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How many milliliters of concentrated phosphoric acdis, 90.0% (g/1OO_gsolution) density 1.90 Bcm? cm"_ 1ml) are required to prepare liter of 0.0500 M solution?0...

Question

How many milliliters of concentrated phosphoric acdis, 90.0% (g/1OO_gsolution) density 1.90 Bcm? cm"_ 1ml) are required to prepare liter of 0.0500 M solution?0 4. 48 mL0 2.87 mL0 6.92 ML0 8.34 ML

How many milliliters of concentrated phosphoric acdis, 90.0% (g/1OO_gsolution) density 1.90 Bcm? cm"_ 1ml) are required to prepare liter of 0.0500 M solution? 0 4. 48 mL 0 2.87 mL 0 6.92 ML 0 8.34 ML



Answers

How many milliliters of 18.0 $M$ $\mathrm{H}_{2} \mathrm{SO}_{4}$ are required to prepare 35.0 $\mathrm{mL}$ of 0.250 $M$ solution?

To calculate the volume of sodium hydroxide that's required to react with the particular volume and concentration of phosphoric acid. We first need to know what the study geometry is for the tight rations reaction. Phosphoric acid reacts with sodium hydroxide, producing sodium phosphate and water. To balance this, we've got three sodium, so we're going to need to put a three in front of the sodium hydroxide and then to finish balancing, we need to put a three in front of the water so the storks geometry is three moles sodium hydroxide for one mole phosphoric acid. So we'll start with the 15 million liters of phosphoric acid, divided by 1000 to get leaders phosphoric acid and then convert our leaders phosphoric acid. Two moles phosphoric acid by multiplying by the molar ity of the phosphoric acid solution. Now that we have moles phosphoric acid, we can convert the moles phosphoric acid into molds. Sodium hydroxide from the three toe One relationship. After calculating mole sodium hydroxide, we can convert the mole sodium hydroxide into leaders sodium hydroxide solution by dividing by the molar ity of the sodium hydroxide solution. We're now in units of leaders, sodium hydroxide solution. But the question asks for milliliters, so we'll simply multiply by 1000 and we get 56.6 mL sodium hydroxide.

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.

To calculate the volume in mill leaders of sodium hydroxide that is required to react with a particular volume and concentration of sulfuric acid. We first need to know what the Thai trey shin reaction is, and it's toy geometry. We have sulfuric acid reacting with sodium hydroxide, making sodium sulfate and water. To balance that, we need to put a two in front of the sodium hydroxide and a two in front of the water. So we know that the stoic geometric relationship between sodium hydroxide and sulfuric acid is to tow one. So we'll start with our 10 mil Leaders of the Let's see Sodium Uh, yeah, sulfuric acid. Divide that by 1000 to get leaders of sulfuric acid solution and then convert our leaders of sulfuric acid solution into moles of sulfuric acid by multiplying by the molar ity of the sulfuric acid solution. Then that will give us mold sulfuric acid. We recognize that one mole sulfuric acid reacts with two moles of sodium hydroxide. Now that we have mold sodium hydroxide, we can use the molar ity of the sodium hydroxide solution to convert the mole sodium hydroxide into leaders sodium hydroxide solution. However, this question once. Milliliters sodium hydroxide solution, so we'll multiply by 1000 and will get 16.9 mL sodium hydroxide.

This question is similar to 63. We know the volume and polarity of one reactive, and we want to calculate the volume of the other reactant, knowing it's similarity, the other reactive being nitric acid at a concentration of 0.25 Moeller. So we'll start out with the volume of for the for part a. The volume of potassium hydroxide in mill leaders converted to leaders and then the polarity of potassium hydroxide so that we can convert the leaders into moles of potassium hydroxide. This, like you metric ratio is one mole of potassium hydroxide reacts with one mole of nitric acid. Knowing the moles of nitric acid, we can use the polarity of nitric acid to convert moles into leaders and then, on the tail end will convert leaders into Miller leaders. We'll do the same type of calculation for Part B. The only difference are the concentrations, and the fact that is, it is a 3 to 1 story geometric relationship. One mole of aluminum hydroxide requires three moles of nitric acid, and then on the last one, it is a 1 to 1 lucky metric relationship. We have 65.9 milliliters converted to leaders of sodium hydroxide at a polarity of 0.475 The reaction is one mole of sodium hydroxide for everyone, mullah nitric acid and then the polarity of nitric acid 0.25 molars. Moeller allows us to calculate the leaders, which we can convert to mill leaders.


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