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4. Is it possible to write a chemical equation for the overall reaction that occurred in the cell? Explain your answerCould an Oscar be electroplated if it were mad...

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4. Is it possible to write a chemical equation for the overall reaction that occurred in the cell? Explain your answerCould an Oscar be electroplated if it were made of carbon (a non-reactive, conductive material) or of plastic? Explain your answer Infer why an Oscar is electroplated successively in the sequence of copper; nickel, silver; and, finally goldCommunicate Your Findings 5 . Prepare a presentation that outlines thekey steps of your demonstration. Your presentation should include an ove

4. Is it possible to write a chemical equation for the overall reaction that occurred in the cell? Explain your answer Could an Oscar be electroplated if it were made of carbon (a non-reactive, conductive material) or of plastic? Explain your answer Infer why an Oscar is electroplated successively in the sequence of copper; nickel, silver; and, finally gold Communicate Your Findings 5 . Prepare a presentation that outlines thekey steps of your demonstration. Your presentation should include an overview of the chemical processes involved in Communicate Your Findings electroplating, including description of the role of Add to your presentation from Part to explain how the electrolytic cell and the flow of electrons and ions the four metals can be electroplated onto an Oscar description that summarizes the electroplating Include brief overview of your experimental design, demonstration, including brief overview ofyour a labelled diagram of the electrolytic cell used for each experimental design; labelled diagram of the metal, and an outline of the procedure_ noting = any electrolytic cell, and an outline of the procedure steps; safety precautions Explain how the chemical reactions noting any safety precautions will take place. Update your literature citation section. description of your observations, including an explanation of the chemical reactions that took place Assessment Criteria during the demonstration Once you complete your project; askyourselfthese literature citation section that documents the questions Didyou._ resources You used to complete your research assess voUI information sources for accuracy and reliability? Part 2: Electroplating an Oscar KIU describe the chemical processes involved in Initiate and Plan electroplating? Use print and Internet resources to research how to KIU describe how you will perform electroplate an Oscar with the four metals described in demonstration of electroplating? the introduction: copper; nickel, silver; and gold_ carry out your demonstration of Write a brief overview of your experimental design. electroplating, meeting all safety precautions? Draw a labelled diagram to show the electrolytic cell draw labelled diagrams and use tables to for each plated metal, illustrating the flow of ions and record your observations? electrons that will take place in each cell Include the evaluate your demonstration and suggest cathode; anode; electrolyte (specify the solution), and ways in which it might be improved? power source for each metal: IU describe how each metal can be Write a detailed, step-by-step procedure that you electroplated onto an Oscar? would follow for plating each metal, including a make = recommendation based on materials list and safety procedures supporting evidence as to whether or not an Oscar could be electroplated if it were made of carbon or Analyze and Interpret ofplastic? 1. Describe any problems that you might encounter with design and how YOu might overcome propose an explanation as to why an Oscar your experimental is successively electroplated according to the them: following sequence: copper; nickel, silver; and, 2. Complete question in Analyze and Interpret Part finally; gold? each metal that was electroplated onto the Oscar: communicate the electroplating process in 3 . Is the Oscar the anode or the cathode during appropriate form, using suitable instructional electroplating? Explain your reasoning: visuals and scientific vocabulary?



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Examine Figure $19.1 .$ Diagram (b) is completely labeled, but diagram (a) has almost no labels. The caption describes both diagrams. Assume that diagram (a) represents one of the earliest known electroplating systems in which both electrodes are copper and the electrolyte is an acidic solution of copper(II) sulfate. The net operation of this system effectively transfers copper atoms from one electrode to the other. a) Write the oxidation and reduction half-reaction equations for the reactions that occur at the copper electrodes in diagram (a). Number your equations (1) and ( 2). (Hint: Read the introduction to this question very carefully. It contains what you need to write these equations.) b) State which half-reaction equation, (1) or $(2),$ represents the change at the electrode labeled " $+"$ in diagram (a) and which is the change at the "-" electrode. c) Which electrode, "+" or "- " is the anode and which is the cathode in diagram (a)? d) What ion or ions are carrying the charge through the electrolyte? From which electrode and to which electrode are they moving? e) Ordinarily, this system operates at close to $100 \%$ efficiency, which means that 100 grams of copper dissolved at one electrode yields 100 grams of copper deposited at the other electrode. Sometimes, however, bubbles appear at one electrode, and the masses of copper dissolved and deposited are not quite equal. Considering everything in the solution, can you (1) identify the bubbles, (2) identify the electrode at which they appear, (3) write a half-reaction equation for what is occurring at that electrode, and (4) summarize in one sentence your answers to (1), (2), and (3)?

