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4/ % Yleld calculation (based on the maximum amount of product obtainable from BOIH starting materials - for all three substrates (anlsole, acetanilide; and anlline...

Question

4/ % Yleld calculation (based on the maximum amount of product obtainable from BOIH starting materials - for all three substrates (anlsole, acetanilide; and anlline) and the bromine- You must Useskx [elevont Dlmenslonolequatlons Twofor each subsrote sccon example uploaded in the "Mscelloneous sub-folder of the "Flles" folder In Canvos) Irs 0 arcot Idco to Use_ tobulorform [bx Insetting _ Loble in WordLto determine thcLR 30 ptsa/ Flrst fInd out the level of bromination by analyzing

4/ % Yleld calculation (based on the maximum amount of product obtainable from BOIH starting materials - for all three substrates (anlsole, acetanilide; and anlline) and the bromine- You must Useskx [elevont Dlmenslonolequatlons Twofor each subsrote sccon example uploaded in the "Mscelloneous sub-folder of the "Flles" folder In Canvos) Irs 0 arcot Idco to Use_ tobulorform [bx Insetting _ Loble in WordLto determine thcLR 30 pts a/ Flrst fInd out the level of bromination by analyzing the three NMRs posted In the Reactivity lab folder the Files menu: b/ Based on what vou find above; write the three balanced equations for the thrce reactlons separately: These equations will give vou the exact stoichiometry that vou can used to find out the limiting reagent for each of the three reactions: cl/ You must start the first two sets (for aniline and anisole) of DA equations starting from the volumes of anillinc (0.30 mL) and anisole (0.35 mL) used and work your DA equations from these quantities: See QStofind out how the bromine solution wos prepored: For acetanilide should be straightforward_ Draw three sets 0f two-row tables for these equations: d/ TebrominatEg mede dissolving [mLeLbromine enough 4824 HBr aqucous folution t0 make total oLIOO mL eL solution Ba carcful aboutheauatititk of broming do bSWed YOu_DAEquutions_ Table for calculation of LR for aniline: (these DA equations must include the volume of anlline used. The volume of aniline 0.30 mL % Yield of the bromlnated aniline product (show work) spts e/Table for calculation of LR for anisole: (these DA equations must Include the volume of anlsole used The volume of anisole used 0.30 ML



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In an enzyme-catalyzed reaction with stoichiometry $\mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}$ is consumed at a rate given by an expression of the Michaelis-Menten form: $$r_{\mathrm{A}}[\operatorname{mol} /(\mathrm{L} \cdot \mathrm{s})]=\frac{k_{1} C_{\mathrm{A}}}{1+k_{2} C_{\mathrm{A}}}$$ where $C_{\mathrm{A}}(\operatorname{mol} / \mathrm{L})$ is the reactant concentration, and $k_{1}$ and $k_{2}$ depend only on temperature. (a) The reaction is carried out in an isothermal batch reactor with constant reaction mixture volume $V$ (liters), beginning with pure $A$ at a concentration $C_{\mathrm{A} 0}$. Derive an expression for $d C_{\mathrm{A}} / d t$, and provide an initial condition. Sketch a plot of $C_{\mathrm{A}}$ versus $t,$ labeling the value of $C_{\mathrm{A}}$ at $t=0$ and the asymptotic value as $t \rightarrow \infty$ (b) Solve the differential equation of Part (a) to obtain an expression for the time required to achieve a specified concentration $C_{\mathrm{A}}$ (c) Use the expression of Part (b) to devise a graphical method of determining $k_{1}$ and $k_{2}$ from data for In versus the pour plot should involve fitting a straight line and determining the two parameters $C_{\mathrm{A}}$ (int the parting of the partating and and the conting are a contation a conting from the slope and intercept of the line. (There are several possible solutions.) Then apply your method to determine $k_{1}$ and $k_{2}$ for the following data taken in a 2.00 -liter reactor, beginning with A at a concentration $C_{\mathrm{A} 0}=5.00 \mathrm{mol} / \mathrm{L}$ $$\begin{array}{|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 60.0 & 120.0 & 180.0 & 240.0 & 480.0 \\ \hline C_{\mathrm{A}}(\mathrm{mol} / \mathrm{L}) & 4.484 & 4.005 & 3.561 & 3.154 & 1.866 \\ \hline \end{array}$$

