already evened out, so we can just go ahead and add one electron to our right side, and that should bring our charge total back to two, plus evening them both out. And so we just scroll up a little. Let's go ahead and add the two together. Before we do that. Notice how we need to balance out. I guess the electrons specifically in order that when we put everything together, they can cancel out um and So again, theologian, transfer the iron. Half of the reaction have been added to the right side, making this half of the equation, of course, the oxidation. It's not forget that. Just always important to know. But anyways, moving on, we need to be able to find the comment the least common multiple between, uh, one and six and that six. So what we're going to do is multiply this entire equation by six and then add that to are chromium half of the equation, so to speak. Let's go ahead and just write that down really quick before we forget. All right, this guy on time six and six electrons already. So let's go ahead and add them both together. This and this, which is actually this. So here we go. This is what we should end up with. Let's make it a different color so we can tell what already this should be Our We're almost there at the last step when the last step before the final overall equation balanced, of course. So we forgot her. We go the two chromium cat ions, seven water molecules and six iron cat ion and six electron. And so in both sides. The six electrons cancel out. And so we're left with an overall equation. It looks something like this. Here we go already. Oh, slightly more complicated, but definitely, um, not too difficult once all the multiplying and and, um, and figuring out where things go kind of gets involved. So let's go ahead and just get rid of all this and we can get started on the third given equation that we have for Okay, All right. So our third given IHS as follows. Go ahead and write that out. All right? And let's go ahead and just identify the two halves of, um, the to half reactions for this overall reaction. So here we have What I I guess we can refer to as the main guinea's portion or manganese have. And then the other is the Florida, Huh? So let's start no. With the manganese, the half thunderbolts making these. So go ahead and write this down and again, we're going to list out everything that we have. So we have one mayonnaise on the left to oxygen's and a zero net charge on the left. One manganese, zero oxygen's and a, uh, two plus net. charge on the right, so let's go ahead and even these out first. So what we can do here is, since we are looking for an oxygen on the right side again, a good rule of thumb is to try and add water molecule. Let's go ahead and try that. And since we need to oxygen's, try adding to water molecules, so that will give us the two oxygen's we need. And this, of course, means that we've now added hydrogen into the mix. So we have to put that we now have four hydrogen H two times, two on the right, and now we need to add hydrogen to the left side as a result. So let's go ahead and add some hydrogen ions. Four to be exact on the left. All right, so now that evens out everything the element wise. But now we need to figure out what the new charges are in each side and even those out as well. So now the charge charge told charge on the left side is now four plus, and meanwhile, the right side charge wise has not changed. So what we can do is just add two electrons to the left side that will bring us to two plus on the left side. And since we've added electrons to the left, that just means that we have production happening in this half of the equation already. Then let's move on to the second half of the equation involving chlorine. You too, Planning gas? So what do we have here? We have one Korean and, uh, Munn minus charge and then on the left side. And meanwhile, on the right side, you have to, Corinne, and because you learn that charge, So all we need to do is just balance out the number of Koreans first. Good. And do that the multiply the left side by two. That gives us two Koreans we need. That also gives us a minus two charge on the left, which means that in order to balance it out, we have to add electrons to the right side. Now, rule of thumb to remember is that one near balancing and figuring out where we need to put the electrons? It's always important to note that we need to put the electrons on the side that is most positive. In this case, the most positive side is, of course, zero since zero is closer to, um I guess, Or more positive than negative, too. So we'll have the electrons again to the right side. That brings right sides a total charge. Too negative, too. And now everything is even and perfect. And since we've added electrons to the right side, that means, but this is where oxidation is occurring is in this. Go ahead and write that down. Always good to make note of that. All right. And then the last step is just to add both of the equations together Now, since, um, both have, um, two electrons on both sides, then they will go ahead and cancel. Once we add our equations together, you'll see what I mean in a second. Um, so let's go ahead and just add them first. So we'll go ahead and add, um, our Korean as well to each toe. Let's are Corinne. Yes. Yeah, and our two electrons. And so these, as you can see, the electrons on this side and this side, the left and the right both cancel out. And so we are left with more tidy equation that doesn't involve electrons, and this should be the overall equation. Um, for this reaction balanced, of course, in acidic conditions. And now we're going to go ahead and work on our fourth given equation, the last finest it. So go ahead and write it out. And we can already see that there are two halves to this equation as well. A silver half in the other half involving C H 20 All right, so let's go ahead and identify the to house. There's one and there's you. So let's go ahead and just write these out and balance them. Do what we've been doing all this time already. So here we just got a list out. Everything that we have and the charges associated with it. Okay, so we have one carpet here to hydrogen, one oxygen and zero net charge. We also have one carbon here to hide regions to oxygen's and a zero net charge as well. So let's see how we can balance this out. So first again, since we are trying to balance out oxygen's, um, we're