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Jmountnickel in some industilal waste Muld Jnatytical chemist adds the @luid and 170 M sodlumn hydroxide (NaOH) solution cllocts the solid nickel(II) hydroxlde 20.0...

Question

Jmountnickel in some industilal waste Muld Jnatytical chemist adds the @luid and 170 M sodlumn hydroxide (NaOH) solution cllocts the solid nickel(II) hydroxlde 20.0 (N(OH) ,) praxluct Wnch mafn M(OH) ' earnpla of nucuced fnce% wasns 386 mg has becn produced. and weighs _ Ind; thatTho balanced chemical equation for the reaction Is; (aq) 2NaOH(aq) N(OH) , (s) ZNa ` (aq)WnablonoTaactionthls?0pucipitatien AcdebaseredorYou sid thla was = prcdpltatiou reaction , McntateMormulaIf You sald this *a

Jmount nickel in some industilal waste Muld Jnatytical chemist adds the @luid and 170 M sodlumn hydroxide (NaOH) solution cllocts the solid nickel(II) hydroxlde 20.0 (N(OH) ,) praxluct Wnch mafn M(OH) ' earnpla of nucuced fnce% wasns 386 mg has becn produced. and weighs _ Ind; that Tho balanced chemical equation for the reaction Is; (aq) 2NaOH(aq) N(OH) , (s) ZNa ` (aq) Wnablono Taaction thls? 0pucipitatien Acdebase redor You sid thla was = prcdpltatiou reaction , Mcntate Mormula If You sald this *as atid-bata [email protected] nactanttnnt Cnie acuno tne biee chen I rou aid Ul tut b oxdted reuott Donon; ta che Calquate the 01.355 percont- cf Ni In the Honllicant dnt Nample



Answers

Element $M$ is prepared industrially by a two-step procedure according to the following (unbalanced) equations:
$$\begin{array}{l}{\text { (1) } \mathrm{M}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MCl}_{3}(l)+\mathrm{CO}(g)} \\ {\text { (2) } \mathrm{MCl}_{3}(l)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{M}(s)+\mathrm{HCl}(g)}\end{array}$$
Assume that 0.855 g of $\mathrm{M}_{2} \mathrm{O}_{3}$ is submitted to the reaction sequence. When the HCl produced in step (2) is dissolved in water and titrated with $0.511 \mathrm{M} \mathrm{NaOH}, 144.2 \mathrm{mL}$ of the NaOH solution is required to neutralize the HCl.
(a) Balance both equations.
(b) What is the atomic mass of element $M,$ and what is its identity?
(c) What mass of M in grams is produced in the reaction?

High River. So in this question, they asked. That is what volume of bms in, um el is required. Okay, having more clarity. 0.0 point 0703 mola. Okay, uh, to press protect and I two plus, I am that is having modularity by in 103 Moller and Volume 10 ml. So, first of all, read the given data. Yeah. Do you want to? So what they've given So they were given m one? Yeah. 0.703 mola then we're even. They ask then. Yeah. Me too. 0.103 Moller. Then we too. Yeah. Okay. Animal. Yeah. No, the reaction is okay. The reaction is mhm. Yeah, Mhm. And I two plus Achilleas. Yes, plus to be MZ use mhm and nine d m Z toys. No, we know that, BMC, you are into a 92 plus. Let us imagine. Um one. We went with the modularity and volume of D. M. C and M. 2 ft to be the malaria and volume appendicitis class. So your morning is no. 0.703 into the one corresponds to, um to is 0.103 into three to its animal. Yeah, mhm. Therefore, we won physical to 0.103 into 10, divided by point, not 703 level. Therefore we run is equal to 1.3 Mhm. Do I dare by 0.703? Yeah. So therefore we one is equal to Yeah, 14 point 64 year marriage. So since we even is called to 14 when 65 for men, this volume corresponds to mhm corresponds two one more love dems and therefore the one x m l. All right corresponds to because mhm two more love BMC and therefore one limb off dems bracket We even that is he called 22 and two We won nowadays 20 to 14.65 and therefore we can say women batch is requiring too 29 point three See it all 23 Mhm And then thanks a lot

