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Select the correct coefficients (0-5) needed to balance the following reaction under basic aqueous conditions:NzH4 Cu(OH)zNz CuNzHaCuOHCu(OHIzNzHzo...

Question

Select the correct coefficients (0-5) needed to balance the following reaction under basic aqueous conditions:NzH4 Cu(OH)zNz CuNzHaCuOHCu(OHIzNzHzo

Select the correct coefficients (0-5) needed to balance the following reaction under basic aqueous conditions: NzH4 Cu(OH)z Nz Cu NzHa Cu OH Cu(OHIz Nz Hzo



Answers

Balance the following net ionic equation by the half-reaction method. The reaction takes place in acidic solution.
$$\mathrm{NO}_{3}^{-}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{NO}(g)+\mathrm{Cu}^{2+}(a q) \quad \text { Unbalanced }$$

Here we have to balance to have reactions in the basic media. So in the first half reaction chlorine or minus, and iron is reduced so core in my chloride and we have to balance this half reaction. Let's do this. So first of all we can place same coefficients of one before each chlorine species. And now let's think whether where well where we need to add high, where we need to add water and where we need to add hydroxide. So as one my guests chlorine plus one is oxidized to chlorine minus one. Therefore on the left side of the equation will likely have electrons. So therefore we need to add hydroxide to the left to the right side of the equation and to balance and we therefore we all need to add water on the left side. So let's share for the number of elections which we need to head. So oxidation number of chlorine changes from plus one, two negative one. Therefore it's two electron reaction. And to balance charge, we have to place to before hydroxide. Now we need to balance hydrogen and this one might see hydrogen are already balanced always oxygen. Therefore we need to place a coefficient of one before water. So therefore the overall the final have reaction looks like the following, wow hello minus reacts with bond water molecule and two electrons and then forms one chloride and two hydroxy and eyes. So this is a first half reaction in the second half reaction um permanganate is reduced to uh manganese dioxide and the basic media. Yeah, let's do the following step. So to balance the church will need to add elections to the left because manganese is reduced from plus seven two plus four. So therefore we'll be eating hydroxide and eyes to the right side of the equation. Mhm and water to the left side. To balance hydrogen. So now we can balance many. Is first on each side of the equation, we have one man Dennis and now we can and we can change the number of elections. So magnus changes its oxidation state from plus seven two plus four. Therefore it's a three electron reaction. And now we can balance uh the church. So on the left side of the equation, total charges uh plus negative four. Therefore we have to place for before hydroxide and I'll so now we can balance hydrogen and oxygen. So on the left side on the right side explores equation there are 44 hydrogen atoms. So then we need to for the coefficient of two before water. And now we can check for oxygen's. So on the left side there are six oxygen's, so is on the right side and everything else is balanced. So therefore this equation, this balance and we've solves a given problem. Thank you very much for Washington's video. I hope I help you buy

To write balanced equations for the following redox reactions and basic solution. Using the half reaction method, we need to start with the half reactions. The first reaction contains a half reaction of permanganate, going to manganese, peroxide and sulphide. Going to sulfur first. We balance everything but hydrogen and oxygen, manganese are balanced. Then we balance oxygen's with water. We have two oxygen or sorry for oxygen and two oxygen's. So we need to more oxygen's or two waters on the right hand side. We then balance hydrogen with hydrogen ions. We introduced four hydrogen is here, so we need four age plus, is on the left hand side. Then we balance charges with electrons. There's no charge on the right hand side. We have a one minus and four plus, so that's three plus. So we need three minus to make it neutral for the next one. All we need is two electrons here. Then we're going to multiply the first reaction by two and the second reaction by three so that our electrons will cancel and then some them up so far. We have balanced it in acidic solution by adding the hydrogen ions To convert to a basic solution. We simply add the same number of hydroxide to both sides of the equation as we have hydrogen ions, namely eight hydrogen ions. And the hydroxide is combined to form water. So we'll get eight waters here. These eight waters will cancel these four waters here, leaving four waters on the left hand side. And then what is left is our balanced chemical reaction in basic solution for the next one. A half reaction to see you going to see you two plus and then cielo minus, going to C L minus. We'll just balance the charges with the electrons because there's no hydrogen or oxygen. The next one will balance the oxygen's with water, putting one water on the right hand side, and then hydrogen is with hydrogen ions, putting two hydrogen ions on the left hand side. Then we'll balance the charges. We need two electrons here, so both sides have a -1 charge. Then we'll sum them up. We don't need to multiply any equation by a constant because the electrons already cancelled. And this would be the balanced equation in acidic solution. To make it basic, we'll add the same number of hydrogen ions as we have had the same number of hydra oxides to both sides as we have hydrogen ions. Over here, the hydrogen ions and the hydroxide combined to form two waters. These two waters cancel this one water, leaving one water on the left hand side and this is that are balanced redox reaction in basic solution.