Explain why neither of those benefits just emphasize border from this starting material will not work, and so just something that will work. So the proposed calculation will place the Ethyl group in tow. Benzedrine, and it's on Ortho pair director. So the Clear Nation which followers will not go to the middle position when you see additional steps are out and first introduced cloying in clothing is still orthopedic director. So on the step off population, the group will not enter the. Does that mean that the position but they don't work, know what real work? So their strategy is for introduce a meta director into oblivion. First, that could be a later converted to the group out when you to perform escalation first. Is it a group? Is it meant to director? It's a business except no conduct coordination and now according intravenous position, and finally were reduced. The actual group toe every group using either Clemmensen reduction or business wolf reduction

In this problem, we are asked to consider burning magnesium in air. One product on burning magnesium air is the product of magnesium and oxygen, and that product is magnesium oxide. The second product is the reaction of magnesium and nitrogen. And that problem problem or that product is magnesium nitrite. Excuse me. When waters met um, added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. Okay, so here's all of our equations. I'm gonna go ahead and balances. I think we have to do these later, but I can't take not having imbalanced and here we have three, two And a three. I think that should be good. Okay, so part A says based on the charge of the nitride ion, right? The formula for magnesium nitride, which I already did Magnesium is a two plus nitride is a three minus. So the formula is M G three N two, part B. So it's right. The balanced equation for already did this here is the balanced equation for part B. What is the driving force for this reaction? Looking at this, I'm gonna say the driving force for this reaction is probably the formation uh And H three, which is a gas. I'm gonna say that that's going to be a driving force for this reaction. And then see, I believe is a long one. So I'm gonna get started with C. On another page. See might actually take two pages. So in an experiment, a piece of magnesium ribbon is burned in air. So we're burning magnesium ribbon. The mass of Magnesium oxide and magnesium nitride is 0.470 g after the reaction. And this is a mixture of M G. O. And MG three and two water is added to the crucible. We're all finished, we add water, we heat it to dryness and we get 0.4 86 grams of M G. O. Only. What's the mass percentage of MG three and 2 In the 0.470 g. That's what we're going to be looking for. Okay, so let's figure this out. I want to get to all of the different things here. So I set this up by a bunch of numbered steps. I'm just gonna go ahead and number of my steps one. My initial mass of magnesium is going to be equal to magnesium oxide mass times. I'm going to call this the molar mass of magnesium divided by the molar mass of M. G. O. Substitute our values in here and we get magnesium. I'm going to call it initial mass. Just previous initial Equals 0.486 g times the molar mass of magnesium is 24 31 g In the molar mass of Magnesium oxide is 40.31 g per mole. When I do my work here, I got 0.293 g. And that was my mass of magnesium and we'll need that further on. Any problem. For the second thing I got ready, I'm gonna say, let X equal the mass of the mg mass in M G. O. And let y equal the mg mass in MG three and two. We know that X plus Y Has to eat my total mass of my product, which we said was 0.470. That's g. My bed zero .293 g of MG. I'm gonna tell you a little later, we're gonna solve this for X. So I'm just gonna go ahead and write X equals 02 93g- Why? Let's do our 3rd step, which my third step was. Find the masses of M. G. O. And M. G. Three and two. So my M. G. O. Mass magnesium oxide mounts. I just took that equal to acts from above times the molar mass of magnesium oxide divided by the molar mass of magnesium. So I'm gonna go ahead and just write those down. That was 40 .31g And 24.31 g. Actually g promote. But those cancel out. And my mm. G three and 2 maths. Well why same thing here? Except it's smaller mass of M. G. three N 2 and the molar mass of mg times three. So in this case that is 100 nine. Mhm. I think I wrote 9-9 Divided by three times 24.31. Okay, if I add for number four I'll keep the same color. If I add 38 Plus three b. and I'm gonna go back moment, Let's call this three a. and three b. If I add those two, I'm going to get my mass of product which we were given. So I'm just going to rewrite this quickly here. Um massive product equals my NGO mass which is X times 40.31 over 24.31 plus. Why times 100 .9-9 times three times 24.31. Now back over here I have this that we were going to substitute in for X. So let's go ahead and substitute this in for X. I'll switch colors and I'm also going to figure out these values. My mask. My product was 0470 g. We're in a substitute for X 0.293 g. Whoops minus X minus why? Sorry about that. Times this number extended is 1.658 plus Y. Times this extended is 1384 So doing my math here. And I got after I I guess I'll just write about. I have a room here 0.4858. So I did negative 0.0 158 Equals zero 274. Why? And I got a Y. Of 0577 g. And this