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TableStundardiaation o un odint Solutlon:mg o LicmL o Iodine Rezgent mg Vit ClmL [2 Reugent Uscd 10.40 Ist Titralion ML Ia blal Ripeat Titration 10.44 9.58 M MmnL A...

Question

TableStundardiaation o un odint Solutlon:mg o LicmL o Iodine Rezgent mg Vit ClmL [2 Reugent Uscd 10.40 Ist Titralion ML Ia blal Ripeat Titration 10.44 9.58 M MmnL Avcragc of tix two Vq IDIII rilmn: buxlnL Note: Ths Value should h "4d Ihe Vit € calculations Whcr Punts "ars exprrnicntTahleVitamin € in Fruil Juico - Calculating mg of Vit-C (show calculations on Ren page)JuiccoljuiceIg Aocorbic ucidlmLjuiceOrange50.00Grapefruit50.00Pineapple50.0050.00Apple50.00Nole: Put u ( br Juurns

Table Stundardiaation o un odint Solutlon: mg o Lic mL o Iodine Rezgent mg Vit ClmL [2 Reugent Uscd 10.40 Ist Titralion ML Ia blal Ripeat Titration 10.44 9.58 M MmnL Avcragc of tix two Vq IDIII rilmn: buxlnL Note: Ths Value should h "4d Ihe Vit € calculations Whcr Punts "ars exprrnicnt Tahle Vitamin € in Fruil Juico - Calculating mg of Vit-C (show calculations on Ren page) Juicc oljuice Ig Aocorbic ucidlmLjuice Orange 50.00 Grapefruit 50.00 Pineapple 50.00 50.00 Apple 50.00 Nole: Put u ( br Juurnsults' Amnt & Iodine Reegant (mL) required totitraeyarr smler" Typed Lide Orange Frs Smple 3 05 mi Sand smple 295M Avgeped two smples Tang



Answers


The amount of ascorbic acid (vitamin $\mathrm{C})$ in fruit juice is determined by a titration using a redox reaction. Iodine is the titrant, but because iodine solutions in water are unstable, the iodine is generated by a reaction in which a titrant containing iodate $\left(\mathrm{IO}_{3}^{-}\right)$ ions is added to a solution containing the juice sample, iodide ions However, any iodine formed by reaction (i) is immediately reduced back to iodide by ascorbic acid from the juice sample: ii. $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(a q)+\mathrm{I}_{2}(a q) \rightarrow$ $$ \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(a q)+2 \mathrm{I}^{-}(a q)+2 \mathrm{H}^{+}(a q) $$
(I $^{-}$ ), and a few drops of a starch solution that turns dark blue in the presence of iodine. Iodate ions are reduced to iodine $\left(\mathrm{I}_{2}\right)$ while iodide ions are oxidized to iodine: i. $\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(a q)$ (not balanced)However, any iodine formed by reaction (i) is immediately reduced back to iodide by ascorbic acid from the juice sample: ii. $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(a q)+\mathrm{I}_{2}(a q) \rightarrow$ $$ \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(a q)+2 \mathrm{I}^{-}(a q)+2 \mathrm{H}^{+}(a q) $$ (FIGURE CAN'T COPY)

Yeah. Then why? Be equal to f f T. NFL T denotes the amount of iodine and the blood at time. T. So given the idea answers and since the blood a rate of 4%/h, the Andean answers to your the your brain Um at a rate of 10 per hour. So the rate of change of the rate of change of iodine iodine levels of blood per hour will be dy over DT. Which is equal to negative for 100. Yeah. Tom's wide minus 10 over 100 times. Why, jesus? This becomes negative 14 over 100 times. Why? And the the initial condition Y. of zero. Yeah, Becomes f of zero mm. Here the negative son indicates that the iodine levels decreasing as time progresses. Yeah.

Question. Number 37 introduces another concentration unit that is, by far the most common concentration unit used in chemistry. That is molar ity. Morality is defined as the moles of solute divided by the leaders of solution. So for this problem, we know the mass of the salute and we know the milliliters of solution. So to calculate mole arat E, we simply need to convert the grams of solute, grams, potassium iodide, two moles, potassium iodide. To do that will simply divide by the molar mass of potassium iodide, which is 166. Well, then, divide that. So this will be the moles. This calculation will give us the moles by the leaders of solution. We weren't given leaders. We were given Miller leaders, but we can easily convert mill leaders two leaders by dividing by 1000. So this denominator now has units of leaders. When we finished, the calculation will have units of moles per leader, which is more clarity, and we get a value of 0.290 Moeller Potassium iodide

So we're given an equation and it is actually balanced. So we're gonna go ahead and work with the equation as it's given. And they've given us the starting part of a polarity and the volume of a substance which is always a great place to start. So we're going to start with arm polarity went 134 moller And make sure you change your volume two leaders. My multiplying this, we're going to get our malls okay. Of the H three A. S. 04 We're gonna go ahead and do them all ratio. So we'll convert moles of that substance two moles of i. two. And then we'll go ahead and change moles to grams and find the mass of ri too. So our mole ratio is 1-1 And our molar mass to 53 8. So this is going to give us five 24 g of iTunes. Now this is the mass of iodine that we had in our 50 millim portion. But originally we had 243 mm. So we're gonna have to do a little ratio to figure out how much iodine was in the original. So if I have this many grams in my 50 ml, just say, how many do I have In 243. So if I work out that math, I'm gonna get to 55 g of itunes. Mhm. And I'm gonna go ahead and change this to mold because we're looking for a polarity. I'm gonna go ahead and divide by the molar mass here hands for more. And this will give us went 0100 malls of I two. And then we'll go ahead and change, We're gonna go ahead and change this to my polarity by dividing by the volume, 243 middle leaders or .243 leaders. And that will give us our more clarity. .04 13 moller

So C six h 806 month is determined by reacting bro mean and then a tie. Trey Shin. So this is illustrated in the following two equations that have included for your reference. So given the information we have, we're determining the vitamin C content and whether it is accurate. So what? We have been informed off. So the moles off c six h 86 is half the moles of H p R due to that moral relationship. So then we have 2.916 minimal where moles of H B. R is 5.832 millimeters just for your reference. So the massive C six h 806 is the moles multiplied by the molecular weight. We perform that calculation to determine a mass off 513.566 mg. So this is greater than the outlined amount that we were initially informed off. So the percentage access is one of 2.71 Therefore, 2.71% is more than listed


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