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The silver iodide formed ( +.96 3) Eeneanuenu silver ittateunknow ionic comipound MI: (2.94 g) [neatcd Calculate the molar mass ofM: 63.5 gmol Emol d) 278.4 Emol 0)...

Question

The silver iodide formed ( +.96 3) Eeneanuenu silver ittateunknow ionic comipound MI: (2.94 g) [neatcd Calculate the molar mass ofM: 63.5 gmol Emol d) 278.4 Emol 0) 243 mol 6) 317.9 gmol ChO.! Which of the follou ing statements Hanier Mcl reucts with nonme{l yield tonic cambaund metal-nonmetal reuction €n aIways assumed be #n oxidation-reduction reaction_ Two nonmetals cun undergo an oxidation-reduction reaction. When two nonmetals Feee covalent compound fonmed Mlelal-nonmetal ruction involves

The silver iodide formed ( +.96 3) Eeneanuenu silver ittate unknow ionic comipound MI: (2.94 g) [neatcd Calculate the molar mass ofM: 63.5 gmol Emol d) 278.4 Emol 0) 243 mol 6) 317.9 gmol ChO.! Which of the follou ing statements Hanier Mcl reucts with nonme{l yield tonic cambaund metal-nonmetal reuction €n aIways assumed be #n oxidation-reduction reaction_ Two nonmetals cun undergo an oxidation-reduction reaction. When two nonmetals Feee covalent compound fonmed Mlelal-nonmetal ruction involves electron share.



Answers

When $1.0000 \mathrm{~g}$ silver chloride $(\mathrm{AgCl})$ is all converted to silver iodide $(\mathrm{AgI}), 1.6381 \mathrm{~g} \mathrm{AgI}$ is formed. You know that an atom of iodine has 3.580 times the mass of an atom of chlorine; calculate the mass of silver in each sample. Also calculate the mass of chlorine in the $\mathrm{AgCl}$ sample and the mass of iodine in the AgI sample.

Yeah. Okay, let us start with the Mueller mass. So mauler mass of silver iodide will be 108 for silver plus 1 27 for id. The molar mass turns out to be 235 g. Para mon. So We know that 127 g out of 235g is denoted for I. D. So the amount of iodine, so the amount of so the mass of iodine mass of ionian in 2.185 g will be 1 27 Molar Mass of the Irene. The molar mass of Total compound which is 235, Multiplied with 2.185g. And the answer turns out to be 1.18 grams of ironing. Hence the same amount of iodine present in the original solution. So now we can say that 1.18 g of iodine is present and 1.545 g of soluble. Uh huh sample the mass percentage of iodine. So the mass percentage of are you dying is equal to 1.18g, Divided by 1.545g, multiplied with 100 And the mass percentage out to be 76.4, of course.

All right. So initially we have an unbalanced reaction. And since I've got to nitrate on the reactant side, I'm gonna place it to here to balance my nitrate, which is gonna give me to sodium. So I'll place it to here to balance the sodium and that gives me to iodine, which I have on the product side. So my coefficients will be 1212 And in part a we're going to be solving for the leaders of mercury to nitrate. So in order to do this, we're going to take our leaders of sodium iodide and we're going to multiply um the leaders by the polarity to give us moles. So if you take the 0.0 to 40 leaders of sodium iodide and we can write polarity as moles per liter. So this will be 400.170 moles of sodium iodide per one leader. And that will allow you to cancel the leaders of sodium iodide. Then we can use the coefficients from the balanced equation. So if we've got two moles of sodium iodide for every one mole of mercury to nitrate, then we can use the polarity of mercury to nitrate Written as .209 moles per one leader to cancel out the moles of mercury to nitrate. So if we put this in the calculator, we're going to round this off to three significant figures. This is going to give you .00976 L of mercury to nitrate. Now, if we go on to Part B where we're solving for the grams of mercury to iodide, if we follow the same format as we didn't say we multiply the leaders of sodium iodide bipolarity that will give us moles of sodium iodide and then we can convert that to mercury to iodide using the coefficients from the balanced equation. So we're gonna start off again with our .0-40 leaders of sodium iodide. And then we have our polarity Of sodium iodide written as .170 moles per liter. These units will cancel. Now we can use the coefficients from the balanced equation. So we have two moles of sodium iodide for every one mole of mercury, too. I had died. And then we're gonna multiply by the molar mass to convert two g. Now the molar mass is 4 54 4 g. And once you put this in your calculator, I would round this off to three significant figures, which is going to be .9-7 grams.