So in problem for match the chemical equations with their descriptions. So we have given some. We have given some equation and we have to match the equations. So first is go see it for a liquid less oxygen gas and it will form toe ch teau or liquid less to watch toe liquid. So this is the first equation and then we can see the next equation that is NH toe C l guess which converts into energy toe si l equals then third equation is toe See it H 14 liquid. Let's 23 or to guess which converts to 16 carbon dioxide gas plus 14 edge toe liquid. Okay, Okay, then fourth equation is h two s 04 equals That's calcium carbon it solar. So it converts into calcium sell food a quest plus kowtow gas plus vegetable liquid than 50. Question is mercury in liquid converts toe mercury in gas than 60. Equation is C 16 aged 30 auto solid this hydrogen gas so it converts into C 16 extracted toe or toe solid And the seventh equation is calcium oxide solid. Let's each toe liquid so that converts into calcium hydroxide solid. So basically from the matching the column. We can see that for the first first equation. It matches with the equation E. Then it matches with their description E. So basically, I can write it as so This is the equation. And this is the description. This is the equation. And this is the descriptions. Okay, so that equation one matches with the descript description e then equation toe matches with the see, then occasion thought matches with the D education. Fourth matches with the mhm equation five matches with F a patient, six matches with G and efficient seven matches with the description be so as we see that in the equation one So it matches with the that when methanol, if ingested, reacts with the oxygen to form formaldehyde, which is toxic water is also formed in this reaction. Then second is see so we can see that second is C So C description is that global mine is added. So see that this is a global mine is added tow our water supply in very small amount to kill bacteria. So third is today. So day is basically when obtained, which is ch 14 and oxygen gets burned in our cars, carbon dioxide and water come out in the exhaust. So then fourth is a so a is basically this three sulfuric acid. So this sulfuric acid dissolved in the rain drops off. Okay, so forth is yeah. So sulfuric acid dissolved in the rain drops off acid reacts with the calcium carbonate with calcium carbon it in this seashell and marble structures. Direction produces calcium self food That is so This is the CSO four and direction producer Calcium self. It dissolved in water and carbon dioxide gas. So this is the reaction for which is a and 4/5 which is F that is, liquid mercury evaporates to produce mercury vapor. So next is sixth is G. So this is the equation. Sixth. So saturated fatty is it Like Paul Medicus? It tend to form long, solid and cloak people's at today arteries. These saturated 30 years. It can be made from unsaturated 30 years by adding hydrogen gas. So finally seven SB So this is the seventh equation. That means toe balance. The effect off the etcetera in solid calcium oxide has been added to many lakes. It reacts with water to form solid calcium hydroxide a strong ways that is only slightly soluble in water. So these these are the equation. And with their description, as we see that the description given in the cautions so we can match it as very easily thank you.

