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In the illustration N is color-coded blue andcolor-cotedWhat kind ofchange represcnted by this illustrution?enetgy absorbed O1 rclenyed when this chunge oecurs...

Question

In the illustration N is color-coded blue andcolor-cotedWhat kind ofchange represcnted by this illustrution?enetgy absorbed O1 rclenyed when this chunge oecurs

In the illustration N is color-coded blue and color-coted What kind ofchange represcnted by this illustrution? enetgy absorbed O1 rclenyed when this chunge oecurs



Answers

Here, the blue coloured complex $(\mathrm{E})$ is (a) $\mathrm{NaFc}\left[\mathrm{Fc}(\mathrm{CN})_{6}\right]$ (b) $\mathrm{Fc}_{1}\left[\mathrm{Fc}(\mathrm{CN})_{6}\right]_{3}$ (c) $\mathrm{Co} \mathrm{Al}_{-} \mathrm{O}_{4}$ (d) $\mathrm{KFe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

The problem for 2151 of chapter seven, We're going to be e uh, doing this question One divisible lies ahead. Ramesh emission spectrum correspond to be an equal six toe n equals two electron transition will coalesce the transition that we're gonna use the rid burger equation. So Lamda is equal to river constant times the difference between in one over end, two squared, minus one over and one square. And that's the reciprocal. So we're gonna do it. Landa is gonna be equal to the writ, Burke. Constant consent, which is going to be the 1.1 times 10 to the 0.1 times 10 to the seventh inverse meters times, times one over two squared nice one over six squared and then in verse. So Landa is going to equal 404 points there. Nine kinds tend to the negative seventh and meters. We're just gonna equal 409 nana meters and so that's going to be in the violet range of the electromagnetic spectrum.

For this problem. We're told that the visible lines of hydrogen emission spectrum correspond to the n equals six, um, to and equals two transition. I'm trying to figure out what color light, um, is going to be produced from that. So the first thing we're gonna do we need to use our equation here where wavelength equals H c over guilty e and for Delta E because we're, um are going to be not necessarily going from an equals one we're going to use our delta is 2.178 times 10. Then I give 18th times one over the final, um, position squared minus one over the initial squared. Um, so go ahead and plugging these values in, we will have that are wavelength is going to equal 6.626 times, 10 to the negative, 34th times, three 0.0 times. 10 to the 10th. Because we've changed this from meters two centimeters. Eso that our wavelength is going to be in centimetres, um, divided by 2.178 times 10 to the negative 18th times one divided by our final position is going to be, and to so two squared, which is four minus one divided by our initial, which is six and six graders, 36. And when we finished solving that, we get that our wavelength is going to be 4.11 times 10 to the negative fifth centimeters, which, when we look at our spectrum, um, for the different colors of wavelength based upon centimeter, we will get that that is violet in color.

Among the given options, among the given options in this problem, among the given options in this problem, DIF six or 2 -2 TIF 6 to minus and see you, too. Sierra two added. The out of the colorless at the colorless spaces are particularly spaces, so according to the absence year option B. H. Correct answer.

In this question, which used on a graphing calculator and here and, uh, I n is defined as a aid in the riding Brian and plus one. So look in the craft, I would say, is excitable sponsored craft in a finger be


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