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4.) (20 pts) There is 3.88 mass % solution of HNO3. Density of that solution is 1.00 glmL. Find: a ) Amount (g) of HNO3 in 200 mL of that solution b:) Molarity (M) ...

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4.) (20 pts) There is 3.88 mass % solution of HNO3. Density of that solution is 1.00 glmL. Find: a ) Amount (g) of HNO3 in 200 mL of that solution b:) Molarity (M) of that solution(show all work)5.) (25 pts) Find T final in (*C) - if 1.23 kg of Fe(s) is heated starting from 22*C , and 33.76 kJ is used. Specific Heat of Fe(s) 0.45 (Jlg *C)(show all work)6.) ( 10 pts) Some of the next statements are wrong, circle them: a. ) Shell 'n=3' never contains 'd' subshell b.) Shell 

4.) (20 pts) There is 3.88 mass % solution of HNO3. Density of that solution is 1.00 glmL. Find: a ) Amount (g) of HNO3 in 200 mL of that solution b:) Molarity (M) of that solution (show all work) 5.) (25 pts) Find T final in (*C) - if 1.23 kg of Fe(s) is heated starting from 22*C , and 33.76 kJ is used. Specific Heat of Fe(s) 0.45 (Jlg *C) (show all work) 6.) ( 10 pts) Some of the next statements are wrong, circle them: a. ) Shell 'n=3' never contains 'd' subshell b.) Shell 'n=3' always contains 'd" subshell C: ) Shell 'n=3' may contain 'd" subshell Subshell 'p' contains max 6 electrons, but it may contain less than 6 e:) Ground State configuration of atom 'Na' has no 'd" Subshell 7.) ( 15 pts) Write: a.) Full electron configuration Valence shell electron configuration c) [Noble gas] Valence shell electron configuration



Answers

A $0.4000 \mathrm{M}$ solution of nitric acid is used to titrate $50.00 \mathrm{~mL}$ of $0.237 \mathrm{M}$ barium hydroxide. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of nitric acid is required to reach the equivalence point? (d) What is the $\mathrm{pH}$ of the solution before any $\mathrm{HNO}_{3}$ is added? (e) What is the $\mathrm{pH}$ of the solution halfway to the equivalence point? (f) What is the $\mathrm{pH}$ of the solution at the equivalence point?

Okay. So we're reacting a strong acid which is nitric acid with a strong base which is bearing hydroxide. So the first thing we're gonna do is write the net ionic equation. So the net ionic equation for strong ass in a strong base is always the same. So h possible O h minus gives us water. The things that will be present at the equivalence point will be the barium ion that came from the barium hydroxide. Okay. And the nitrate ion that came from the nitric acid. And of course our water. Since we're gonna do some strike geometry here, I'm gonna write out the complete reaction rather than net ionic equation. It helps us see that there's a 2-1 ratio here between the barium hydroxide and the night nitric acid. I find it useful to go ahead and write that out rather than using a net ionic equation. Well they've given us some information about our barium hydroxide. Okay so I've got the polarity This .237 More. So if I multiply by its volume and leaders, I'll get the malls of barium hydroxide 119 Malls. So if I want and then find out how much nitric acid would react with that. I'll just do our moral ratio great moles of barium hydroxide to nitric acid And that's 2: one. So we'll see that we needed Or 237 moles of nitric acid. We'll have reacted. So I'm gonna go ahead and take those malls 0- 3 7 Malls. And I have it's polarity someone to divide by its polarity which is moles per liter. So that will give us the leaders that we needed to add 2.0593 leaders or 59 3 ml of our nitric acid was added at the equivalence point. Okay So the ph before anything is added, all we have is 237 Moeller, barium hydroxide. We haven't added any nitric acid yet. So I'm going to multiply that by two to get the polarity of my hydroxide ion. My O H minus 2.474 Moller. Right? Because each barium hydroxide has 20 H- is in it. If I take minus the log of that treatment, bye Take minus the log of this. Since it's O H minus, I'm going to get the P O. H. And that's going to be .3-4. And then if I subtract from 14 14 minus 140.3 to 4, I'll get the ph The 13.68. Okay, so we're going to try to figure out now what the ph is when we're halfway to the equivalence point. Cool, so barium hydroxide was reacting with two H N 03. So, if I started with .0119 moles of this, We're going to divide that by two And see that .00595 moles of barium hydroxide remains. Okay, so one half of it remains at the half equivalence point mm. Then I'm going to multiply that by two because I've got the moles of barium hydroxide but I want the moles of hydroxide. So I've got .0119 bowls of O H minus. So what's my total volume? So I can find the polarity? Well at the half equivalence point I have the 50 ml of barium hydroxide. I started with Plus half of the 59.3 ml that it took me to get to the equivalent point. Okay, we're not at the equivalence point, we're only halfway there. So that's a total of 79.65 mm. That's my total volume. So now my concentration of O. H minus is my malls divided by my total volume and leaders of course. And that will give me my concentration of my hydroxide, which is .149 moller. So if I do minus the log of that, I get the P O. H. So P O. H. Is 00.8 to six. And then if I subtract that from 14, I'll get my ph. So my ph pathway to the equivalence point Is 1317. It's still quite high because I still have pretty much all hydroxide there. Okay. And then the last step asked us, what's the ph at the equivalence point, any time we take a strong acid and react to strong base. The ph at the equivalence point Is going to be seven. So no calculation is necessary.

