What Is The Enthalpy Change When 4.25 Kg Of Nitroglycerin Detonates?

HO 0) N (g) 02 (m)2285.82.) Calculate the enthalpy of reaction for the decomposition of nitroglycerin: Is the reaction exothermic Or endothermic?` ( (41'` `l Vpal " Sc7 |. " H 40 (IOM}4' 0 < 'grand 3.) What is the enthalpy change when 4.25 kg of nitroglycerin detonates?

HO 0) N (g) 02 (g) 2285.8 2.) Summate the enthalpy of reaction for the decomposition of nitroglycerin: Is the reaction exothermic Or endothermic? ` ( ( 41'` `l Vpal " Sc7 |. " H 40 (IOM}4' 0 < 'yard 3.) What is the enthalpy change when four.25 kg of nitroglycerin detonates?

The explosive nitroglycerin $\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{three} \mathrm{O}_{nine}\right)$ decomposes rapidly upon ignition or sudden impact co-ordinate to the balanced equation: $\begin{aligned} 4 \mathrm{C}_{3} \mathrm{H}_{five} \mathrm{~N}_{iii} \mathrm{O}_{9}(l) \longrightarrow & 12 \mathrm{CO}_{2}(\mathrm{~g})+x \mathrm{H}_{two} \mathrm{O}(g) \\ &+6 \mathrm{~Due north}_{ii}(\mathrm{~thou})+\mathrm{O}_{2}(g) \quad \Delta H_{\mathrm{rnn}}^{0}=-5678 \mathrm{~kJ} \end{aligned}$ Calculate the standard enthalpy of formation $\left(\Delta H_{\mathrm{f}}^{0}\right)$ for nitroglycerin.

No, he was just looking at Nitro glycerin. I have a reaction off it breaking downward Here we're full C 3 h five and 309 which is converted to 12. CEO to tonnage to work 6 and two on 02 R and B. Modify of reaction is negative. 5678 Killer jewels on wages Calculating our standard entropy off germination. So we need to exist referring to our standard and announced germination tabular array head. To summate this value, we add upwardly all of the and appear formation of our product values which are multiplied by our story geometric values. So this streak geometric values are the numbers that come up before each off the structures and elements upward in my equation, which is why we need to have it balanced. Having a balanced equation is vital in getting the right reply when these sorts off scenarios. So we don't have tracked the entropy of germination reaction values, which again they have been multiplied by their strike. Geometric values on this allows united states to generate a standard cease announced germination value off negative 366 kg joules per mole

So we're given in an equation for ah, the decomposition of nitro glycerin. We're told that it decomposes in the following products on. We're too told the empathy of that reaction. The question then is what is the in flippy of formacion when we want What's the entropy? Information of the nitroglycerin? Uh, that's what we want to effigy out. And so Ah, we practice have some data we can apply, so we know. Ah, only based on thean flippy of reaction standard entropy interaction that information technology'southward going to be equal to the in Philippi of formation standard atmospheric condition of the products minus entropy of formation of the reactant. Great. Then the reactions in this case, that's the nitroglycerin. This is the bit that I want to notice. That'south the number that I want. I'm gonna rearrange this equation slightly and you'll meet that nosotros get Temple P of formation of the reactant is equal to in puppy of formacion of the products minus the in Philippi of Formacion of the history of the entropy of the reaction. Okay, and so we tin can write this out a petty bit more. Ah, this will exist the, uh and Bill P of formation of the nitroglycerin, and I'chiliad going abbreviate that with n g times formals. That's the number of moles involved here, and that's equal to entropy. Information of the products. So, empathy, information of the products that is in Philippi of formation of the CO two, uh, times 12 moles, plus the envelop of formation of the water water vapour specifically times the x moles and so, ah, all the remainder of these. So the nitrogen, that doesn't affair. The oxygen. That doesn't matter because they're already in their standard state. Uh, and so then we just take to subtract off the impala p of the reaction. Ah, and that he is given to us. Okay, so let's take this. Ah, bottom equation. Rearrange it a bit. Ah, and I'g gonna go alee and divide both sides by before and so that we only have to worry most the united glycerin on the left hand side. So let'southward go ahead and do that. So, in Philippi of react, her sorry formation of the nitroglycerin is going to equal. So we've got the co two, um, go and do that. So that's negative. 393.5 killer jewels Permal And at that place were 12 miles plus the water vapor, which is negative. 241.8 killer jewels. Permal. And there were 10 miles. Good. Ah, and and then I'll go ahead and divide that by the four. Her, actually. Permit's allow me see if I can squeeze this in over hither. Ah, and that'southward gonna be and then minus the entropy of reaction. But because the entropy of reaction is a negative number, that's gonna go a positive 5000 678 killer jewels. Lovely. And then you dissever the entire thing by the formals from the night Douglas fir inside. Okay? And once yous plug that in your reckoner, what you go is that Thean flippy data for nitroglycerin is negative. 480 6.4 killer jewels Permal of nitro glycerine.

