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Table Heat of neutralization data and calculationsSampleSample 2[1] Volume of HCI (mL)50,.249.9[2] Temperature of HCI (mL)25,024.6[3] Volume of NaOH (mL) [4] Temper...

Question

Table Heat of neutralization data and calculationsSampleSample 2[1] Volume of HCI (mL)50,.249.9[2] Temperature of HCI (mL)25,024.6[3] Volume of NaOH (mL) [4] Temperature of mixture after reaction ("C)49.949.931,.831A4Temperature difference ("C)6,8 06.80[5] Number of calorles evolved (cal) enter # positive value680,68 0678,640[6] Moles of H* that were neutrallzed (mol)3753 036710[7] Calorias evolved per mole of H" (caVmol)494,930496 410

Table Heat of neutralization data and calculations Sample Sample 2 [1] Volume of HCI (mL) 50,.2 49.9 [2] Temperature of HCI (mL) 25,0 24.6 [3] Volume of NaOH (mL) [4] Temperature of mixture after reaction ("C) 49.9 49.9 31,.8 31A4 Temperature difference ("C) 6,8 0 6.80 [5] Number of calorles evolved (cal) enter # positive value 680,68 0 678,640 [6] Moles of H* that were neutrallzed (mol) 3753 0 36710 [7] Calorias evolved per mole of H" (caVmol) 494,930 496 410



Answers

A dilute solution of hydrochloric acid with a mass of $610.29 \mathrm{~g}$ and containing $0.33183 \mathrm{~mol}$ of $\mathrm{HCl}$ was exactly neutralized in a calorimeter by the sodium hydroxide in $615.31 \mathrm{~g}$ of a comparably dilute solution. The temperature increased from 16.784 to $20.610^{\circ} \mathrm{C}$. The specific heat of the $\mathrm{HCl}$ solution was $4.031 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ that of the $\mathrm{NaOH}$ solution was $4.046 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$. The heat capacity of the calorimeter was $77.99 \mathrm{~J}^{\circ} \mathrm{C}^{-1}$. Write the balanced equation for the reaction. Use the data above to calculate the heat evolved. What is the heat of neutralization per mole of $\mathrm{HCl}$ ? Assume that the original solutions made independent contributions to the total heat capacity of the system following their mixing.

We can't innate the number off the mall for Noor Shank. I said they're in a harsh have system. Really made off Anil Dash multiple by the murder of Fennel Horse, which is 1.5 it was too. 52.5 more for its sale. It is 25 minutes, Monty paid. Why 1.86? Because you 46 point point Morning. We're going to know that How much Machu is the released for the neutralization off the sister. We have the collusion caramel on. We know that How much more they are minus 50. Boy point 84 Collusion pair moon off the hostile The model partially closed toe to the moon off the Australian But we know that this mall off price retail is in one million. This really divided by want was on, but it becomes the mole. So what? A multi good boy 40 sees 2.5 divided by 1000. They want off. The heat that is released is minus 2.59 two news. We also know that Culio hosts to emcee Delta T here take you on. The hit that is released is 2.59. But it is in collusion with the multiple. It 1000 that it becomes a parachute equals Tow em. We don't have the mass off the system. What? We have the density. The density here is three point 98. Then, since we multiply by the volume wishes the volume off the Exhale and William the off the harsh 50 plus 25 Because so 75 sold this port because I m multiple boy, see one point zero to multiple boy identity, which is the coin on temperature t minus the initial temperature which waas 24 point 72. From here, we can calculate the T equals two 32.23 13 three 0.2 degrees centigrade.

So in this problem we're gonna go ahead and racked some nitric acid with some strontium hydroxide. So let's go ahead and start with a balanced equation. All right. So we're going to react in a 2-1 ratio to nitric acids for every strontium hydroxide. Got 20. H. is in it. So that will make strontium nitrate and two waters. Okay, so the first part of this problem is some stock geometry. So we've got some information about our nitric acid. We've got our polarity and our volume. So let's use those to calculate the moles. So we'll take more clarity Times. Leaders 405 leaders. And that will give us with 03 72 moles of our nitric acid. Oh, do a more ratio here. Yeah. Will change moles of HNO three, two moles of strontium hydroxide. That's a 1- two ratio. And that will give us .0186 moles of our strontium hydroxide. And we want to know what volume we have to add. And since it's one moller, we'll just go ahead and divide this by its polarity 2.0 More. So that will give us point 0186 leaders. Or if you prefer 18.6 ml of the strontium hydroxide is needed to combined with that nitric acid there. Okay, so now we're going to go ahead and look at the heat, they told us Delta H is minus 52 killer jewels per mole of our nature gas. It so we're gonna go ahead and multiply by our moles of nitric acid which we calculated above. Mhm. We'll see that we're gonna give off 4.93 Killer Jewels. I'm gonna go ahead and multiply by a 1000 To get to jules so -19 34.4 jewels. And then let's see using Q equals negative M C delta T. Let's see if we can find our change in temperature. So we'll plug our heat in here and we want it in jewels because our specific he capacities and jewels and then we need the mass. So the mass is a little bit of a challenge. Okay, We're gonna have to combine the two solutions. So we had 50 ml of the nitric acid And we saw that we had add 18.6 mm of the strontium hydroxide. So that gives us 68 .6 ml of solution total. All right. And they said that the density, we should consider it to be one. So that's going to be 68.6 g of solution. So we'll go ahead and plug that in here 68.6 g. Um We use the specific key capacity and we'll try to solve for delta T. So if we go ahead and do our math, We're going to get 6.75°C.. So our temperatures going up by 6.75°C because it's positive. So we'll go ahead and add that to our initial temperature, which was 27.3. We'll add those together and we'll see that our final temperature Will be 34.0°C..

So here we're looking at the neutralization reaction of a solution of h. n. 0. 3 with S. R. O. H. Two. So this is an ex Islamic process. We have to H. N. 03 dot S. R. The wage to, that gives us S. R. And 03- two h. 20. And so the moles of S. R. O. H. Two divided by the moles Of H. and three is equal to 1/2. Where the malls of S. Are their wage to Is 9.9372 divided by two. So that's not point not 186 moles. So we can look at the volume of solution which is the molds divided by the morality divided by juan mall polyta, Multiplied by a 1000 Milita divided by one l Gives us 18.6 mm. And so the heat released. Hence Poppy. Here He released H. R. is equal to negative 52 times 10 to the three jewels per mole, Multiplied by 9.937 two moles. That's equal to negative 1.93 times 10 to the three jewels. And so the final volume after mixing Is equal to 68.6 mil. And then we multiply that by one g The mill to get a mat, which is 68.6 g so we can determine delta T, which is equal to cute with calculated over em. With calculated multiplied by C, which we know one. And so what we have is 6.73°C. Where delta T. Is equal to T. Final. Subtract the initial, Where T. final is equal to 34°C.

He'd of mutilation is less than 13.7:00. Hillary Permal for the reaction. We know that in the strong acid base neutralization. The reason here is that 13.7 kilocalories per mole. Right? But if we take any weak acid or weak waste than Hito no relation will be less than 13.7 by because we have to give energy in the organization of the weak acid or base. That's where he'd of mutilation will be Less than 13.7 kilocalories per mole. So we have to check in which option weak acid or base is using. So we can observe that in that day, option that day option ascetic acid plus any words ascetic acid plus any porridge in this mixture. Ascetic acid is a weak acid. They so weak acid. Is that responsible for the less energy from 13.7 the low calorie para moon. So this is the correct awesome. Okay.


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