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Wrre tne balanced equaton tor Ihe reacton aalecus Pb(CIOslz wilh aqueous Nal: Indude pnases Pb (Ciosk;leq/-2Nalq/ Pbyls}+ 2Naclos(aq)WunaiMass preapiata Iform If 1,...

Question

Wrre tne balanced equaton tor Ihe reacton aalecus Pb(CIOslz wilh aqueous Nal: Indude pnases Pb (Ciosk;leq/-2Nalq/ Pbyls}+ 2Naclos(aq)WunaiMass preapiata Iform If 1,50 L of highly concenbraled PbICIOskeis mixed wilh 0.450 L 0.1J0 M Nal? Assuna Ihg reacuan gces cometonLncThrorahn #uabnDrdnt0r"Upchin(ca AnEtaer

Wrre tne balanced equaton tor Ihe reacton aalecus Pb(CIOslz wilh aqueous Nal: Indude pnases Pb (Ciosk;leq/-2Nalq/ Pbyls}+ 2Naclos(aq) WunaiMass preapiata Iform If 1,50 L of highly concenbraled PbICIOskeis mixed wilh 0.450 L 0.1J0 M Nal? Assuna Ihg reacuan gces cometon Lnc Throrahn #uabn Drdnt 0r"Up chin (ca AnEt aer



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Click on the site (http:Ilopenstaxcollege.org/l/16PhetAtomMass) and select the "Mix Isotopes”tab, hide the "Percent Composition" and "Average Atomic Mass" boxes, and then select the element boron. (a) Write the symbols of the isotopes of boron that are shown as naturally occurring in significant amounts. (b) Predict the relative amounts (percentages) of these boron isotopes found in nature. Explain the reasoning behind your choice. (c) Add isotopes to the black box to make a mixture that matches your prediction in (b). You may drag isotopes from their bins or click on "More” and then move the sliders to the appropriate amounts. (d) Reveal the “Percent Composition" and "Average Atomic Mass”boxes. How well does your mixture match with your prediction? If necessary, adjust the isotope amounts to match your prediction. (e) Select "Nature's”mix of isotopes and compare it to your prediction. How well does your prediction compare with the naturally occurring mixture? Explain. If necessary, adjust your amounts to make them match "Nature's" amounts as closely as possible.

In this problem I am writing the reaction for. So just look at it carefully. The U. S. 1/4 Into five page 2. When heated at 110° integrated. It will change in to see us a four As to percentage of lost water here is equal tool Called by five multiplication 100 which is equal to 80%. Now I can write DDX an edge to see us so forth. Plus four K. I will give eight K two S. +04 Plus c. U two i 2 plus I too. In further reaction I can write see us of full Plus two KCN. We're changing to c. u. CN two plus. Get us a fall and to see you CN hold to we're changing to see you too. CN hold two plus c. n. All too. Now the final reaction can be written ahead. See You too. CN All 2-plus 6 K C. N. Will give who get to the see you C. N. Food to option C. It corrected.

So in this problem we are given a reaction. It's a geometric reaction between elemental sulfur and chlorine gas to form die sulphur chloride and sulphur chloride. Now this is actually a two step reaction, which I've representatives two steps here. So first I sulphur chloride is formed and then any excess chlorine reacts with the silver chloride, a form sulfur chloride. So our first step is to figure out how much dye sulfur deployed is formed in the first step and then we'll figure out how much is it used to form sulfur deployed in the second step. Considering that are final answer of interest is the amount of dye sulphur chloride that is formed. So we'll start by converting everything the molds. We have 50 g of sulfur over its molecular weight. 256.52 gives us 0.195 moles of sulfur. It's a terrible eight. Let's try that again. There we go. Uh and we have 102 g of chlorine. So 102 g of chlorine Over its molecular weight. See it 69.94 gives us 1.458 moles of chlorine. Now it says that we are reacting completely. So we are going to assume a reaction of to completion. So we're gonna do a limiting reagent calculation. The easiest way to do this is to take them animals of each reacting and go, okay, well, for every one mole of sulfur, we produce formals of products. So we'll multiply this by four to get equivalent product, which gives us 0.7797 Moles Product. Whereas the chlorine, basically chloride is 1-1. So 1.458. So obviously we have some excess moles of chlorine that we're going to have to deal with. So we can figure out the initial amount by saying that this is the initial amounts to 0.7797 bowls product. Initial, Yeah. And then we're going to react here. So we have 0.7797 moles. And we have The amount of chlorine. We started with the 1.458 moles minus whatever was used in a 11 reaction. This is the amount we had. So 7797, Which that's going to give us 0.679 moles. And so then we know that if this reacts completely, then the final of sulphur chloride, not die. Sulphur chloride is going to be the amount of dye sulfur dioxide we have minus the amount of sulfur, like or amount of chlorine gas we have left over which is going to give us. And we might uh it would give us some number of moles, notably this number of moles is the same as this number of moles that we have left over because we can't react anymore at a sulphur chloride with chlorine gas. And we might say that we take this number, we'll call it x for now, uh and we can say that 0.7797 minus zero point 679 the number of moles of diesel for declared. We have times. It's more weight here. 135 0.4 Would equal the number of g of diesel for deployed we have, which in this case is 13.6 three g.