In this problem, we are given to structures were given the structure of nitroglycerin, which you come across right here and the structure of TNT, which you run into right here nosotros are asked to do 2 things first, we're asked to find the energy energy released. 11 m of nitroglycerin undergoes explosion, an explosion. Then I'm gonna run to this page and I'g going to write downwards the bonds and you lot can flip back a piddling earlier or observe information technology into text if yous want and look at the bonds, but I couldn't fit them all on i folio. So let'south do the bonds for nitroglycerin. I'm going to write the chemical equation, which we are also provided in the text. I should have probably washed that already. Okay, then now I'm going to write downward all the bonds that we accept in T. North. T. And actually write downwards the bonds for everything. And then permit me get started on that. And over again, you tin reference that earlier diagram or go look at your text to follow along. So I'm going to look up. I already did when write down what each of these private bonds are? These john killer joules per mole. These have just looked upwardly at a table in the text now for nitrogen which are six, don't forget nitrogen has a triple bond and C. E. O. Carbon dioxide has each carbon dioxide. And I'll write these downwards here. There are 12 of them and each one has to C double bond does h2o. I'thousand going to write correct here and each water has to oh H bonds and there are tin can a chugalug. So 10 times besides times the O. H. Bail and last only not least we have our oxygen. And that is 4 95 and R. C. Double bond, those are seven 99. So this will be 24 times 7 99 in this, I should have left a little more room there I am going to this will be twenty times the old H bail is for 63. And once again my oh to bond is 400 95. So we can take a look here and run into if we have any bonds that. And I don't believe we do on this looking looking. No we don't. And then nosotros cease up having to practice all of these. So we're going to take five times. Have to multiply each of these so add together them up and when you do that I got I have to take my four. Then I'm gonna take four times 68 62 and I'k going to decrease from that. I might move this down a fiddling bit. I'll only exercise it down hither. I saw I had 20 seven. I'm going to even move that downwards farther. 4 48. Again, these are killer jewels. And so I'm going to subtract each of these four and then I'll decrease from that 6. Don't forget to multiply the coefficients 6 times 9 41. Run into if I wrote this down anywhere, I didn't merely when I did each of these I got 34 5 77 34,000 577. What did my math here? I believe I got 66 49. But I need to get double check that. And so hang on, just a moment while I'm in the process of double checking my work hither. Sorry well-nigh that. I shouldn't do these things to y'all. No, I got negative 7129. Killing joules per mole at present I'm going to use and actually that's killer jewels per formals. Since I have a coefficient of four here perform als of nitroglycerin, which was my C three H five Due north 309 So let's move that to the next page. We were given one thou of nitro glycerin and I'm but going to abbreviate it N. G. For hither I just cheated and looked up the tooth mass which is 227.nine grams per mole. Then I multiplied that past the number we got on the final page, negative cy 71 29. Kill a jules perform ALs. When I was done, my answer was 7.8 five killer jewels, which means that 7.85 killer jewels of energy is released. Last office was easy were given this construction as the chemical construction for TNT. And so I'm simply gonna write downwardly the chemical formula for this as C six H ii, no two and I have three of those CH three and that decomposes into six and two. I'1000 going to go out usa off Plus seven CO 2 plus 10. I shouldn't have put my, well I will 10 H 20 plus 21 sees carbons and of class this would be a G. I'm non certain what TNT is probably a solid. It's pretty large. But anyhow we were only asked to remainder this already, residuum that terminal part. And so let me go through and highlight the coefficients hither. You tin can double cheque that and there is a counterbalanced chemical equation. Nosotros did it.

Hullo. Students in this question The reaction of sina might with the dye oxygen is carried out and that irresolute internal free energy delta you. Information technology is found to be minus 7 42.7 kg per mold at temperature T equals 2 to 98 Calvin. So we have to summate the changing and summate for this reaction. Okay. And we have given the reaction and we know that the irresolute mentality delta H. It is equals to changing and changing internal energy plus changing number of gaseous smalls. My pleasure. Bye. Artie. So we can calculate the number of guesses smalls delta N. G. This is equal to the number of guesses small Cindy product minus number of jesse smalls in the creative. So from the given options from the given reaction nosotros accept nitrogen and carbon dioxide in the product. And then nosotros have two moles and minus reactant. Has uh Number of modes equals ii. i.5. Okay, so nosotros get from here, this is equal to 0.5 mold. Okay, and then now we tin substitute the values so nosotros will obtain that Delta H. It is equal to Delta U. Which is equal to this -742.7 plus delta N. G. Which is equal to this 0.5 players are which is universal gas abiding, that is viii.314 multiplayer beta. And to the power minus three in the kaluga because this value is in the collegial per mole, so this is converted into collegial per mole, kelvin per mole, kelvin. Okay. And the temperature which is 2 98 Calvin. So from here afterward solving the changing and therapy, it is obtained as minus 77 41.5 kg jewel per move. So this becomes the answer for this question. Okay, thank y'all.

v answers

Utilise the Wronskian t0 determine the given functions are linearly independent on the indicated interval.f(x) = thirteen,chiliad(x) = 4x hlx) = 5x2 the real lineSelect the correct option belov' and, necessary, fill in the answer box to complete your pick(Simoliiy your answver )The Wronskian W(f; k, h) =As Wis identically on the eal line nine) g(x) and h(x) are linearly dlependentThe Wronskian W(f; chiliad, h) =As W is never on the existent line ix) m(ten) and hx) are linearly independentThe Wronskian West(f; g, h) =As W

Use the Wronskian t0 decide the given functions are linearly independent on the indicated interval. f(x) = 13,thousand(x) = 4x hlx) = 5x2 the real line Select the right choice belov' and, necessary, fill up in the reply box to complete your pick (Simoliiy your answver ) The Wronskian W(f; grand, h) =...

5 answers

Solve for the unknown quantities in the triangle. Round lengths to the ncarest tenth:8 = 50",a = 10.4 m,c=74 m B=45",a= 10.4 grand,€=seven.4m 8 = fifty",a=74m,c= 10.4 m 8 = 55" a=74m,c= ten.iv chiliad

Solve for the unknown quantities in the triangle. Round lengths to the ncarest tenth: 8 = 50",a = 10.iv m,c=74 m B=45",a= ten.4 one thousand,€=7.4m 8 = l",a=74m,c= 10.4 chiliad 8 = 55" a=74m,c= x.4 m...

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