This is the answer to chapter 20 to problem number 47 fromthe Smith Organic chemistry textbook. Ah, this problem asks us to draw the product formed when fennel acetic acid is treated with the tree agent. And we're told that with summary agents, no reaction occurs. Ah, and there are 12 of these. So I'm just gonna go ahead and get started because this is gonna take a little bit of time. Okay. S o for a Ah, we have sodium bicarbonate. Uh, that's gonna act as base. Um, and we have a car back so weak acid that we're starting with. Oh, and I drew fennel acetic acid at the top of the screen here. Rather than redraw the aromatic ring 12 times, I'm just gonna use a pH to abbreviate it. Since the ring itself is not participating in any of these reactions. Okay, s O sodium bicarb here is gonna act as a base. And so we're gonna get an asset base reaction and will de protein, eight hour carb oxalic acid. Ah, and we'll get the sodium ion as the counter I on here. Okay, s so that's a for be, um, again be Is a base. Ah, and so the exact same thing is gonna happen as in a we will de protein eight. Ah, and we'll have the sodium ion as our counter ion. Okay. Ah, see, we're using signing chloride. Remember, signing chloride, um, takes a carb oxalic acid to the acid chlorides, and so we'll have an acid chloride for a product there. And C D is just salt table salt. There's gonna be no reaction here. Okay, s so then e um and actually, let me move these up a little bit. Okay? So in e our using one equivalent of ammonia. Ah, and so that is going to just act as a base. And so again, we're gonna get D prone nation. Ah, and no further reaction. So Deep Throat Nation, and then we'll have the ammonia my eye on as the counter ion. Okay. Ah, for F. We're using ammonia, followed by heat. Um, and so that is actually going to drive this reaction. And so, rather than just a pro nation here, we're actually going to make the A mite. Okay, so there we go. There's f moving on to the next page here. Fergie were using methanol and then sulfuric acid. Um, and so we're going to get the Esther product here on DSO Since we're using methanol will get the methyl ester for H. We're gonna try to do the same thing, but instead of using acid, we're gonna use base. Um, and so that's not going to work. Ah, and so we will end up just deep resonating. Okay. For I AA were using sodium hydroxide, um, followed by on acid chloride. And so, uh, this is going to create the mixed and hydride. So we d pro Nate. Ah, and then our deep throat nated oxygen is able to act as nuclear file Ah, and kick out the chlorine. Um, in the acid chloride that were using is the second re agent in this sequence. And so we end up making the next anhydride. So then, for j r. Using Mesilla mean ah, in D c c. And remember D c. C. Is gonna act as a dehydrating agent here, and so that is going to allow us to form the A mine. Instance for using meth will mean our product is going to be metal. A my okay for kay, we're gonna use. Sign your chloride. Um, and then we're gonna use problem meid. Um and so the final chloride is gonna make the acid chloride. Um, And then, uh, the pro chloride is going to, um, add to that. And so we end up with the A mine. So 123 Okay. Ah, and s o l is going to be similar. Um, but we're not using a mean, so we're not gonna get in a mine. Ah, we're using an alcohol. And so we'll get an Esther. So the Esther will be the Pro Bowl. Esther. Okay, so there we go. Um, yeah. So that's all of the products to these reactions. Ah, and that's the answer to chapter 20 to problem number 47.

