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Molecules A and B react to form a product according to the equation2A+1B+3PA minimum volume of mL of 0.100 M B is required to completely react with 16.8 ml of 0.100...

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Molecules A and B react to form a product according to the equation2A+1B+3PA minimum volume of mL of 0.100 M B is required to completely react with 16.8 ml of 0.100 M A,and this combination will produce a maximum of moles of product P.2.52 X 10 -32.52 X 10' 1.68 X 109Submit1.68 X 10Do you know the answer?know itThink soUnsureNo idea

Molecules A and B react to form a product according to the equation 2A+1B+3P A minimum volume of mL of 0.100 M B is required to completely react with 16.8 ml of 0.100 M A,and this combination will produce a maximum of moles of product P. 2.52 X 10 -3 2.52 X 10' 1.68 X 109 Submit 1.68 X 10 Do you know the answer? know it Think so Unsure No idea



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Consider the equation $A+2 B \rightarrow A B_{2} .$ Imagine that 10 moles of $\mathrm{A}$ is reacted with 26 moles of $\mathrm{B}$. Use a scale from 0 to 10 to express your level of agreement with each of the following statements. Justify and discuss your responses. a. There will be some As left over. b. There will be some Bs left over. c. Because of leftover As, some $\mathrm{A}_{2}$ molecules will be formed. d. Because of leftover Bs, some $\mathrm{B}_{2}$ molecules will be formed. e. Even if $A$ is not limiting, $A_{2}$ molecules will be formed. f. Even if $B$ is not limiting, $B_{2}$ molecules will be formed. g. Along with the molecule $\mathrm{AB}_{2},$ molecules with the formula $\mathrm{A}_{x} \mathrm{B}_{y}$ (other than $\mathrm{AB}_{2}$ ) will be formed.

In part first a part We have been given one more of A. That is a reacts went one more of the So this is one more. This is also one more and it leads to reaction or let us product it be okay let assume that the product is be here. So they all react in a volatile vessel So one letter volume. This will be also one letter. So you can say that Mueller concentration, one Mueller concentration. It's the same for both. Both A and B are the re attempts and both have same concentration. Same Mueller concentration, that is one Mueller. Now for the park two moles of a. Yeah two Reacts with two more of B. The do more In two little vessel. If I'd invite to later it will be $1 and it will be one more. This is why because the Mueller concentration or we can say modularity, Mueller concentration or more clarity is equal to the number of moles of solute number of moles of Hollywood presenting one Letter of vol. So here we have for this for this reaction also we have a number of moles and the volume. So we find out the molar concentration and similar in a similar way for the beep part. We also find out the molar concentration. So now for sea park It has 0.2 more low key. Is there a point to move and little point to more of the B is 0.2 more And it leads to in the reaction in 0.1 letter weather. So this is 0.1 later and this is also 0.1 later. So that is it is to Mueller. And this is also storm warning now or any reaction or any reaction. I suppose they have A plus the gives P product. Now for any reaction, the rate of reaction is directly proportional to the concentration of the reactant. That is rate of reaction rate off reaction. So this is be here. Rate of reaction depends on depends on concentration of reacting concentration. Als reactant that is the more the concentration of reactant, the mold. A concentration of the extent. The more will be the rate of reaction done more well. Me. Uh huh. Leg off reaction. So now, since in the part two little realism has to more of a cool mole of A And two more of B. Well, more of B. That is the maximum mule, maximum moles of reactant, their dance. So the more the number of maximum number of reactant. The more will be done. Yeah. The more will be their tendency to react faster. They will react faster. They will react faster. In see part since the vessel has maximum Mueller concentration. The vessel as maximum smaller concentration, mullah concentration of the attempts while the unit volume. The reaction in vessel. The reaction in vessel well proceeded entirely. We'll see. Okay, hi

