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Part 2: This is & measured composition of = an amphibole from the Azores islands in the Atlantic. SiOz 40.67 TiO; 5.38 AL,O,11.56 FeO 13,99 CaO 10.77 MgO 1141 M...

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Part 2: This is & measured composition of = an amphibole from the Azores islands in the Atlantic. SiOz 40.67 TiO; 5.38 AL,O,11.56 FeO 13,99 CaO 10.77 MgO 1141 MnO 0.35 Ko 0.84 Nao 2.87 Total 97.84Write the chemical formula of this amphibole; placing all of the atoms on the correct site. (30 points)To do this: first convert this composition to molar units by dividing by the molecular weight of each oxide, then calculate the number of cations of each element present and the number of oxygen at

Part 2: This is & measured composition of = an amphibole from the Azores islands in the Atlantic. SiOz 40.67 TiO; 5.38 AL,O,11.56 FeO 13,99 CaO 10.77 MgO 1141 MnO 0.35 Ko 0.84 Nao 2.87 Total 97.84 Write the chemical formula of this amphibole; placing all of the atoms on the correct site. (30 points) To do this: first convert this composition to molar units by dividing by the molecular weight of each oxide, then calculate the number of cations of each element present and the number of oxygen atoms contributed by each cation sum up the number of oxygen atoms you have accounted for; then renormalize both so that there are 22 oxygen atoms in the formula by multiplying each oxide by the same factor; then write the formula in this format where A, B, and C are examples of the cations (there will be 9 cations in the final formula) (Si 1.48 T; o." Ald.34 Fc 0.35+ Alg D.46 MnDusu E oelina ABCO



Answers

Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
(a) The number of moles and the mass of Mg required to react with 5.00 $\mathrm{g}$ of HCl and produce $\mathrm{MgCl}_{2}$ and $\mathrm{H}_{2}$ .
(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 $\mathrm{g}$ of silver(I) oxide.
(c) The number of moles and the mass of magnesium carbonate, $\mathrm{MgCO}_{3},$ required to produce 283 $\mathrm{g}$ of carbon
dioxide. (MgO is the other product.)
(d) The number of moles and the mass of water formed by the combustion of 20.0 $\mathrm{kg}$ of acetylene, $\mathrm{C}_{2} \mathrm{H}_{2},$ in an
excess of oxygen.
(e) The number of moles and the mass of barium peroxide, $\mathrm{BaO}_{2}$ , need to produce 2.500 $\mathrm{kg}$ of barium oxide, BaO
$\left(\mathrm{O}_{2} \text { is the other product. }\right)$
(f)

So in this problem we need to explain how to convert from grams of one substance into moles and then grams of another. Using a balanced chemical equation? The first one they give us, we have sodium plus chlorine. Remember chlorine is die atomics with cl two producing an A. C. Elena synthesis reaction. They tell us the mass of the sodium and we need to convert this into moles and then grams of cl two. Before you do anything, you have to balance this equation. The only thing we need to do is to put A two in front of the N. A, C. L. And then a two in front of the N. A. That will let us balance that equation. So we're in grams of N. A. In order to get two moles of N. A. We're going to have to divide bye the molar mass. And I'm going to use this a lot. So I'm going to refer to this as capital mm. When I write it from now on, we're gonna divide by the molar mass of N. A. To get to malls from there, we need to use coefficients. Those are from our balanced chemical equation. In order to convert um from moles of N. A two moles of the cl two. That answers the first part of the question from here. I'm going to multiply By the Molar Mass of Cl two in order to get two g. You will see as we go these are all going to follow the same basic pattern in B. We have mercury to oxide and that is going to decompose, it's going to break down into H G plus O two. Remember oxygen sai atomic balance it out by putting twos in front of both mercury containing compounds. So first thing we're gonna do, they gave us um the mass of HBO. So I need to divide the mass of H G O by its molar mass. That gets me to malls. Okay, after I'm in moles, I'm going to use coefficients. This will allow me to convert from moles of H G O two moles of 0. 2. And lastly, I'm going to multiply by the molar mass of 02 to get that into grants same exact pattern as the one above it. This is going to be the way you have to manipulate all of these problems in order to solve them. The next one, we have an A and oh three Decomposing to form NANO two and oxygen. We have four oxygen's on the product side. We have three on the reactant side. You need to get this to be even because otherwise the three is never gonna work with the even on the other side. So start by putting a two there. That gives me six oxygen's, it also gives me to sodium into nitrogen. So I'm going to put a two here that fixes my sodium and my nitrogen. It gives me four oxygen's in the N A N A N A N 02. It gives me I still have to oxygen's from the 02. So four plus two gives me six. And it's balanced. This is going to be the same exact method, divide our mass by the molar mass of And what was given here is 02. This is the difference. They give us a product instead of a reactant, but it doesn't matter for the way we're going to go about solving it. It doesn't matter which direction you're going in next. We need to use our coefficients, we're going to convert From moles of 02, two moles of NAN 03. And it really doesn't matter what this is. It depends on what the question is asking for. At this point we are then going to multiply that answer by the molar mass of the NANO three and that's all you need. D and E are both. It's very slightly different. Um There's one extra step in each of these. So indeed, we have carbon plus oxygen gives me carbon dioxide. It's already balanced. The trick here, is that we are in kilograms. So I need to take those kilograms And multiply by 1000 To get two g from there. I am going to follow what I've done all along. I need to divide by my molar mass. Um and that's just the molar mass of carbon. Use your coefficients to convert from moles of carbon to um moles of co two. And then multiply by the molar mass of CO two. Yeah, so these these last three steps are going to be the same in all of these problems. He actually looks exactly like C does. Except we have a different reaction. The reaction here Is CUC 03. This is decomposing so it yields CO two plus CUO. Again, we're going to take the number that were given and we're going to multiply it by 1000 To get two g. We're then going to divide that number. Um bye. The molar mass of see you. Oh, you're going to use your coefficients. You're converting this to Moles of CUC 03. I never had a space. The last thing we need to do is just multiply by the molar mass of what we're trying to get to, which is the c u c 03 mm

