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Volume H2SO4 titrated = 30.00 mL NaOH titrant molarity 0.35 M Initial buret volume 3.10 mL Final buret volume = 18.36 mL7 . What volume of 0.35 M NaOH reacted with ...

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Volume H2SO4 titrated = 30.00 mL NaOH titrant molarity 0.35 M Initial buret volume 3.10 mL Final buret volume = 18.36 mL7 . What volume of 0.35 M NaOH reacted with the 30.00 mL of HzSO4 solution? 15.26 mL Computer's answer now shown above. You are correct. Ypur receipt no. is 151-308 Previous Tries8. How many moles of NaOH reacted with the 30.00 mL of HzSO4 solution?molSubmit AnswerIncorrect. Tries 2/3 Previous Iries9. How many moles of HzSO4 were in the 30.00 mL of H2SO4 solution?molSubmit

Volume H2SO4 titrated = 30.00 mL NaOH titrant molarity 0.35 M Initial buret volume 3.10 mL Final buret volume = 18.36 mL 7 . What volume of 0.35 M NaOH reacted with the 30.00 mL of HzSO4 solution? 15.26 mL Computer's answer now shown above. You are correct. Ypur receipt no. is 151-308 Previous Tries 8. How many moles of NaOH reacted with the 30.00 mL of HzSO4 solution? mol Submit Answer Incorrect. Tries 2/3 Previous Iries 9. How many moles of HzSO4 were in the 30.00 mL of H2SO4 solution? mol Submit Answor Incorrect Tries 2/3 Prevlous Tries 10 _ What Is the HzSO4 concentration (molarity) in the 30.00 mL of HzSO4 solution? Bubmil Mnumur Incorrecta Tries 1/3 Prevlous Tres



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In a titration of $\mathrm{HNO}_{3},$ you add a few drops of phenolphthalein indicator to 50.00 $\mathrm{mL}$ of acid in a flask. You quickly add 20.00 $\mathrm{mL}$ of 0.0502 $M$ $\mathrm{NaOH}$ but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add 30.00 mL of the acid, and the solution turns colorless. Then, it takes 3.22 mL of the NaOH to reach the end point.
(a) What is the concentration of the $\mathrm{HNO}_{3}$ solution?
(b) How many moles of $\mathrm{NaOH}$ were in excess after the first addition?

So our thai tradition is between a strong acid and a strong base. So our net ionic equation it's just gonna be H plus plus O. H minus gives us water. And at the equivalent point other than water we'll just have our two spectator ions also present. Okay and then we're gonna see if we can figure out how much. Anyway H. We need to get to our equivalence point. So let's start with our information about our strong acid. 5.1988 Moller multiply that by its leaders And we'll see that we had 7.36 Times 7 -3 Malls of our strong acid. Therefore we must have also reacted that same amount of our strong base to R. O. H minus. And then if we take that and divide it by the polarity will get our volume. Okay because polarities moles per liter. So that will give us point 03457 leaders or 34 .57 ml. Okay so before the titillation occurs, all we have is our strong acid R. H. Plus concentration Was given as .1988 Moller. So if we take minus the log of that, we'll get our ph Which is 7016. Yeah, mm. Yes, halfway to the equivalence point one, half of the moles of each plus have reacted. So we're left with half of our moles left and we've added half the volume of R. O. H minus. Okay, so the moles of Rh plus that are going to be present Is going to be 73, 6 Times 10: -3 divided by two. So we'll have 3.68 Times 10 to the -3 moles of O of H. Plus that remain. They haven't reacted yet. And we used 34 .57 ml. Well, at this point we have only added half of that. So 17.28 ml. So if we add that to the 37 ml that we had of Rh I who will get our total volume at that point? Which is 54.28 ml or .054- eight leaders. Okay, so our concentration of H plus halfway to the equivalence point Is our moles 368. I'm Santa -3. And we'll divide that by our total volume and we'll get our more clarity as 0678 Moller. That's our age plus. So we take minus the log of that. We'll get our ph as 1.17. Okay. And then finally they ask us, what's the ph at the equivalence point, while the ph is seven, Okay, anytime you have a strong acid and a strong base, the ph of the equivalence point is going to be seven