Calculate several thermodynamic properties of the synthesis of nickel tetro Carbondale. So the first thing we're gonna do is calculate our NDP of reaction. So to do this, we're going Thio, subtract our reactions from our products using our standard enter a piece of formation. So starting with our products for our nick for a nickel touch a carbon eel were given in the problem a Delta H of formation of negative 633.0 kill jewels for more for our reactions, here we have that are nickel solid is going to be 0 kg promote because it's solid metal. We're gonna add that to four times our carbon monoxide, which is going to be negative. 100 0.5 killed goals per mole. This gives us adults h standard value of negative 191 killed jewels. Permal. So we're going to calculate our delta s value in a very similar way. We're going to use our standard Delta s is for each of these molecules products minus reactions. So we're given the problem that our standard delta us for nickel tetra carbon eel is 313.4 jewels promote Calvin subtract that from our reactant here, where we have for nickel 29.9 joules per mole Kelvin plus are four times our carbon monoxide, which is going to be 197.6. Tools for more Kelvin's. This gives us a delta as standard value of negative 503. Excuse me, 506.9. Jules, Permal, Calvin or negative 0.569 Kill a Jules Permal. Okay, so now that we have both of these values, we can go ahead and calculate our adult Aggie. We're going to use the equation here that are adult EGY. Standard is equal to our adult h standard minus T Delta Standard. Thank God. And plug in some numbers. Here we know the negative. 191 killed jewels per mole. We're running at a temperature of 298 Kelvin. So we can do that for a standard temperature, and then our entropy is going to be 0.569 killed. Jewels promote Calvin. Okay, this is going to give us a delta G standard value of negative 39 0.9 kill jewels criminal. Okay, So now that we have this value, we can do a couple different things. First, we're going to calculate the temperature. I wish the three action becomes non spontaneous. So in a reaction becomes non spontaneous. That means the delta Gee of the reaction will become positive. So we're going to set our delta Gee here, equal to zero, so we can see the exact moment when we go from a negative Delta G value to a positive Delta G value. So we have Zero is equal to R H standard minus t dealt us standard. We're going to go ahead here and solve for R T, which is going to be Delta H over our adult US standard, which is going to be negative. 191 Killer jewels Permal account for Permal Divided by my gosh, I went crazy. More time. Here we go. Divided by negative 0.569 Kill jewels. Criminal Calvin. Okay, this is going to give us a temperature value of 376.8 kelvin. So at 376.8 Calvin, this reaction will become non spontaneous. The last thing we can do here is try to calculate our delta G standard of formation. Four our nickel toucher Carbondale To do so, we're going to recall our equation here that are standard Delta G of reaction is equal to the sum of the delta G of formation of our products minus the sum of the delta Gee of formation of are reacting. So we already know where we can find out the Delta G of formation for our react mints. And we already know our delta g of reaction so we can solve for R Delta G of product here. So let's do some algebra and rearrange. So we have our Delta G formation for our nickel touch. Your carbon deal is equal. Thio our Delta G standard minus the delta G of formation of our nickel solid plus the Delta G of formation hoops forgot the GI plus the delta G of formation of our carbon monoxide. Let's go ahead and plug in some numbers. We know that our delta G standard value is equal to negative. 39 0.94 killed jewels Permal, subtract that from our react INTs here we know that the standard energy information of a solid metal is zero. So we're just focusing on our carbon monoxide here, and we know that is going to be four times negative. 137 point to kill a Jules Permal. So this will give us a delta G of formation value for our nickel tetra carbon eel equal to 97.3. Kill jewels, perm all.

Sodium borough hydride reacts with silver nitrate to give us silver and a few other products here. So three, either you sure we've been once, and then we can figure out how much silver weekend produce. Ah, depending on the given amount of silver nitrate and sodium or a hydrate. We start with 1.7 grams of silver nitrate. It's a seven there. You convert that to moles using the mouse, and there are two moles in the reaction. There are also two moles of silver on one mole of silver as equal to 107.9 grams. We multiply and divide there. We find that if this is how much silver nitrate we start with, we can get 0.68 grams of silver. If we start with 13 grams, we'll have any BH four. We get the mass of any BH four that is not quite right. Sodium is 35.5 gram or I'm sorry, 22.9, trying to 0.0 plus 10.8 plus or so you have 37 point e grounds, Sigulda. One more and we have three moles have any beach for in our reaction, and then we're still looking for silver. So the last part of our chart here will be same information. That's grams of a GI that's equivalent to one more. We multiplying divide there we get 24.7 grams of silver, so the most silver that we can produce with this information is 0.68 grams.


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