So this video, we're gonna go over question 94 from Chapter four, which says balance the following oxidation reduction reactions that occur in basic solution, eh? So how do we balance oxidation reduction reactions in a basic solution? Well, first we take the reaction and we split it into an oxidation half reaction and a reduction half reaction. Then, for each of our 2/2 reactions, you balance Adam's other than hydrogen or oxygen first, then wants those atoms are balanced new balance oxygen atoms by adding water molecules. And then you balance hydrogen atoms by adding protons, and then we balance charged by adding electrons. So you need the net charge in your reactant ce to be equal to the net charge of your products. So wherever those aren't equal, you add electrons to make them equal. Um, and so far everything is exactly the same as if we were doing this an acidic solution. We just need to add two more steps to make it adapted for basic solution. So then, if you added protons, you add just a cz many hydroxide ions to both sides of the half reaction. Then you can simplify protons and hydroxide ions that occur on the same side of 1/2 reaction into water molecules. And if that results in water molecules on both sides of the half reaction, then you can cancel those out. Then that all that's left to do is to make sure you have the same number of electrons and attack reaction and add them together. So let's go ahead and do an example. So when a were given CR plus c R +04 minus two form C R O H three. So let's go ahead and assign oxidation states. So for solid chromium, we have an oxidation state of zero because it's in the pure elemental form. Then for oxygen, we have an oxidation state of minus two. S Owen auction responded to Lexa lives for negative Adams. It takes all of the electrons in the bond, giving an oxidation state of minus two on, and then the oxidation states on each other. Adam's need to add to equal minus as do so. There's four oxygen atoms, so minus eight plus what will equal minus two. The oxidation state on chromium must be plus six and then we have cr Ohh three. So this is an ionic compound with our hydroxide. Ion is the anti on eso since there's three of our hydroxide I owns and we know where hydroxide ion takes a minus one charge that's minus three, which means the oxidation state on chromium must be plus three. Then the oxidation states on oxygen and hydrogen are the usual minus two and plus wine eso where the oxidation states changing. Will we see an increase in oxidation state going from zero to plus three. That increase in oxidation state is due to a loss of electrons or oxidation. And then we see a decrease in oxidation state going from plus six two, plus three so that decreasing oxidation state is due to a gain of electrons or reduction. So we have 2/2 reactions notice we have the same product in our half reactions. So when 2/2 reaction share the same product or reactant, meaning the same species is being oxidized or reduced, we call that a disproportion nation reaction. Um, but anyway, we don't know where to half reactions, so let's write them out. So first we have CR to C r. O H. Three and then we have CR or to CR Oh, age three. So first thing that we need to do is to balance Adam's all they've been oxygen or hydrogen, which is just chromium. So in this case, are chromium. Adam's already balanced. Then we need to balance oxygen atoms by adding water molecules. So when I first half reaction, we need to add three water molecules in our reactions and our second half reaction. We need to add one water molecule in our products. Then we need to balance hydrogen atoms by adding proton. So we have 600 rounds over here and three over here. That means we need three crow Johns. Then we have five hydrogen atoms over here and none over here. So we need five protons now. We just need to balance charges by adding electrons. So over here I have a plus three charge, and over here I have a net zero charge. So I'm gonna add three electrons to my products. And then over here I have a net charge of plus three. That's plus five minus two Internet charge of zero. So I'm gonna add three electrons to my reactions. Now, I need to adopt this for a basic solution. So my first type reaction, I added three h plus, which means I'm gonna add +30 h minus to both sides on then my three age plus plus my +30 h minus simplified to three h +20 But since I already had three h 20 on the other side of my equation, this cancels out my second half reaction. I added five protons, so I'm gonna add 50 H minus to both sides of my reaction. When I do that my five h plus plus my +50 age minus simplifies to five h +20 um, than I already had one water molecule over here. So my five h 