In this question, we need to determine the modularity of sodomite drug side which reacts with nitric assets. Both volumes of these re agents are provided as well as the polarity of nitric acid. And the calculation needs to be done according to the reaction equation which is given in the question and also presented on this page. All right. So for us in order to calculate polarity we know that this is um moles per liter. So we need to calculate the number of moles of sodium hydroxide. And we already have the volume even though in milliliters we can just convert that to leaders. So first of all we will need to calculate the number of moles of sodium hydroxide. And this we will calculate from the number of moles of nitric acid. And then you just use this trick geometric ratio to calculate the number of moles of sodium hydroxide. And from there we can easily then calculate irregularity. Right? So our first step is to calculate the number of moles of city. My drug side sound oh sorry not sorry my drug side of a nitric acid. Um We have the volume in terms of ml. So we need to convert that two l because we're going to multiply that with the modularity of nitric acid which is in terms of Malls per liter. So this is no 0.0 342 moles. The one liter. Yeah. Right. And if we do that, you can also see that the units cancel out and we end up with molds. I'm not going to do the calculation here. Gonna do that later on as part of a bigger calculation but now and we need to calculate the number of miles of syria. My truck side according to the strict geometric ratio given by their psychometric coefficients in the equation. And this is really simple for this equation. It's just one over one. So, this week geometric ratio, it's one of the one. This makes library simple because that means that the number of moles of sodium hydroxide is going to be equal to the number of moles off. Um nitric acid. Right now we know that more narrative, the mill aridity of sodium hydroxide is equal to the number of moles of sodium hydroxide divided by the volume of the solution. All right, so um this is going to be um the number of moles off. Sorry, my drug side which I showed a top at the top, how to calculate its one times 0.375 liters time. The polarity Which is 0.0342 moles per one liter. And then we can just divide that by the volume of the sodium hydroxide, which is given in terms of milliliters, which is convert that two leaders Which is 0.0 414 L. And then when we do this calculation We end up with a value of 0.0310 Mola. So that is the more clarity of sodom hydroxide in this question. Just quickly to recap, we needed to calculate the clarity of ceremony drug side. We need reacting on 1-1 mile ratio with um nitric acid. Both volumes of these re agents were given as well as the polarity of notched acid. So we were able to calculate the number of moles of not acid And then use this together with historic geometric ratio of 1-1 to calculate the number of moles of certain hydroxide. And now we know that clarity is a number of malls the volume. So it just um then took the number of moles of sodium hydroxide and divided it by the given volume of sodium hydroxide. In terms of leaders.

In this question, we need to perform historic geometry, you figure out how much for the volume of sulfuric acid solution. This need to prepare 37.4 g of aluminum sulfate through the given reaction. So we're given the density of a solution and we're told it contains 15% of sulfuric acid by mass. So we are going to use the mole ratios from the balanced chemical reaction to obtain a mole ratio between the reactions and products. And then we will use the molar mass to convert from mass two moles of a compound. So we'll start with what we know. We want to prepare 37.4 g of aluminum sulfate. And then we use the molar mass to convert to the moles of the all of them selfie. And then we see from the balanced chemical reaction that we have three moles of the sulfuric acid for every one mole of aluminum sulfate. So so we can use that as a conversion factor. And if we have the same units on the top and bottom, they will cancel out. Then we usually have the mass of the sulfuric acid. So we are going to use the molar mass to convert to the mass of sulfuric acid. This is a massive Sofia gases that will read Act to Form 37.4 g of the aluminum selfie. So we're told that we have a solution that contains 50% of the sulfuric acid by mass. So we need to find the mass of the solution before we can find the volume of the solution. So 15% of X, which is the massive solution is equal to 32.16 g of the Sofia gas id Because it's 15 So fear kassid. So we would then solve for X. And we find that X is equal to 214.4 g of solutions. So that's the massive solution. Then we can convert two ml of solution using the density because density is mass divided by volume. So we can flip that to use it as a conversion factor. So the mass on the top and bottom will cancel. And then we get milliliters of solution and that is how we would solve that problem. Now for the second part we're going to do the same thing. So now we're dealing with a different equation and we need to know the volume of sodium hydroxide solution. So we approach is the exact same way we're going to first convert the mass of the product into the mass of sodium hydroxide needed and we need to use the balanced chemical reaction to get the mole ratio And then we're going to solve for the massive solution. Knowing that 12% of the solution is sodium hydroxide. And then we get the massive solution and we used the density to convert to the volume of solution like we did in the previous part

This question is a tie, Trey Shin problem where you need to understand this, like geometry between a strong acid and a strong base. And then also recognize what conversions air involved when you're going from milliliters of a based solution to milliliters of an acid solution. So for the first part of this problem, part A, we have 45 million leaders of K o. H, which we can convert toe leaders. Then, knowing the mole arat e of K o h, we can convert leaders two moles K o. H. And then this is where we need to recognize the stoy geometry between K o H and H N 03 One mole of K o. H reacts with one mole of h N 03 because K o. H can take on one hydrogen because it has 10 h and H and 03 can donate one hydrogen because it just has one hydrogen. So the stoy geometries oneto one. Then, when we know the moles of H n 03 we can use the mill arat e of h n 3.1 Mueller to get leaders h n 03 solution and then convert leaders to mill leaders. 300 mL. Nitric acid. For the next one, we have 58.5 mil. Leaders of a 0.1 Moeller aluminum hydroxide solution will convert the mill leaders, two leaders and then the leaders. Two moles by multiplying by the molar ity. 20.100 molds aluminum hydroxide per leader. Then, when we know the molds aluminum hydroxide, we recognize that the stoy geometry is 3 to 1. Hydrogen could only donate are in sorry. Nitric acid could only donate one hydrogen, but each aluminum hydroxide has three hydroxide so it can accept three hydrogen. So for every one mole of aluminum hydroxide, we need three moles of nitric acid. Then, when we have moles nitric acid, we can convert to leaders nitric acid solution. Using the polarity of the nitric acid. 0.1. Mueller and then leaders to Miller leaders. 17.6 million liters. Nitric acid. For the last one, we have 34.7 mil. Leaders of a 0.775 Mueller sodium hydroxide solution will convert the mill leaders two leaders and then leaders two moles by multiplying by theme Olara Ity of sodium hydroxide. We recognize the stoy geometry between sodium hydroxide and nitric acid is one toe. One when we have moles, nitric acid. Then we can convert toe leaders nitric acid solution by dividing by the molar ity 0.1, and then go from leaders to mill leaders by multiplying by 1269 mil leaders nitric acid.


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