This question asks us to calculate the polarity of each of these solutions, and here have written the salt of the amount of salt that's been dissolved. And, uh, they also give us the volumes of each of the solution so we can figure out the polarity. So let's start with the 1st 1 We have 5.623 grams of N a. H C 03 or sodium bicarbonate. First, you need to convert this two moles because for all these will need number of moles over number of leaders. So one more, uh, and H CEO three is about a four grams. They need to convert this, um, or divide this by the number of leaders and tells us it's 250 milliliters solution. So I'm gonna put one over 250 mils. It's the same thing is dividing by that lumber mills, but polarity is in mulls over leaders, not moles over milliliters. Real quickly. Convert this to leaders putting the conversion between milliliters and leaders. 1000 no leaders equals one leader. All other units should cancel grams grams. Oh, I'm sorry. We should be left with moles. Uh, Mills Mills we have now. Moles over leaders should what we want and we get answer of 0.26 77 Moeller this dismal back to 677 Mueller for the next one, we have 184.6 milligrams of K to see our 207 or potassium die crow mate again, We're gonna convert this to Moulds, but this is an milligrams. Normally, the molecular weights of compounds were listed in grams. So we're gonna do that conversion real quick. One grab. It's equal to 1000 milligrams. Now. We can, um, divided by the molar mass. One more of potassium die. Crow mate is equal to two large compound 294. About 294 0.2 grams were divided by the volume multiplied by one over the volume tell it's a 500 mil solution. But just like she write, this was part, eh? The and see in part A. We had to tow convert back to leaders will do the same here. If we cancel our units, we should be left with moles over leaders Milligrams cancels, grams cancels and middle leaders cancels great. And this is 1.2 55 times. 10 The negative third. But where great and for the last one, it tells us a little more. Complicated. 10.1025 grams of copper metals dissolved in 35 milliliters of concentrated H and a three to form See you two plus ions than water is added to make the total volume 200 mils Avon's as Catholic similarity for See You two. Plus. That's just a lot of extra words in this problem that gives us the exact same problem. This is the massive copper volume is 200 and we could just figure out the polarity first. Convert to moans one more of copper is 63.54 grams. Divide by the volume, says says 200 mils milliliters and then we to convert that leaders. 1000 middle leaders her leader. Grams cancel. Mill leaders cancel. We're left with moles per leader, which is what we want. And this is eight point zero six five times 10 negative. Third, Moeller

For this question, we will be figuring out the polarity of solutions made of solid amounts of certain substances. Let's start with the first substance. We want a 250 milliliters. Solution made a 5.6 to 3 grams of NHC. Oh, free Now remember that mill Arat e? They're small over leaders. So we need to find how many leaders we have and how many moles we have. Start with. Moles already have grams. We know grams can convert two moles. Let's do that. We start with our 5.6 to 3 grams of any H CEO through multiply by the molar mass, which we confined by calculating to calculate you just add thesis odium the high shin carbon, the oxygen Matthews together. Now that Mueller mess, we know it is for everyone. Mole, there are 84.7 grips That's going to give us 0.669348 moles of this sodium bicarbonate that's gonna go in this portion over formula for polarities now to find the leaders. Fortunately, we already have our mill leaders. We just have to convert from milliliters into leaders. So 250 mills. I was going to be for everyone later. There are 1000 milliliters. That's our conversion factor and that will be 0.25 zero leaders. No, we're going to plug everything into our M equals Moldova leader formula. So the moles have 0.669348 moles over 0.250 leaders. She's going to have a similarity of 0.2677 molder. Now, this next part wants us to find the polarity of a 500 militar solution made with 184.6 milligrams of K two c r 07 I do want to highlight that This is milligrams and not grams. So we're going out to convert it into grams so it can do milligrams. Two grams, two moles. Okay, so 184.6 milligrams and everyone Graham has 1000 milligrams, cancel other milligrams. And now we see that we have 0.184 six grams of the substance. Now we can convert to mold. So, Moller mess, we have 294 0.1 for five grams and everyone No. Notice how many grams we have here to Philip One mole and here we don't even have a single gram. This is going to be a very small polarity at the end. That's going to give us a value of 0.6 to 7 two and so on, which is about six 0.272 times 10 to the negative. That put in scientific notation. You count back, Skip. 123 fours Faces back. Needed for now. When you defined how many leaders we have, we know we have 500 milliliters. So we know for every leader, we have 1000 milliliters. So this is 0.500. Uh, sorry, leaders. No, we will calculate moles. Over the years, we have our 6.272 times 10 to the negative four moles over 0.5 year their leaders. That's gonna equal very small polarity. It comes out to zero point 001255 which, when you convert to scientific notation, is roughly 1.255 times 10 to the native three. That is our polarity. Now this last bit is different. We need to find similarity of this money. Grams copper, solid copper metal that was dissolved first in 18 03 solution and then diluted with the H 20 Let's start with the grams for since that's pretty standard grams to moles. Okay, so that means we'll do 0.10 to 5 Kreme's couple two plus multiply by the molar Mass. So for everyone mold copter, it is full of 63.55 grams of copper. There's going to give us value of 0.16 12903 A very small number of moles. Now for the volume, we have to do two things. First of all, it's a little leader, so we'll have to convert to leaders. But also there's two volumes because you're are putting the cover metal into the 35 milliliters of H and 03 to dissolve it. Then you take the whole 35 milliliters of the Agent 03 and add it to the 200 milliliters of water. So 200 milliliters plus 35 millimeters with a total of 235 millimeters. So that's our total volume. We're gonna be plugging into the leaders. So now 235 million years quivered until leaders. So for everyone, leader, it's 1000 milliliters. We have a final volume of 0.23 five liters Now to do our final calculation of the polarity. Take our 0.16 to 903 Moles are copper divide by our leaders, which is 0.235 That gives us a small mill. Araji. It will end up being zero point zero for six. Moller. That's our solution.