In this video, we're going to be solving the story geometry problems. The first step to any store archaeology problem, did you write the balanced chemical reaction? So this would be our balanced chemical reaction between before the reaction between chlorine gas and sodium metal. We want to know how many moles and how many grams of the chlorine Gas is needed to react with 10 g of sodium metal. So what we were going to do is, first of all, we need to make sure our equation is balanced and now it is. So what we have to do is start with the value that were given. So that's the massive sodium. And then since we have the massive sodium and we want to convert to the moles of sodium, what we will need to do is use the molar mass or atomic mass of sodium in this case, since it's an atom to convert two moles of sodium and then we're going to take the mole ratio from the balanced chemical reaction, which is two moles of sodium girls, one more of Corinne, And to get the mould of chlorine, we can stop right here. We just have to calculate this. It would just be 10 divided by 2.99 Times two. However, to convert to the mass of the chlorine, we need to use the molar mass of the chlorine gas taken out. There are two chlorine atoms. So we're going to multiply The atomic mass of chlorine by two and this is what we get. And then we would just solve that in order to get the mass of the chlorine. So it's just one more step for the massive chlorine. Okay, now let's do another problem. The next problem is asking us to find the bowls and massive oxygen that's formed by decomposition of mercury to oxide. So first of all, let's start with writing gap mercury too oxide. So the roman numeral indicates the charge on the mercury and oxygen always has a charge of negative two. So therefore this is mercury oxide. We're told that this is decomposing. So the composition of this is going to give us the following and we're going to balance this using coefficients and that's what we end up with. And then we are going to be Decomposing 1.2 5 2 g of mercury to oxide. Yeah. And we're looking for moles and mass uh, oxygen. So we're going to do is we have to follow the same process as in the last problem, we need to find the molar mass of mercury oxide. So we add up the atomic mass of mercury. Now the atomic mass of oxygen. And then this is what we end up with. Okay. And then using the story geometric coefficients from the balanced chemical reaction, we convert From moles of the Mercury Oxide, two moles of the oxygen. And this is how we get the moles. If we stop right there, however, we need the mass. We need to use The molar mass of oxygen, which is going to be 16 times too. So that's going to be 32 grams of oxygen per mole. And that's how we would get the mass of the oxygen. There was just one more step. Mhm. That's it. Mhm. Now, the next problem that we're going to do is we're going to find the number of moles and massive sodium nitrate required to produce 128 g of oxygen. So, the reaction that we're looking at right now is The sodium nitrate, which is written at NANO three sodium nitrate. It's going to give us two products so we have to balance this equation. Mhm. All right. So double checking the elements on both sides. We can see that this is now balanced. Now, we're told That 128 g of oxygen are produced. We want to know the moles and the mass of the sodium nitrate. That would be used to produce this amount of the oxygen. So therefore, let's start with what we're given. 128 g of oxygen. We use the more mass of the oxygen and then from the balanced chemical reaction, get the mole ratio. And then, as we did with the other problems afterwards, if we want to get the mass, then we have to use the molar mass, uh the sodium nitrate. And we can calculate that using the periodic table. And I'll let you do that on your own. No. The next problem we're going to do is find the number of moles and massive carbon dioxide formed by combustion of carbon with and access of oxygen. We're told the oxygen is an access so we know that carbon is going to be the limiting reacted this is his combustion. The carbon will react with oxygen to form carbon dioxide. So this is already balanced. We're told that we have 20 kg of carbon in order to use lower mass to convert the carbon too, from grams to moles. We have to convert kilograms two g in order to convert kilograms two g, multiplied by 1000. We move the decimal .3 places to the right. So that's going to be that make grams of the carving. Now, what we need to find is the moles and mass of the carbon dioxide. So this problem after converting, we will follow the same process. So we start with the mass of the carbon from periodic table. We obtain the atomic mass of carbon, Convert the carbon two moles from the balanced chemical reaction, Convert moles of carbon two moles of carbon dioxide. And if we want to get the mass of the carbon dioxide, then use the molar mass of carbon dioxide, which we obtained by adding up the atomic mass of carbon and to oxygen. And that's what we get. And solving that, we will get mass. Oh, carbon dioxide. Okay, now let's do another problem. We're going to find the moles and mask of copper to carbonate. Need to produce carbon to oxide. First of all, we need to write down the formula for copper to carbonate. Carbonate has a charge of negative too. So it's going to combine with one atom of the copper in formula. So that would be our formula. And then it's going to give us the copper to oxide. And that would be the following. The other products that we are going to get is carved been dioxide. So this is going to be our reaction. Now we need to check that this is balanced and we see that it is so we can move on to the next step. Now we're going to be producing 1.500 kg of the copper to oxide. So it was right that down here. Now we need to convert this two g. So again, multiply Bye 1000. And that's how many grams of the copper to oxide that we have. Now we need to find moles and mass of the reacted. So let's once again startup dimensional analysis like we have been doing again use the molar mass To convert two moles of the copper to oxide. And then use the mole ratio between the product and the reactant that we're looking for to convert two moles of the reactant. That's what we get. And in order to get from the moles of the reactant to mass of the reactant, use the molar mass, as we did with previous problems, and that would be how we would do that last problem.

