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16. Give the numerical (a) 3P, (2) 2s, (c) 4f values of n and | corresponding to each of the following orbital designations:17. Sketch the shape and orientation of...

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16. Give the numerical (a) 3P, (2) 2s, (c) 4f values of n and | corresponding to each of the following orbital designations:17. Sketch the shape and orientation of the following types of orbitals: (a) Px , (6) 2sfor each of the following electron configurations: 18. Indicate the number of unpaired electrons (a) 1s* 2s2 (6) 1s* 2s2 2p' (c) [Ar]4s' 3ds each of the following electron configuratior 19. Identify the 'specific . element that corresponding to (a) 1s?2s1 () 1s' 2s*

16. Give the numerical (a) 3P, (2) 2s, (c) 4f values of n and | corresponding to each of the following orbital designations: 17. Sketch the shape and orientation of the following types of orbitals: (a) Px , (6) 2s for each of the following electron configurations: 18. Indicate the number of unpaired electrons (a) 1s* 2s2 (6) 1s* 2s2 2p' (c) [Ar]4s' 3ds each of the following electron configuratior 19. Identify the 'specific . element that corresponding to (a) 1s?2s1 () 1s' 2s* 2p* (c) [Ar]4s' 3ds magnesium hydroxide is dis- A 0.5895-g sample of impure solution The excess acid solved in 100.0 mL of 0.2050 M HCI '0.1020_ M NaOH for neutralization. then needs 19.85 mL of = by mass of magnesium hydroxide Calculate the pereningehar {as the only substance reacting in the sample, assuming with the HCI solution:



Answers

A 50.00 -mL sample of solution containing $\mathrm{Fe}^{2+}$ ions is titrated with a 0.0216 $\mathrm{M} \mathrm{KMnO}_{4}$ solution. It required 20.62 $\mathrm{mL}$ of $\mathrm{KMnO}_{4}$ solution to oxidize all the $\mathrm{Fe}^{2+}$ ions to $\mathrm{Fe}^{3+}$ ions by the reaction
$$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \stackrel{\text { Acidic }}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q) \text{(Unbalanced)} $$
a. What was the concentration of $\mathrm{Fe}^{2+}$ ions in the sample solution?
b. What volume of 0.0150$M \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution would it take to do the same titration? The reaction is
$$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Cr}^{3+}(a q) +\mathrm{Fe}^{3+}(a q) \text {(Unbalanced)} $$

This problem asks us to do some analysis of a tight tray Shin of is losing containing the iron to I on and with it solution containing a permanganate ion. Here's the information we're given. We have a 50 of the iron two. Plus, we have 50.0 mill leaders and we're tight trading it. And we've got K aumento four with a zero 0.216 pullout polarity Page Epps I can see the rest of problem and it took 20.62 mL of this toe oxidize all the FB to to MP three. And here is our unbalanced chemical equation. This is conducted. This is an acidic solution, and we're going to balance this chemical equation first. We're going to balance this. Um, I'm gonna write that down here, bounce the equation. They were going to get the concentration of iron two plus ions, and then we're gonna do another Titra ation to see how we would do this with a see or to Die Chrome eight solution. Okay, so let's begin. So we're basically doing the same reaction with two different solutions. So first, let's concentrate on this. So I have an M N o for and mn two plus. So I'm going to have to add four aged rose to this site. Eight h plus is to this side, and that will take five electrons. As you've done more of these, this is one of those you just get to know. Okay, five and next. We've got our we'll switch colors for the second one. So there we are on this, and we're going to have Teoh. Likely we're just going to multiply this equation by five. So this will be five, five and five all the way across there. Let's add these together and we get eight h plus, which is a quiz permanganate, which is a quis by a five iron twos, which is our Equus, Equus, five f e three plus. And for waters, which, of course are liquid. Okay, there's this equation which we just need so we can get our We've done that. Now we're going to do our second part of this. Um, we're going Teoh figure out polarity equals bulls per leader. I know. My I can figure out my leaders. I've just got to figure out my moles, so I think it really raised this piece and I'm figuring out my molar ity of what? Here? Iron. So I'm gonna have to figure out. Since I know this information, we can figure out that I have zero point 0 to 62 leaders of my permanganate, and its concentration is 0.0 216 moles per leader. And then I can use my mole ratio of my f e to my permitting an eight to get my moles of iron to. And that will be four 0.46 times 10 to the minus two moller F E two plus Let me go to the next day it Paige. Oh, that was moles. My bad after will fix it on this page to in case I did that. Yep. It's not Moeller. It's moles. Okay? And molar ity is moles per leader. So that's going to be four point. Did I do that wrong? Yes, I did. I I typed on a wrong answer here. I got ahead of myself. This is not the correct equation or that's correct equation. But I didn't write down the right thing. I wrote down her second answer. There we go to point to three times 10 to the minus three. So if you're mixed up, you're right. That was me. And 2.23 times 10 to the minus three. 2.2, three times 10 to the minus three Moles of iron per our Miller leaders were 50. So with that 0.500 leaders, that equals there's our answer 4.46 times 10 to the minus two moller F B two plus And that's our answer for the first part of this problem. The second part of this problem has its using the exact same equation as our first. I think I'll go to the next page just to make it all fit. So I'm just going to copy this down from the first page epi F e two plus plus and Effie three plus e minus in the second one. Ah, the second equation for this. We were given the dye chrome eight and CR two. Plus, I'm going to go back to our first page. We're doing this right now, and I'm going to see if they gave us that equation because I might be getting a little ahead of myself. Yes, I'm going to write this equation down in the bottom right here. Then we'll work a little bit backwards. It is C R. Two 07 There we go. So that's what we're doing right here. And so what? We're going to dio I've given you the basics here. We need to add seven waters to this side to get my owes. It means I have 14. H plus is on this site. And if I have 14 age plus is I'm also going to need six electrons. 6 14 1 I needed to Right there. There we go. So now I need to multiply this equation by six giving me 66 and six. So I end up with the following 14 h plus six F p two plus CR 207 Tu minus six f e three plus to see our two plus and seven waters. Now we are going to find the Let me go see what it says, what we're doing here Now it says, What volume? So that'll be in mill leaders. We'll probably find leaders first of 0.150 Moeller que two cr 207 for the same tight rations. So we're doing the same titrate Asian. We've just got a different tie Trahant on here. So that means I'm going to have Let's figure out my molar ity of the iron. So I have 50.0 mill leaders and that is 0.500 leaders of iron. We figured from her last problem that we had similarity of 4.46 times 10 to the minus two moles per leader. And we have a 1 to 6. And that is my CR 207 and my FB two plus. But I do that I get Oh, I got one more step here that gives me to malls. And then I have 0.150 bulls per leader. Okay, there we go. And that when they punched this into a calculator, I get 0.2 for eight leaders equals 24.8 mill leaders. You we did it. These durn redox take a little time, but we did it