Hi there. Let's look at how to determine the identity of the cat eye on and the anti on. In problem number 138. We're told that we have a cat eye on that. We're giving the symbol M and it has a two plus charge. And that is reacting with an anti on that has a negative one negative charge to produce the compound, the ionic compound. And next to so the first thing I want to do is balance this. Um to balance this, I would need a two in front of the anti. R. Next let's list the information that we know, we know that we have 1.92 g of the cat eye on. We have 0.158 moles of the anti on. And just to keep things interesting, they gave us a percent composition for the compound. It is 86.8 percent X. That means that for em the M must be 13.2% because there are only these two elements in here and the percents have to add to 100%. So 13.2% of the cat island. So if we think about that 13.2% times the total mass of the sample has to be equal to 1.92 g. And I know that because I had 1.92 g of the cat. I am reacting. So all of that went into the products. So 13.2% is 1.92 g. So, I can find the mass of the total sample. Now by using this equation 1.92 g divided by 13.2% gives me the total mass of the sample. Yeah. Or if the product I guess let me say product I like that better. It is also our sample. But we'll call it the product. So, the total mass of the product calculates to be 14 0.55 g. Okay. All right now we're getting somewhere because we have 14.55 g of this mX two. Now that we have that we can think about the law of conservation of mass. If the total mass of the product is 14.55 g and the mass of the cat eye on or the metal in this product is 1.92 g. The difference between 14.55 and 1.92 What must be the mass of our non metal? So subtracting or finding the difference between these two gives me 12.63 gramps. So the non metal, the an ion has a total mass of 12.63 g. That's good. We can work with that because we can now determine the molar mass of X. Because Mueller mass is grams divided by moles. So I take 12.63 g and divided by the malls. I get a molder mass for X to be I lost my spot. Here. There we go 79.90 g per mole. Looking at the periodic table. The element that has a molar mass of 79.90 g per mole is brewing. So bro mean his ex annex with a negative one charge or one negative charge is the bromide ion. All right. So, we've determined what taxes and now we have. Now we can determine what Emma's, what I'm going to do with. That is I have a balanced equation and I have the number of moles of X. So I can use the mole ratio to determine how many moles of the cat eye on em that we have because for every two moles of the anti on, we need one mole of the cat eye on this tells us that we have 0.79 moles. Yeah. So, we already have the mass. And now we have the moles molar mass is grams per mole. So, I am going to go ahead and figure out what the molar masses of our metal. Yeah. Mhm. By taking 1.92 g and dividing by 0.79 moles. This gives me a molar mass of 24.30 grams per mole. The element with that Mueller mass is magnesium. So magnesium is M in the magnesium ion would be M with the two plus charge. All right, So, are two elements here I am this magnesium ex's br or brewing. Thanks so much for watching.

So atomic Lee iss The some off mass times abundance which is in percentage off each ice. It hope so if Nora and Variables and we have atomic wheat equals to mass one and the percent abundance close and to plus a percent abundance. Now the percent abundance for isotope two would be the left over percentage because of their only two isotopes there. And the isotope one is a certain percentage. What is left? There will be the 100 minus X one over 100. Now we can simplify it us even further. So atomic wheat, Because I m one x one over 100 plus m two because ah 100 over 100 is one minus M two x one over 100. Now what we can do next, then ISS rearrange So a atomic weight minus them too equals to m one x one over 100 minus them to x one over 100. And this was give us X one, um one minus m two over 100 not weekend put numbers in. So if atomic weight is well have atomic weight, it's equal to eight w, and we now know that them to IHS on the other side of it all. Then we can do 107. It's just all atomic weight, minus the mess which oft I set up to, which is 100 8.9047 Then we put X one, which is unknown over 100 and Mass one, which IHS 106.9051 a minus must 208.9047 Then, since you're trying to find what x one iss, we can divide it and what you get. It's around 51 0.84% and this with the around 50%.


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