Hi, guys. Let's own problem. 14.1 21. The precursor Acid on the permission off Die. I hadn't been docks. ID's. I began Sid and the former ladies h i old tree The reaction. Want information off the ad in Pentax ID's do more so by the DNC and undergoes the reaction. And 202 to 40 different since years and read it their ardent dockside and watch her the loosest structure. All that hadn't been Doc Sandy's this one. Okay, for the second problem, why the bones to the terminal? Oxygen on shorter, then the bonds to the bridging oxygen. Here we see that the original excision I don't forms single bond with Adam and the terminal oxygen Adams. Their form doubled on the organ. Now we know always the double bones and shorter than the single bond. Therefore, the bonds to the terminal oxygen are shorter than the bombs to them bridging oxygen. Now we need to read the reaction auf die. Arjun went topside with garden one upside to the reaction leaves. It gives carbon dioxide. And I didn't

So in this question we need to write the balance reaction to show the formation of die iodine tent oxide by dehydration, conversation of oxo acids. So the oxo acid is going to be I OTIK acid. And you can see basically it will react to form you die iodine into oxide and also water as a byproduct. But so that would be the balanced chemical reaction. So this species H I L three as I OTIK acid, because I owe three negatives in Ohio date ion. So when it's an asset, we change the eight to an IQ. So its I. O. Dick. This is a lewis structure for the diagram can't oxide. So you can count that we have five oxygen's and two paradigms looking at the structure, we can answer the next question. The bonds to the terminal auction are shorter than the bonds to the bridging oxygen. The reason is because these are double bonds and double bonds oh, shorter than single bonds. That is why these are shorter than the ones in the middle. So for the last one, we're going to right the reaction between the Dieudonne Pataki side and cover monoxide. So we're just going to right the products as iodine and carbon dioxide three put a five in front of carbon monoxide, A five in front of the carbon dioxide And then you can see that we balanced the oxygen because now we have 10 oxygen's on the left and right. So that's how we would balance this equation


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