So between the lithium night treat, the nickel to night tree and the strong team nitrate, the one that we should actually used to precipitate the maximum out of the carbonate ions from the solution should be the strong team nitrate. And the reason for this is has the lowest ks beef value asking here. And so they gave us a case be table with the lithium carbonate, nickel to carbon and the strontium carbonate. And so adding strong human nitrate trying him carbonate is actually what we'd be making. So it would make sense for look at these case p values. And so when a case be value is closer to one, that means that the, um substance dissolves very readily and be squeak. It's only ions and it doesn't like to stay in the forms of for instance, lithium carbonate. The case be was one then what we would have is only the theme ions and only the carbonate ions. We would actually have the lithium carbonate, but here so lithium carbonate has the value closest Thorne and strong from carbonate has value closest zero, which means that it likes the state s surround him carbonate and so once we put the strontium nitrate into the solution and it reacts with the carbon e ions, we get the strong carbonate and have a lot of precipitated. Then for one beer, what we're looking at is for the carbon compound that contains the cattle on chosen in part a, which we chose as strontium nitrate, giving us drunken carbonate determine the concentration of each eye on of that compounded solution equilibrium. So this is the equilibrium, expression or equation. And so what? We have your strong him carbonate solid, solid in equal air be in, which is what these double arrows are with strong team two. Plus, it's the cat eye on and it's a quest. And then the carbon, eh? I unders and own. And that's also Equus. So then, um, what we need now is chaos P is actually just an equation, like it's just a formula that you follow. So they gave us the KSB and Cass be equals the concentration of whatever is a quest on the right side of your equation. So we have the strong tea, um, cat eye on multiplied by the concentration of the carbon A an eye on. So now since they gave us our case, P value, which is just 5.60 times 10 to the negative 10th. We're setting that equal to basically just the concentration of the strong Thiam times, the concentration of the carbonate and what we can do there is that's just x Times X giving us X squared. So now if we just take the square to both sides and we saw for eggs X equals the concentration of the strontium I on which is also equal. So the concentration of the carbonate ion, which overall is equal to 2.37 times 10 to the negative fifth Moeller. And so that's the concentration off each eye on in solution. So now moving on part C one mixing the solution should the student ensure the carbonate solution or the nitrate solution is an access. So for this, what we have to think about it, Which one of these ions would actually help us more? So we are Actual precipitated is strong team carbonated, which is as our CEO three. So we don't need more carbon ian solution because it doesn't matter how much carbonate we have, because that's just gonna stay as an eye on but for the strong tea Um, the way that we got the strong team is from the strong tea of nitrate, which is this So if we don't have enough strontium, then we can't make the strong him carbonate. So the nitrate solution should be an access because in order to create the maximum amount of precipitate, you need enough strong team I owns to react with all of the carbonate ions and having the access strontium ions in the solution doesn't affect the master of the carbonate in the virginal example. So we have a fixed amount of carbonate, but we I don't know, having more having the access won't give us the precipitate but having the excess of strong here Mayan means that we can react all off the carbon e into precipitously and then having the extras drawn to him doesn't affect anything. So then moving on to part dear after titrate and sufficient solution to precipitate out all of the carbonate ions, the student filters the solution before placing in the crucible and heating it to drive off the water after several beatings. The final mass of the precipitate it remains constant is determined be 2.2 grams. So the first thing we wanted to do is determine the number of moles of Pacific it. So what we d'oh is we take our 2.2 grams of our strong team carbonate, and we multiply that by the molar mass, which is basically just a ratio that allows us to go from grand symbols or moral stra graham. So 1 47.63 grams Uh, s or C 03 her one more and overall, that gives us okay after I put it into my calculator one point 37 times tend to the negative seven Negative, second more. And now, um, moving on determining the mass of the carbonate present in the precipitate. What we do is we take these moves that we just got and then we multiply that bye. You have one more of, um, you have one mole of the carbon e ion her one more of the strong him carbonate substance. So then, since it's just a wonder one we end up with 1.37 times tend to the negative second more of the carbonate and then what we need to do is take the Moler Mass off the carbon e ion and multiply that by the moles. And so once the molar mass of carbon e is so carbon is 12 plus the 40 the 48 from the three oxygen. So it's 60. So then once we take 60 times so 0.137 times succeed, which gives us zero point e to two grams of this carbon ee eye on. So then, um, poor E. What we need to do is determine the percent by massive carbonate in the original sample. So the 0.822 grams of the carbonate weakness around that 20.82 grams and just carry the 26 eggs and then you divide that by don't know why I did that. Um, so it's the zero point 82 grams divided by the one point 39 grams from the original sample. And then if you multiply it by 100 that's how you're gonna percent and we end up with 43.4 percent of the carbon E who of the, um carbon e i aunt in the original solution so that moving on too. Oh, you know, I did that. So then moving on to part, as is the original compound most likely lithium carbonate, which is ally to Theo. Three, um, sodium carbonate ing any to Theo three or potassium carbonate, which is you if you go through here. So what we do now is so it's the, um 60.1 for all three, um, on the numerator because what we're looking at for all of them is the carbonate. But then the only difference is the total math. So we're either dividing it by 73.89 because you had the mass of the carbonate plus the mass of dilithium or 16.1 divided by one of 5.99 or 60.1 divided by 1 38.21 Oh, and so you multiply all these by 100 and you enough with anyone 0.2% for that 56.6% for this and 45 0.4% for the pizza, potassium carbonate and


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