is grams of magnesium in MG three and 2. So this is how many g I have at M. G. three and 2. Just one more quick step before we can calculate our percentage. Let's figure out our mass of MG three N 2. It's going to be equal to 0.057 seven g. Time is the molar mass Of this will be again three times 24, g per mole and 100 0.99. Do the math here. And I got 0079 nine g of MG three and two. And last but not least. We're going to take our 0.0799 g magnesium nitride divided by the total mass of our sample. of course, times 100 because it's percent And I got 17.0%. MG three M 2. Okay, next. Yes, there's announced that was part C. Part D. Asked us to um write a balanced chemical equation and do a quick strike geometry. Super easy. So the equation we're given is we're not given the whole equation but I've balanced it for you. We're told that magnesium reacts with ammonia gas at high temperatures. The product is products are magnesium nitrate. Perhaps that's an S. And hydrogen gas. Okay, so I'll need a three right there. Let me shoot 3213 Okay. And were given that we have 6, 3 g of magnesium. We have 2.57 g of ammonia. I'm going to divide each of these by their molar masses to give moles because I like to have balls And I get 0.259 Most of Magnesium and 0.151 moles of hydrogen. Or excuse me, ammonia. We're asked to find which is limiting. And then we're also asked to find grams Of H two produced. Let's do it. I just wanna is never going to do this. But I decided I'm going to I'm going to do something called the mold box here. And for the mailbox, we're going to write down what we're given. I was given magnesium and I was given ammonia in the mold box. We put coefficients First we had three and 2. How many moles we have, Which I just calculated zero 259 And 0.151. How many moles we need. And then we'll figure excess wall boxes are sort of fun to find the number that goes right here. We start with this, Multiply it by three And divide by two. So we take 0.15, one moles of NH three. Figure out what we need. We put our two moles of n H three in the denominator and our three moles of magnesium on the top. So mathematically you can see what we're doing there. And I got zero two 265 Which means I have an excess of 0.0 2 to 5. Then I'm going to switch colors to find the number that goes here. We start multiply it and divide. So we take the same thing here. But we're gonna switch this switch this factor And I got 0.173. I don't have enough. So this is my limiting reactant. Then next we are going to do a simple story geometry. I'm going to start with the moles of what my limiting is do simple strike geometry using my mole ratio for hydrogen and ammonia, Which was two and 3, which were two and 3 and my molar mass for hydrogen. And when we're done, I went to three decimal places because I had three there and I got 0.458 g of hydrogen. And then last but not least e any were given that the delta H. It's not a very nice delta. The heat of formation for magnesium nitride Is negative for 61 0.8 Kill jobs per mole. Using that equation that we just wrote down. I'll write it down again here and I better put states on since we're doing these. And I'm gonna write down the anthem peas For each one of these. This one we were given as -461-08. And this is in colegiales promote And I looked up ammonia and got negative 46 19. Remember that Hess's law is products minus reactant. So for the reaction, I'm going to get no coefficient there. So I'm just gonna write one times negative 4 61.8 I'm sorry please, I can't start this over again. I can't start this over again. Whoa. Uh huh. No. And I got negative 3 68.70 killer jewels per mall.

Predict the falling borders off penetration. Hey, for each group is Ah, Ortho PETA director, and she is free group that's met the director. So it's a case off consistent orientation. So that letter group enters with this Brothers Bay like trial is that met the director and this will free H is a metal director. Let's again consistent orientation in the latter group interests this position. See him in the struggle. Is it met the director. It directs into this position and horses figure is the offer. Period Director directs the order positions and the prayer position. The donor Rene's over except when it comes to orientation. Substitution to the Ortho position. Prevent Letter Group is not favorable because over stone edit from accepting activity or from in a group, so the product would contain the latter group into this position. They can be really a very minor impurities within matter. Group entering the close positions

Fennel visited, which can be produced from funeral, is less reactive than funeral in automatic. There are a few exceptions fearful. Explain what it's this, but the group is still Order Pereira director. Just like hydroxy group, you know, So I don't see a group in. Fino is a force of credit director because over too long in the pants, which supplies elections, most it through the orphan and better positions, this group still has to look natural in prayer on oxygen, which pretty X the same way as the age group. In funeral, however, the compound is less reactive because for conjugation off long and compare with adjusting Gruber. No, which most away park the church former our since one So the little there compares on oxygen become less active.


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