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MckcjzloCotege Phyulcs /Coure Homo(hontthont208 5Problem 13.25Pan AUhatna&ncud Deann Joaoda: can prpbtalra dma Frequaiy €l# Erton hah Even 4 Ia #7oiicn dv...

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MckcjzloCotege Phyulcs /Coure Homo(hontthont208 5Problem 13.25Pan AUhatna&ncud Deann Joaoda: can prpbtalra dma Frequaiy €l# Erton hah Even 4 Ia #7oiicn dvinticn #cnb tucm Dammett acoutlaton cithe Laim cn Be #rere Orcu'€ Iin cnlrtale 00u/6EEanurlvibuucua Mltampuit 4021Fiela ofAean fumr cle 6niIre Marinum acoulerauor Venions per scoord Exprets Youf enswar Lalng two #ignlficant riqurotSuaulJ0 qurAntatt 3 JuetintaerUncormect;Aqain; ? attaipls rmainingpan B0r5iJul entnttullelu JUr Ftieee

Mckcjzlo Cotege Phyulcs / Coure Homo (hontthont 208 5 Problem 13.25 Pan A Uhatna&ncud Deann Joaoda: can prpbtalra dma Frequaiy €l# Erton hah Even 4 Ia #7oiicn dvinticn #cnb tucm Dammett acoutlaton cithe Laim cn Be #rere Orcu'€ Iin cnlrtale 00u/6 EEanurlvibuucua Mlt ampuit 4021 Fiela of Aean fumr cle 6niIre Marinum acoulerauor Venions per scoord Exprets Youf enswar Lalng two #ignlficant riqurot Suaul J0 qurAntatt 3 Juetintaer Uncormect; Aqain; ? attaipls rmaining pan B 0r5iJul entnt tullelu JUr Ftieeet elenlleani nuuich 0nrer Ceten



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A ccrrain pale grocn subsrancc, (A) becomes dark brown on axkting $\mathrm{NaNO}_{2}$ in prcscncc of dil. $11, \mathrm{SO}_{4}$ irs aqucous solurion gives prccipitarcs wirh (1) $\mathrm{BaCl}_{2}$ and (II) $\mathrm{NaO}] 1$ The larrer NaO1 I in scparare resrs. The larter prccipitare (B), gradually changcs colour from grcen ro brown, on cxposure to air. Whar is A? (a) NiSO $_{1}$ (b) $\mathrm{FeSO}_{+}$ (c) CuSO $_{4}$ (d) $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

In this problem, just look at it carefully. BB I too, will react in pageants of BSE here, too. ECU was to give compound Are we change which aid 80 cl here, which it A d c L. Here? So this is white PPT. They said white PPT. Therefore, according to the option option, age, correct and said for this problem, A d c l a will be the white PPT in this problem.

So seven and 56 cancel out to eight. Why? To the fourth over 12 the third. We just have to subtract the explain and values of four and three. To get one was kind of like saying Why? To the one which is the same thing. That's just why again, I'm just going to subtract the five in the three to get Z to the second power and then 45 5 cancel out to nine. Why squared and why? It's kind of like I have an imaginary one there again. So two minus one is one, and then the Z Square, there's only on the bottom. So I'm just gonna rewrite that. At this point, you can see that this and this are the same. So you just have to work with the coefficients. So eight minus nine is negative. One who have negative one. Why Z squared

Is this reaction commentator or district? Later? And what does it stay? A chemistry off the product of it, each in turtles involved into this reaction, it responds toe that enter our image transition stick. To make it on a magic will need to introduce another Lord to Homer, which is achieved by the color tater process. So those two parts the erupted in the same direction. It will bring this Mitchell job up and business or group down. So that's trance relationship. Same reaction under photochemical conditions in this case, it electrons to respond to the on a magic transition state toe. Keep the number off notes. We need to rotate those professions in the upper the directions, which is just prepared to refresher. It will bring Mr Woodroof one up and missile group. Then up it responds to the cysts stood a chemical relationship

Part A. We're told that a 0.1450 g sample of a solid he'll a cane is combusted. An heir to give 0.5 or 63 g of carbon dioxide was the empirical formula. This hill a cane so we can figure out the the weight of Carbon Way have 0.5063 g of co, two times 12 g of carbon of 44 grams of CO. Two. This will yield 0.1381 grams of carbon, and we can calculate the, uh we'd of hydrogen in the combustion. Um, so we have played a total mouse is 1450 g. Um, of the, um killer Kane minus 3.1381 grounds of carbon. This will be equal to 0.0 69 g of hydrogen. So we have 0.1381 g of carbon. It's convert this two moles, so yield 0.115 moles carbon and 0069 g of hydrogen, 1 g per mole. This would be 0069 moles, hydrogen divided by 0069 on and seriously then we go zero 0.69 to give me one. And, uh, one 0.67 What kind of multiply this by three to give me five and three. So my empirical formula for hill a cane, like with the C 58 three? Yeah. For part B, I have a 0.938 g sample is dissolved in 12.5 g of solvent to give a majority of 0.175 Mueller, what is the molecular formula of this hill? A cane. So we can solve for the, um, molecular weights or the Mueller mouse, which would be equal to, um so we have the most pincer on some five. Mhm 0.0 938 g. Um, times. Um sorry. Mm. Times a look. Wait, sorry. Let me back up here. We're going to solve for the morality is equal to the weight of the salute. Comes 1000 times the Moller mouse times the wheat of the solvent. Rearranging this, we consult for the molar mass, just equal to, um 0.938 g times 1000 over and 0.0. Once open five more times, 12 0.5 g and This would be equal to 248 0.8 g per mole. If I go back here, three empirical mouse is going to be 63 grams per mole. So n is at Mueller Mass over empirical mass to 48.8 g per mole, divided by 63.0 g per mole, and we find that this is equal to close whole number closest whole number would before. So the molecular forming of ah, that's equal to four times C five h three, which is equal to C 20 each 12 and the balanced reaction for the combustion of hell a cane C 20 h 12 plus 23 +02 would produce 20 c 02 and six h 20


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