20 simplifies took four aged. Well, so now both of my half reaction they're balanced and conveniently, they both contain three electrons, so we can go ahead and write them out and add them together. So then I have 30 age minus plus C r yield C R O H. Three plus three electrons. For my second half reaction, I have three electrons plus C R. +04 minus two plus four h +20 um form c r O h. Three plus five o h minus. Um, and that's everything in our 2/2 reaction. So now I can add these together. And when I do, my three electrons cancel from both sides just like I wanted them to. But then I have three hydroxide I owns that are gonna simplify my five hydroxide ions to two. Um So then I add these together and I get CR plus C R four minus two plus four h +20 yields to C R O H 31 from each reaction plus two o h minus. And that's my simplified reaction for part A. Let's move on to Part Beach. So when b, we have him in oh, poor minus. Let's s minus two forms M in us. Buzz s So what are my oxidation shapes? And Emma No. Four minus Ivan oxidation state of minus two times for that's minus eight. Which means my oxidation state and manganese is plus seven. Then my oxidation state on this mon atomic eye on is just the charge on the ion and M and s button ionic compound and ionic compound sulfur likes to take a minus to charge. Um, and then that means my oxidation state on Megan Eases plus two and then for solid soul for the oxidation state zero because that's the pure alone mental state. So where my oxidation states changing? Well, they're changing for minus 2 to 0 s o that increase in oxidation status from a loss of electrons or oxidation. And then there are also changing from 72 plus two. So that decrease in oxidation state is due to a gain of electrons or reduction. So now I have my 2/2 reactions, so let's write them out. So for my oxidation half reaction, I have s minus two. Tow us and then for my reduction, half reaction. I have Emma no. For to him and us. Um, so how could I balance these s o? First? I want to balance elements all there than oxygen or hydrogen. So my first half reaction is fine, but my second half reaction I have a soul for Adam that appears my products, but not my reactive. So I'm gonna go ahead and add as minus two. And you might be wondering, How did I know the ad silver in the form of s minus two. Why didn't I Just add, for instance, just Esther Solid silver. Um, and the reason is because we already had s minus two is a reactant, so it's safe to use. It is reacted when we're balancing our half reactions as well. Um, so then now that we've balanced Adams are there than oxygen or hydrogen. We bounce our oxygen atoms by adding water molecules. So I have four oxygen atoms. My reactions, but not in my products. So I add, um, for H 20 which means that I have or hydrogen atoms and my products, and none in my reactant. So I'm gonna add H plus, um, so now I need to balance charges. So in my first half reaction, I have minus two over here and zero over here, so I'm gonna add two electrons and my second half reaction. I have plus eight minus two minus one. So plus five on this side and zero on this side. So I'm gonna add five electrons over here, and now I need to adapt this for a basic solution. So since I added eight protons to the reactant in my second half reaction, I'm going to add eight hydroxide ions of both sides of the reaction. Um, and the results of that is that my 80 h minus combined with my age plus to form a TSH to O. But since I already had four water molecules over here, this is gonna simplify to four h 20 s. So now both of my half reactions or balanced. But I have two electrons in my first half reaction and five electrons in my second half reaction. So my second half reaction gets multiplied by two. And my first reaction gets multiplied by five. So what do I get when I do that? Well, then I have five s minus two forms, five s plus 10 electrons. And then in my second half reaction, I'm going to have 10 electrons plus to as minus two plus two women and a whore plus h +20 And that's going to be forming to M in us. Thank you. Um, plus 16. Oh, age minus s. Oh, those are my 2/2 reactions. No, I can add them together, and my 10 electrons will cancel from both sides of the equation. But that's all that cancels. So when I add them together, I get seven s minus two plus two in Minot four plus h +20 yields five us close to him and us plus 16 hydroxide ions. So that's my balanced reaction for part B. Let's we want apart. See? So in C I have seen minus was Emin or four minus form CNO minus plus M N 02 Okay, so I think this will be a little bit easier identifying oxidation states if I just go ahead and glass the Lewis structures for two of these ions so