This question. We've been actually Copley, the minority of each of the following solutions. So in order to Coakley, the polarity is important to note exactly what the melodious to the polarity of a solution is given by the moon to ready by the volume. So for each of these solutions, we just need to find out what the Moelis and divide by the bowling. So for the first question, this is relatively easy because we've been given both quantities, the more we've been given this 0.4 44 well, we just have to divide this by the bowling with a solution, which is my given as zero point 654 leaders to this thin results in similarity off their point 679 Moeller. That's similarity of a solution. We could do the same thing for Part B, but important be we have to now calculate the mole because we only being given the mass. So, like I stated previously, the morality yes, going by the more divided by the volume and for the more that is given by the massed divided by the molar Mass. So given the masters of the masses, 98.0, Gramps, you can divide this by the moon and Mars off. Let's wreak acid. And the mill Imus A calculated was 97 point 994 grams per more. So that would give us the mall. You're the numerator in this for Miller and I would previously more patrolling. Now all we have to do is divide this quantity. But they've only which is 1.0 leaders. This one sold should give a value approximately 1.0 zero Moeller. So current forward for part C. We're going to do the same thing as we differ. Pride be because I've been given a mass and we have to calculate the we're already here. So the polarity, like I stated, which I'm going to just represented by Capital M for part C, it is The mole is the moon of William, right? So the move is the mass directed by the more mass here the mass has been given, which is 7.2074 gramps. But we want the more for the numerator like we've been doing in past examples. So we derided by the molar mass of causing my drug side, the mullahs most of calcium hydroxide when I calculated that was 74.92 So it's still only 4.0 92 This obviously asked units grams per more. We couldn't divide this by the volume the volunteers for the familiars. We can cover that two leaders because it's the polarity issues that given units off more per liter. So 40 millimeters is this, um zero. All right, well, looks like I can write that underneath there, but 40 milliliters iss Well, go ahead by zero point zero for zero leaders. So this would result in a concentration wolf. Sorry. Yeah, concentration. Which is the polarity off zero point zero 69 nine eight cooler. So we just do similar thing for part D. E and F here. I'm just going to speed up a little bit. So for party the polarity, which I'm gonna just indicated by m here, I couldn't give him by, like, I've been doing more prevalent. And the move here have been given the mass. We have to find the mole a mass. So the masses one, there's 10 point five kilograms to cover that two grams. That's just turns into the part three grams. We divide this by the well, a mass for sodium sulfite deca hydrate that Mullah must went calculated. I go to Valley off 3 to 2.186 three, right, and I was just divide by the bowling. The ruling be given his 18.6 leaders. It's when sold. This gives the value 1.75 more in Part E. This is really simple because we've been given, the more I have been given the volume but head of loneliness and familiars yourself to cover that's leaders. The melody is the moon just 7.0 times tens of the poor, minus three? No whether by volume, 100 milliliters, sis 3.1 leaders. So this essentially uses of early a zero point They're 70 Moeller. And lastly, 1.8 times since of the bar four mainly graham off each seal in 0.75 liters of solution. So one point, sometimes a powerful milligram is 18 grams. Right? So, um, polarity we're going to be 18 grams. Do right by the mullahs. Mask off. Easier because we need to calculate the malls. The movements of a Chino is 36 point for 58 Graham Spoon, you divide this by the volume just zero point 075 leaders who insult this gives us the value off 6.5 neat war. I always feel this is how you Coakley similarity when given you the mass or the mole and the volume as well.

Here we are identifying each of the unbalanced reactions as either Precipitation Asset Base or Redux things First. Example. We have part A. That is a reduction reaction because we have the transfer of electrons. B. This is an acid base because we have the presence of the acids and bases within the chemical reactivity. So part C. There's an oxidation reduction reaction. You have the transfer of electrons party. This is an acid base reaction. E can take a look at next. So this is a precipitation reaction. So when we have a precipitation reaction, we're forming a solid product. And then f is also a precipitation reaction because we're forming a solid product. Next, we're looking at G, so this is an acid base Reaction. H is a precipitation reaction again with forming a solid product, and I is an oxidation reduction reactions


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