In this problem were given to different molecular scenes depicted and diagrams, and be to describe the equilibrium of this reaction system. We're told that the molecular scene depicted in diagram A represents this system at equilibrium. We want to determine how many moles of each species will be present in the diagram and be once that system reaches equilibrium so we can begin by using the fact that diagram a represents equilibrium to solve for the equilibrium. Constant. Casey. We know that we can raid out an expression for K C in terms of the molar concentrations of all the species to be the concentration of mn squared, divided by the concentration of em to time, See concentration of end to and since the molecular scene and diagram A is known to be an equilibrium, whatever the concentrations are those species which we can calculate once we plug them into this expression, we can solve for that value of K C, which will be constant at a given temperature. So we find the concentrations of mn at equilibrium and diagram A as well as M two and n two. We're told that each one of the molecules that are present in the molecular scene corresponds to 0.1 moles of that specific chemical substance. So if we count them all together, how many of them are there in multiplied by zero point when we get the total number of moles and we're told that this is a one leader volume so that would correspond also to the molar concentration of each one of these species. So we know that the m are are black. In the end, molecules are orange. So if we begin by counting how maney m ends we have we're looking for how many? How many in this scene depicted in diagram A. We see one black and one orange molecule bonded together. If we count those out for diagram A, we see that we have a total of four multiplied by 0.1 molds for each one of them, divided by a one leader. Volume comes out to a concentration of 0.4 Moeller and now we look for diatonic m two, which would be to those black Adam's bonded together. We see that there are two of them and we also see that there are two of those diatonic and two molecules, which are the two orange atoms bonded together, divided by that one leader volume that comes out to 0.2 moller for both of those equal agreeing concentrations. And now we can plug those values into our expression. 0.4 Moeller squared, divided by 0.2 Moeller time. 0.2 Moeller. And that means that the equilibrium constant is equal to four. And now we know that the molecular scene depicted in Diagram B is not at equilibrium. We can write out the reaction quotient using the same expression, the concentration of M n squared, divided by the concentration of em to times the concentration of end to, and we can count up each of these molecules again in the diagram and in part B to find what these concentrations are. So when we do that again, we discount each one of these species and diagram B and the multiplied by 0.1 moles and then divided by the one leader volume we should get are concentrations for mn to be zero Moeller cause there are no and and molecules found in diagram B, and we should find that we have 0.6 more m two and 0.3 Moeller and to And so now we conform a reaction table for this system. Remember that we have m two plus and two going to two mn the initial concentrations or what we just solved for which is what is shown in in diagram B. You have 0.6 smaller 0.3 Moeller in zero. Based on this street geometry, we have minus X for both of the reactors and plus two x for the product. So an equilibrium we have 0.6 minus X and 0.3 minus sex in two X. And now we know based on the expression for K C, which equals four, which is what we determined from diagram A that this is equal to two x squared based on the equilibrium concentrations from the reaction table time 0.6 minus X time 0.3 minus x And now we can multiply these two together and then multiply that result by four and I will be equal to four x squared. And so when we do that and rearrange it, we see that we have 0.72 minus 3.6 x plus four X squared equals four x squared so we can cancel off the four X squared terms we're left with 0.72 equals 3.6 x, and so X equals 0.2. Now we plug in their value of X into our equilibrium expressions. So at equilibrium, the concentration of each species can be found using those expressions. And since we have a volume of one leaders, those would also correspond to the number of moles of each species president at equilibrium. Once the molecular scene depicted in diagram b reaches equilibrium. And that is what we we're ultimately trying to determine in this problem. So for concentration of m two at equilibrium, this is also equal to the number of moles of em to since we have that one leader volume and that comes out to 0.4 moles for the number of moles of end to we have 0.1 mole and for the number of moles in equilibrium for mn, we have 0.4 moles. So those air in the total number of moles of each species at equilibrium after molecular your seen in I agree and be reaches equilibrium

We have a plus to be In Equilibrium with three C plus D. Going to make an ice box to kind of keep track of what's happening here. Start with 1.18 And 2.85. Okay, those are given to us as moles and we have none of these to start with. Mm And then when we're done we have 376 moles of D. 4.376 moles Of D has formed. So I must have added so plus .376 the same time. I must have added three times that for my C. So that means we're going to have 1.128 malls. At equilibrium, I must have used 376 moles. So I'll have .84 moles And I must have used two times .376. So I'll end up with two .10 Malls. So at Equilibrium I have 2.10 malls be present. And thats answer D.

So business question says that we have ah, to a reaction would be to produce a to B and, uh oh, my mortal a is reactive. Be on. But the question is how ski nose are minimal or it should be produced. To answer this question, you have to understand the concept of giving to the region. So the military James is basically the region started. You stop first in the chemical reaction and definitely tell me if the amounts of products down before looking at this equation Ah, we can see that two mortal A needs formal of B to form one of you to be, which basically transferred, translates to the facts jets. Um, well, every more B deaths, uh, it's needed in this reaction. We need tools off, eh? In, uh, what we have here provided your horse won't buy their full of B in one of my more off Hey, Dusty reactions. Number of more. Basically, it means death. We have less a month off a in this reaction because for every mole of being, we need to mold for it. But we have normal B and warm off a instead of two. Most of it not to me. His death, it would be usedto first in this reaction, therefore, is the limiting regent. And because of deaths, the amount of aid that is present with Tommy's de amounts hope it to be home. Which means that we have to calculate the amount or into VW produced, based on your months off a dice present, Joe to most off, eh? Where do you produce one? Oh, yeah. Juvie like you, Mexico. Now we have just one more. I'm minimal. Will need to be with prettiest from so Oh, should one's ill. Oh, it should be. Dobby was to set up like five moles. Oh, it's you. It's day 60. A man's home. It should be doubly produce, Jennifer. Five more.


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