foreign my cyanide eye on that Looks like Arvin triple bonded to nitrogen and then each carbon and nitrogen you get on the loan pair of electrons and then my CNO I on my sion eight ion is goingto look like nitrogen. Triple bonded to carbon single, bonded to oxygen. Okay, so let's go ahead and assign oxidation states. So for carbon in this, all of my electrons are gonna go towards nitrogen because nitrogen is the more elector Negative. Adam s O carbon gets four minus two. So the oxidation state is just plus two. And then for nitrogen, it's, um it's five minus two minus six. So that's five minus eight is minus three. Um, and that makes sense because these add two people the overall charge on the ions, which is minus one. Um, so then, for imminent or minus, I get my usual oxidation state of minus two on oxygen because it's the more Electra negative. Adam. Um And then I have four votes. That's minus eight, which means vaccination state on Magan eases plus seven, then in CNO minus. So for carbon, I have four minus zero. All of my electrons are in bonds to more Electra Negative Adams. So I have plus four. And then for a nitrogen. My my oxidation state is five minus two minus six. That's minus three again. And for oxygen, it's six minus six minus two. So that's minus two. And then an immuno, too. Austin takes its usual oxidation state of minus two on, and then, since there's two of them, that's minus four, which means they're oxidation. State on manganese is plus for so where the oxidation states changing? Well, we're going from a plus to shoe a plus four oxidation state on carbon, so that increase in oxidation state is due to a loss of electrons, which we call oxidation. Um, And then I'm having a decrease in oxidation state from plus seven two plus four on Mani's, which is due to a gain. And electrons, which we call reduction. So now we know are 2/2 reactions. Why don't we write them out? S o r oxidation. Half reaction is C and minus to see an old minus. And then my reduction half reaction is m ine 04 minus. Send him in 02 Um, so those are my 2/2 reactions. So first, I need to balance Adam's all there than oxygen or hydrogen zone. My first reaction, carbon and nitrogen already balanced. And my second half reaction manganese was already balanced. Um, then I balance oxygen atoms by adding water molecules. So in my first half reaction, I'm going to add one water molecule on the left and my second half reaction. I'm going to add to water molecules on the right. Um, now I balance hydrogen atoms by adding protons. So I had to h two over here, so I'm going thio on to H plus on my products. And then over here I had four hydrogen atoms. I'm gonna add four H plus. Now all that's left is to balance charges. So I have minus one over here and minus one plus two is plus one over here. So I need two electrons and then over here I have plus four minus one. It's plus three on one side and zero on the other side. So I need three electrons now. I need to adopt this for a basic solution. Eso I take So my first reaction, I added two protons, which means I'm gonna add to hydroxide ions to both sides of my equation. My second half reaction. I'm going to have four protons added on the left hand side. So I'm going to add four hydroxide ions to both sides of the equation. Um, so now I can simplify some things. So my two h plus combines with my +20 H minus to form to H +20 And since I already had one h two over here, that's going to cancel out. And then in my second half reaction, I have my four age plus plus +40 age minus yields for H +20 I'm sincerity had to h 20 that canceled out too. It's too. Um, and now my 2/2 reactions are balanced. But, um, I have three electrons and 1/2 reaction and two in the other half reaction, which means this half reaction gets multiplied by three and this half reaction gets multiplied by two. So let's go ahead and write out what we get when we do that. So then we get 60 H minus plus three c n minus yields three cno minus plus six electrons plus three h +20 And then, for my second half reaction, I have six electrons plus to emit no whore plus four h 20 forming to Emma. Know too. Plus eight O. H minus. Um, so that that's correct. OK, so now we can go ahead and add these two together and our six electrons cancel from both sides just like we wanted them to. But now we can cancel some other things as well. So I have three h 20 and four h 20 So I'm left with just one h 20 Um and then I have 60 age minus and ate a wage minus, which leaves me with 20 H minus. So when I add all this together I end up with three C and minus plus two women +04 plus h +20 yields three cno minus plus two image or two plus two Oh, age minus. So that's our balance. Redox reaction for part C.