Chapter 15 problems 16 tells us about a nitric acid solution that has a peach of 2.7 and asks us to determine a few different things about So we have our pH of 2.7. So a is asking us to determine the concentration of hydrogen ions. So remember that we can convert from P H two concentration of ions using the Formula 10 to the power of negative peach. So that's going to be 10 to the power of negative 2.7, which gives us a concentration of two times 10 to the negative three. Moeller now be asks us for the concentration of hydroxide ions, and there's a couple of ways to go about it. Remember that if we do one times 10 to the negative 14 divided by concentration of hydrogen ions will get the concentration of hydroxide. There's another way to solve it, though, and we can do that by determining that p o. H equals 14 minus pH, which in this case would be 14 minus 2.7, which gives us a value of 11.3. Then we could do the same processes, calculating the I H ion concentration where we have 10 to the power of negative P O H, which is 10 to the power of negative 11.3, which is equal to five tens, tend to the negative 12 more. All right now, problems See tells us that he's determined the number of moles of H I know three that were required to prepare a 5.5 leader solution. So notice that in H and 03 each bowl of a Jenna three has one mole of hydrogen that is the mole of hydrogen ions. We calculated earlier. He's the same polarity has the nitric acid here, so we know that nitric acid has a two tens tend to the negative three concentration. So how many moles have been required for 5.5? Leaders will remember that mill Arat E is the same as moles per leader. So if we dio polarity times the number of leaders that will get us two moles, which is what we're looking for, So that's two times 10 to the negative three times that 5.5 leaders, and we find that that gives us 0.11 Smalls, now D is asking us for the mass of this acid, and this is pretty straightforward. We'll just use the moulder mass here. So 0.11 moles and will use the molar mass here where one mole is equivalent to 63 grams and we'll see that. That gives us 0.693 grams of our acid. Okay, Now, the last question is asking this to determine how many milliliters of a concentrated solution will we needed to give us this many moles. So for e of the reminder, we're looking for 0.11 moles of our acid. So of R H a No. Three. And what we have available to us is a 69.5% solution that has a density of 1.42 grams per mil. So that's her concentrated solution. How many milliliters of that will get us 0.11 molds. So the way that we do this is all we have is this percent by mass and the density of from there we can get how many grams per mil of asset that we have. It's just that much per cent of the 1.4 till once we have grams per mille. We can convert that to malaria t two moles per liter. Then we can use our dilution formula to go from are concentrated acid toward delete passage. So let's go ahead and start that process. So first we need to determine the density of just the nitric acid. So our density of 1.42 grams per mil well, multiply it by this percent or 0.695 and we'll find that the density of Onley the acid in this solution here is 0.9 869 grams per milk. Now we can convert this to polarities, so malaria's moles per leader. So let's convert the leaders part first. So we have 1000 mills is equivalent to one leader and then to go from grams to moles will use the molar mass where again when wool is equivalent to 63 crams and this tells us that we have a polarity here in are concentrated solution of 15.7 Moeller. Now we can convert this to the number of milliliters that we need by using M one V one equals and to me too. So what? We're looking for here is V one, the milliliters of concentrated acid. So we can rewrite vests as V one is equivalent Teoh m two V 2/1. So here to interview to our dilute solution. And remember, we had previously calculated molar ity of our hydrogen ions to be two times 10 to the negative. Three more. And we have a 5.5 leader solution and the concentration of our very daily concentrated acid here was 15.7 more. So calculating this out, we get 0.7 leaders, which is equal to 0.7 milliliters of our constant.


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