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The discovery of hafnium, element number $72,$ provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements $58-71 )$ compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnum comes from the Latin name for Copenhagen, Hafnia).(a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group 4 $\mathrm{B}$ , can be produced as a metal by reduction of solid $\mathrm{ZrCl}_{4}$ with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation-reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, $\mathrm{ZrO}_{2},$ reacts with chlorine gas in the presence of carbon. The products of the reaction are $Z r \mathrm{Cl}_{4}$ and two gases, $\mathrm{CO}_{2}$ and $\mathrm{CO}$ in the ratio $1 : 2 .$ Write a balanced chemical equation for the reaction. Starting with a $55.4-\mathrm{g}$ sample of $\mathrm{ZrO}_{2},$ calculate the mass of $\mathrm{ZrCl}_{4}$ formed, assuming that $Z r O_{2}$ is the limiting reagent and assuming 100$\%$ yield. (d) Using their electron configurations, account for the fact that $\mathrm{Zr}$ and $\mathrm{Hf}$ form chlorides $\mathrm{MCl}_{4}$ and oxides $\mathrm{MO}_{2}$

In this problem, I'm writing the reaction. Just look at it carefully. Any energy for at be your ford four H two will react to form any and at four at people for plus four as to and and ME energy for add bill for will react to form and the beauty plus an STD plus as to So according to the option. In this problem option. See each correct. Was there not that we have any Beauty which reacts with metallic oxide to give colored or to post fails.

This question was sort of interesting in this question. Were given the following information. The discovery of half million, which is element number 72 was a controversial time in chemistry, A French chemist, I think his name is pronounced urbane claimed in 1911 to have isolated 72 from a sample of rare earths That were in the 58- 71 compounds. Niels Bohr predicted that half the um was more likely to be found over by zirconium. A couple of scientists coster and I think it's Hennessy working board working in boards laboratory in Copenhagen showed that element 72 was present in a sample of Norwegian Zircon, which is an ore of zirconium. And the name half name comes from the name fur coat. Copenhagen half mia. How could we use question a says how could we use electron configurations to confirm Poor's thoughts on this whole thing? Well, first of all, let's write what we know today. The electron configuration would be for first we'll do zirconium and that would be crypt on. Yeah. and then five years to Yeah, four D two. You can see those written either way. I prefer to draw on this place. So then let's look at half the um which would be Xena then success too. 40 14 five D two. So the D two will highlight these. The S- two D 2 similarity is proof enough that those would be similar, but those behaviors would be likely in the same group and we know from the periodic table. They are indeed in the same group here. I'll ask my question then. Okay in question be let me see if I have enough room for me here. I think I do shit. Uh huh. Game. Okay. In question be we are asked to write a chemical equation for the reaction between meconium can be produced by reduction of solid we call him chloride, tetrachloride and molten sodium. And to balance this I'll have to go like that and I think I'm good. This is a solid and this is that's an R. That is also a solid. This was molten my bad there. I just noticed that okay, there we go. So we wrote the balance coming from equation then were asked is this an oxidation reduction? And of course it is. The oxidation number for sodium is zero in here. It's one plus. The oxidation number is four Plus and zero. So we can see that this is being reduced and what color do they do that with sodium is being oxidized. That is deep question. C says solid. Zirconium oxide is reacted with chlorine gas in the presence of carbon and the products are zirconium, pork, chloride or meconium tetrachloride. I'm not sure what the best name for that one is. Plus co two gas and C. E. O. Gas. And these occur in a one 22 ratio. When I balance these I just Quick did a 1- two and I did one of these. I did a 1-2 like that, making this one half. So then I did a 3/2 for my carbon and two right here and that was a balanced. But then in order to make that fully balanced I'm gonna change this to a green. I got a two a four a three a two A one and the two for my coefficients. So those are my coefficients for the equation. Then we are given that we're starting with 54 55 .4 g sample of Zirconium oxide. I should have written that over here. 55.4 g sample of zirconium oxide. Yeah, Getting the molar masses for this. I had 1, 23, The save room. I'm gonna write 02 for both of those. My mole ratio, I forgot to tell you. Here's what we're finding which you would have known in a moment. And the boulder mass for the product was to 33.04 grams Z. RCO four. Personal, Doing the math on this and reporting to three significant figures. I had 105 grams CRCL four and last but not least part D. Using electron configurations which we already did predict or account for the fact That the chlorides are the metal seal four and the oxides are the metal oh two. And this is pretty simple here. I'll start with chlorine, which is the And well, I should just go ahead and do it. But this would be true for any chloride. I'm just gonna put the outermost three us to three P five and this is to s to two P four. So this one will accept two electrons and this one will basically, I'm using except in a term one electron. So if you want to think of it in terms of gaining or losing electrons, that's what that does. We already wrote the other two. Zirconium had a I'm just going to I guess I could do it. It wasn't that big. Krypton. I got to pull my periodic table up here. Krypton lives too four D two. So it wants um the tendency will be for those electrons to be donated or they're available to share. Which means it has four electrons from zirconium Would require four chlorine And it would require two oxides. Should my negatives on that side, I shouldn't. What was I doing? I had it right okay. And then have them. These are the two electrons right here, and these two Also four electrons are available, Which would require four. c or to oxygen's oxides. And I think we have answered everything in that question.

In this question were given 53 data points for the price of a box of macaroni and cheese. And the woman who collected the data wants to test whether the average price of a box is at most $1. So we can write down what she thinks, as the no hypothesis we can say that is the mean is less than or equal to $1. So an alternative that we would test is that the mean is greater than a dollar. So she has taken a sample from grocery stores and samples of Size 53. And if you collect all these data points that are given in the question and put them in this mathematical spreadsheet or a calculator, you can find that the sample average his 1.35 dollars. The sample standard deviation is 0.747 So let's assume a significance level for this question of 0.5 The sample averages are distributed according to the T distribution here, because we do not have the population standard deviation. We only have the sample sample standard deviation that was calculated here so we can calculate t our test statistic. And if you want to see the details of the calculation, you can pause the video right here, moving on if t is equal to zero point 341 and so you can see that if we plot 0.341 here, RPI value will be it will be in the upper tail because this is an upper tail test. So the T value is all the area in red here. And as you can see, this is going to be a big P value. We can be pretty sure that we're going to fail to reject the no hypothesis. So being a calculation or in a tea table, we can determine that the key value is equal to 0.368 This is based on our test statistic of 0.341 and so the P value is much bigger than Alfa, and therefore we failed to reject the no hypothesis and we can conclude by saying that there is insufficient evidence to conclude that the average cost of a box of macaroni and cheese is more than a dollar. So therefore the students hope that a box of macaroni and cheese still cost less than the dollar is reasonable


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