In each part of this problem were given an unbalanced Redox reaction. And for each one of the given reactions, we need to balance it using the half reaction method. So starting with the reaction given in part A, you first need to split up the unbalanced total reaction into 2/2 reactions. The way that we do that it was, is that we split up the oxidation and reduction reactions so chromium is oxidized in nitrogen is reduced. So that's how we know that these are the two half reactions. And now, in order to balance each one of the half reactions and then add them together, we first need to balance all of the atoms that are not oxygen or hydrogen. And then we need to balance the oxygen's with water and the hydrogen is with H plus. Since we're in acidic solution, then after that we need to balance the charges and then we can add the 2/2 reactions together to get the total balanced reaction. So starting with the oxidation half reaction above, we see that the chromium zehr balanced and we have no oxygen's or hydrogen, so we just need to balance the charges the reactive side has a charge of zero. The product side has a charge of three plus, so we need to add three electrons to get a charge of zero on each side. And now for the other half reaction, we see that we have one nitrogen on each side. So those air balanced and we need to balance the oxygen's with water. So if we add to liquid water on the product side, we can see that we now have a total of two oxygen, a total of three oxygen atoms on each side of this reaction. In essence, we're in acidic solution. We need to balance out the hydrogen is that we introduced from water with H plus. So we have four. Hydrogen is from water, so we need for equals H plus ions, and now we need to balance out the charges. On the left side, we have a total charge of plus four plus negative one from an +03 so total of plus three. And on the product side, we have a total charge of zero. So we need to add three electrons to get the charge on the left side from plus 3 to 0. And now that we have balanced both of the half reactions, we can add them together to get the total balanced Redox reaction we can cancel out. There's three electrons on either side of the reaction arrow. We can get our final balanced reaction to be for each plus yanquis. Plus I know three minus equally ists plus chromium solid goes to chromium three plus Aquarius, plus a no guess plus two liquid water. Now we can see that all of the Adams and the charges are not are now balanced. So this is the final answer for the balanced Redox reaction. Now we use that same method to get work through Parts B and C for Part B. We split up the 2/2 reactions again, and we start by bouncing the top one, and we see that we have one carbon on each side, so that is balanced. And now we need to bounce out. The oxygen's with water and we can add liquid water to the left side, and that will give us a total of two oxygen atoms on each side. So now the oxygen's air balanced, and now we have a total of two plus three plus one. So six hydrogen ins on the left side. So we need six h plus against. Since we're in acidic solution, we need to balance the hydrogen is with H plus and now we need to balance out the charges and we see that on the left side there is a charge of zero in on the right side gave a charge of six plus. So we need to add six electrons to the product side to get aid charge of zero on each side. And now for the other half reaction, we just have CE and so the Adams air balanced. And now, in order to balance the charges, we need to add an electron to the four plus charge to get a charge of plus three on each side. And now we add these two reactions together. But notice that we need to cancel out the electrons and we have one electron on the reactant side from the second reaction, and we have six electrons on the product side from the first reaction. So we need to multiply the second reaction by six, and then we can add these two together. So if we multiply that second reaction by six. We now have six elect electrons that we can cancel out on each side, and we can begin to write out our overall balanced reaction each to oh Liquid was ch three. Ohh, Equus plus And remember, Since we multiply that second reaction by a factor of six, we now have six C four plus acquis and for the products we have again from multiplying the second reaction by six, we have six C three plus Equus Plus Co. Two Equus plus six H plus Equis. So that's the final balanced Redox reaction for the given reaction in Part B, and now we do the same thing for part C. We have our reaction given in the problem, and we split it up into the 2/2 reactions. And so, starting with the the first reaction, we can see that the sulfur zehr balanced and we need to balance the oxygen's with water. If we add water to the react inside, we have a total of four oxygen's now on on each side of the reaction, and we have to hydrogen on the left side that we need a balance out with to H plus ions on the right side and now we can balance out the charges. We have a total charge of minus two on the left side and a total charge of minus of of zero on the product side. And so we need to add two electrons to the product side to get a charge of minus two on each side. And now for the other half reaction. See that man Younis is balanced and we need to bounce out those four oxygen's with four waters and now we have 888 Hydrogen is from the four. Water is that we need to bounce out with eight h plus ions and now we bounce out the charges we have on the left side eight plus and minus one. So a total charge of plus seven on the reactant side, On the product side, we have a total charge of plus two should get from plus seven to plus two. We need to add five electrons. And now when we add these two together, we need to cancel out the electrons so we can multiply the first reaction by five and the second reaction by two so that we have 10 electrons cancelling out on both sides. And so now we can go through and write out the final reaction. So we want to play that first reaction by five. So we have five s 03 tu minus acquis. And for the H plus ions, we see that we have a total of eight on the reactant side, times two so 16 and to times five So 10 on the product side. And so all of the products will be used up and we'll be left with a net of six h plus on the react inside and from the second re actually multiplied by to sue two mn 04 minus Equus. And then on the product side, we have to and then two plus after multiplying that second half reaction by two. And for the waters, we see that we have four waters times two so total of eight on the product side and one water times five. So a total of five waters on the reactant side. So we have eight waters on the product side and five on the reactive side. So all of the reactant waters are gone and we're left with three waters for a product. So plus three each to a liquid and then was five s 04 tu minus a quiz. And that's from that first reaction, multiplying it by five. So that is a balanced Redox reaction. Were they given reaction and part C.


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