This question is basically a strike geometry question. First we have an unbalanced chemical reaction so we need to balance it. We have mercury to nitrate reaction with sodium iodide producing mercury to iodide and sodium nitrate. To balance that, we need to put a two in front of the sodium iodide and a two in front of the sodium nitrate. Then, if we have 24 mil leaders of the sodium iodide solution, we can convert those mill leaders into leaders by dividing by 1000 and then convert the leaders into moles of sodium iodide by multiplying by the polarity of the sodium iodide solution. Now that we have mold sodium iodide will use the stock geometry of the balanced chemical reaction, namely to mold sodium iodide are required for every one mole of mercury to nitrate. Then when we have moles mercury to nitrate will use the polarity of the mercury to nitrate solution to convert the moles into leaders and then last of all convert the leaders into mill leaders. By multiplying by 1009.76 mil leaders of the mercury to nitrate solution Is required to react with the 24 mil. Leaders of the sodium iodide solution next for part B. To determine the mass of the precipitate. Well again, start with our 24 ml of the mercury to nitric. I'm sorry, the sodium iodide convert the mill leaders to leaders, then convert the leaders two moles of sodium iodide. Then seeing the stock geometry, we have one mole sodium iodide, creating one mole mercury to iodide, Sorry to M.nolds to one Mole. Then when we have moles of the mercury to iodide, we can multiply by the molar mass, mercury to iodide to get the grams of the mercury to iodide precipitate. That will form 0.927 g.

All right, so in this video, we're gonna be looking at various reactions of Ah, potassium iodide, which is okay, I underlying it on, and we're gonna be just practicing with our Redox reaction skills. So identifying what's the reaction? Ah, what is the Redox? What is the reduction? What is the oxidation and what is the reducing and oxidizing agent in each case? So, to start off, we have this 1st 1 And just to be clear, I'm going to use read for my reductions, and I'm going to use blue for my oxidation. Excellent. So the first thing we're gonna need to do that, which I will do in green, is nup. Label oxidation numbers for each compound slash Adam. So chlorine gas we know is a zero. It is non combined, k I we're gonna break down by ions. So we have I, which is gonna be minus one in potassium, which is gonna be a group one element. It's that ionic plus one charge. And then we have Casey O which is going to be negative one plus one and I to a solid zero. And let's look at what got reduced. What? Got oxidized right So we have Klein here at a negative one state. That's already because its a negative one. That's only a pretty good sign. It probably got reduced. And indeed, it did start at zero. So we know that this is our reduction. Chlorine is getting reduced because it's oxidation. State is reduced. Next, we're gonna look for out for that oxidation. So here we notice k. It stayed the same. It stayed plus one left side and right side of the reaction. However, iodine started his negative one and jumped up to zero. And that is gonna qualify as an oxidation. And so in this case, are reducing. Agent reducing agent is itself oxidized. So that's gonna be iodine. And then our oxidizing agent, which is itself reduced, is the US going to be chlorine? I know it seems a little bit confusing, but as you get yourself more familiar with these terms is going to be like, no problem. So we're gonna move right along here. Um, we're going to look at this case. So iron chloride, iron. Try chloride, plus testy. My died iron die chloride, plus, um, potassium chloride, plus iodine solid. Great. So let's take a Look here we have three Corinne's that's gonna be a negative one. Times three Kansas to negative three. Leaving our iron with the charge of plus 30. Same thing here. We have a negative one and a plus one. And to analyse the second part of the equation, we have a negative one times two, that's gonna lead us to negative two. Making our iron take on a charge of plus two. We want neutrality. R K C l will break down to negative one and plus one again. That's exactly what we expect. And I two solid in un combined state is zero now, looking out for that reduction in that oxidation. So looking out here for some negative numbers, I see chlorine and negative one, but it also started as a negative one. So chlorine did nothing. Chlorine just watched in this case, Um, and again here, negative one. It did nothing. Let's look, take a look at iron. So iron started as a plus three and it got to a plus to that is a reduction of charge because it fell from plus three two plus two. So iron is getting reduced, and even then we can just go ahead and say that because Iron is getting reduced, it is itself the oxidizing agent. So why don't we just go ahead and write that? Because then that will help us link those two terms together. The thing that is itself reduced is the oxidizing agent. And finally, let's find our oxidation. So we looked at Ironman, looked at chlorine. Let's take a look at potassium, but I'm seeing no changes from a plus one to a plus one. However, if we look at iodine, we went from a minus one to a zero. That is gonna be our oxidation and therefore are reducing. Agent is iodine, and you'll see that that's the same as the example above. Finally, our last example today. Ah, we have copper chloride, potassium ideye going to copper iodide, potassium chloride and I two solid. Great. So let's go on with this Like we've been doing. Negative one times two is negative. Two, leaving our copper as a plus two oxidation state. Okay, I breaks down the same way it always does. Negative one and plus one copper iodide. Here we have iodine as a negative one copper as a plus one k c L breaks down the same way and I to stay insane. Great. So let's check out. What is our reduction? What is our oxidation? I am looking at iodine, and I see it ends up as a zero solid and it started out as a negative one. That is an increase in oxidation. States of iodine is getting oxidized. And because iodine is itself oxidized is going to be our reducing agent. And again, that is the same as the examples above. So we're starting to see a pattern. Emerge here and finally, let's look out for our oxidation. So let's take a look at copper. Copper is a plus to charge, and it ends up as a plus one charge, not as a reduction in ah, in oxidation state. And therefore it's our reduction. Sorry if I said oxidation before I met reduction. And so here we see copper is itself reduced. Therefore, it is our oxidizing agent, and I hope this video is helpful


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