going to add, go ahead and add a water molecule to the side that's missing an oxygen, and that would be the left side. So let's move this down a little bit on, go ahead and add one watermark, so that brings our number of oxygen's to two. But that also means that we have now added two more hydrogen za as well. So that brings our total number of hydrogen is on the left side to four. And, um, we can just not worry about that for now because one thing that we can do is even out. The number of hydrogen is on the right side now, by adding to hydrogen ions that will bring our total member of hydrogen is on the right side, up to four. Smoke this a little bit closer this way. There we go for symmetry, and now that everything is even out, all we need to do is balanced charges. So as we can see, the charges now on the right side have gone up to two plus. Meanwhile, nothing has changed on the right side. So in order to bring our right side back down to zero charge wives, we're going to add two electrons and so that automatically brings that two plus back to zero. And now everything is even. Let's go ahead and move on to the silver half. So with silver, we have once over cat eye on on the left side and regular solely so around the right, it's list out what we have and the charges. So we have one silver and a plus one charge on the left, one silver and zero net charge on the right. And so the easiest way to balance this out is simply to add one electron to our left side already. And so moving that back up really quick. Um, since we added electrons to be right side of this half equation or half reaction, we know that this half is where oxidation is occurring. Go ahead and write that down oxidation have. And since we added electrons to the right side of this half reaction, you know that this is where reduction is the grains. We'll go ahead and put that down as well. And since we're adding these two together and trying to get rid of our electrons were going to find the least common multiple between the 22 and one, and that is obviously too. So that just means that we're going to multiply our bottom equation by two and So the new equation for the bottom it's going to be as such. Okay, And when we add both of these together this half and this house, we're going to get this We go and finally and here's where we see our electrons cancel out. And finally we're left with this overall equation now balanced. And so, uh, as you can see, it's not too difficult once you get used to just finding the house that, um, makes sense that are similar and then listing out what you have. And then after that, just, you know, going through the steps, balancing, you know, either hydrogen zehr Oxygen's first, then balancing the rest and then balancing your charges, that's all it really takes.


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Boat i5 pulled into dock by rope attached to the bow of the boat and passing through pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at rate of m}s how fast is the boat approaching the dock when itis 5 m from the dock? (Round your answer to two decimal places:) MC
boat i5 pulled into dock by rope attached to the bow of the boat and passing through pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at rate of m}s how fast is the boat approaching the dock when itis 5 m from the dock? (Round your answer to two decimal places...
5 answers
[0/5 Points]DETAILSPREVIOUS ANSWERSMYNNCFind one implicit function defined by the given equation7x2 9y2State the domain of the function. (Enter your answer using interval notation:)Submit Answer
[0/5 Points] DETAILS PREVIOUS ANSWERS MYNNC Find one implicit function defined by the given equation 7x2 9y2 State the domain of the function. (Enter your answer using interval notation:) Submit Answer...
1 answers
Use linear approximations to estimate the following quantities. Choose a value of a to produce a small error. $$\tan \left(-2^{\circ}\right)$$
Use linear approximations to estimate the following quantities. Choose a value of a to produce a small error. $$\tan \left(-2^{\circ}\right)$$...
5 answers
Polarizer axis For the polarizers the figure suPP ose the incident light linearly polarized, the transmitted intensity (through both polarizers) 0.1SxIo, and the axis of the second polarizer makes an angle 8=420 with the axis of the first polarizer: What is the agle (in degrees) Unpolarized the initial direction polarization makes with the first polarizer? (Insert the number of degrees without unit ) lightSubmit Answcr Tries 0/8
Polarizer axis For the polarizers the figure suPP ose the incident light linearly polarized, the transmitted intensity (through both polarizers) 0.1SxIo, and the axis of the second polarizer makes an angle 8=420 with the axis of the first polarizer: What is the agle (in degrees) Unpolarized the init...
5 answers
J9 calculalc { Two masses horizonta the I rotational "'819 centimeters Should 1 10 kg and m2 Tto 0l m S be placed t0 suspended 1 and the right or to You the for the V [0 place rod to be 1 E mas5 The rod of tbe rod $ 1 hung from suspension point? Explain your 1 ceiling 3 1 1 whicbCalculate the exact [ocation 1 should oe placed
j9 calculalc { Two masses horizonta the I rotational "'819 centimeters Should 1 10 kg and m2 Tto 0l m S be placed t0 suspended 1 and the right or to You the for the V [0 place rod to be 1 E mas5 The rod of tbe rod $ 1 hung from suspension point? Explain your 1 ceiling 3 1 1 whicb Calculate...
5 answers
Calculate the limit for the given function and interval. Verify your answer by using geometry: lim LN = where f(x) = 2x +4 over the N-0 interval [0, 4]lim LN N-0
Calculate the limit for the given function and interval. Verify your answer by using geometry: lim LN = where f(x) = 2x +4 over the N-0 interval [0, 4] lim LN N-0...
5 answers
2 2 conducting The room 2 metal balls; Duldedeb 'separation , Wre. Nhed B distance Athey Hsolated wihe d [ 1 Goforeatcact 3 Entropy this electric charged objects 3 way, Dividedied chaarges elllecurons 33 3 tripled, Is dow 3 Jirom 1 onee eeh 8 2 54 cxerts anothsthbysfcr 0
2 2 conducting The room 2 metal balls; Duldedeb 'separation , Wre. Nhed B distance Athey Hsolated wihe d [ 1 Goforeatcact 3 Entropy this electric charged objects 3 way, Dividedied chaarges elllecurons 33 3 tripled, Is dow 3 Jirom 1 onee eeh 8 2 54 cxerts anothsthbysfcr 0...

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