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Roachnt~r TCCFmesHCI ItnkL9ic1WcCCTaro) Z4Tc 3556 KES 1L64 32c 4L6 '€042 605 Vaote} (OU Ls uu CALCULATIONS (work ALONE) Caiculata the hent bansloned Ralutan ((aeae) by Ehe rexaton ct "a nIn HCL 3M . (CO "37 47 4a '3.0} $187.1} ]33.3 m SH Cakulale AH Flmol for reacton #1; 5787.13T 24,305 0 ,32c0 * i000431.42 kdlmclCakulate the heat transferred t0 the soluton (Qeueen) by the reacton of Ugo with HCL Ci U( I6 | . 0r74 e > 3m (M 00GuoioU 30333M Calculate AHrE In kJlmol for r

Roachnt ~r TCC Fmes HCI Itnk L9ic1 WcCC Taro) Z4Tc 3556 KES 1L64 32c 4L6 '€ 042 605 Vaote} (OU Ls uu CALCULATIONS (work ALONE) Caiculata the hent bansloned Ralutan ((aeae) by Ehe rexaton ct "a nIn HCL 3M . (CO "37 47 4a '3.0} $187.1} ] 33.3 m SH Cakulale AH Flmol for reacton #1; 5787.13T 24,305 0 ,32c0 * i000 431.42 kdlmcl Cakulate the heat transferred t0 the soluton (Qeueen) by the reacton of Ugo with HCL Ci U( I6 | . 0r74 e > 3m (M 00 Guo ioU 30 333M Calculate AHrE In kJlmol for reacton 02: 4400 2 740.2 66 uellmel 112 61 7ioco Use Hess' Law 0 determine AHa for reaction & (the bunalon cN3O) Stow your work Use your AHim from #5 cakulate Guzm when 1,75 kg ot Mg teact Win ercess axygen:



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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.

Oh, were to show that the specific heat oven element is inversely related to its atomic weight. So for part A, we're gonna use the data from the front cover to plot a straight line graph relating atomic mass and specific heat, and showing that this is an inverse relationship will pick several elements here. Let's create a data table, Theo. Elements that will choose are magnesium, aluminum, iron, copper, mercury and let the atomic weight. Yeah, magnesium is 24.3. Aluminum 26.98 iron 55.84 Copper 63.55 Mercury 200.59 and lead 207.2. Now we'll calculate one over the inverse of the atomic weight. So one over the atomic weight. This would give me 0.0 4115 0.3706 0.17 91 went 01574 0.0 499 and 0.483 The specific heat for each of these abbreviate. That s H 1.23 for magnesium 0.897 for aluminum 0.449 for iron 0.385 for copper, 0.14 for mercury and 0.13 for lead. Now, from this data table, we're going to plot thes specific heat versus the inverse of the atomic weight. So we're going to plot specific heat versus one over the inverse of the atomic weight, and we're going to show that this is a, um that specific heat is inversely related if we get a linear plot and our plot would look like this here. So there is our plot of specific heat versus the inverse of atomic weight. And we do get a a linear plot the equation of this street line using the proper variables. The equation of the line is thes specific heats Z equal to you. Yeah. 2 24 0.9091 over a W plus 0.131 That would be part a for part B. We're gonna use the measured, specific heat and equation derived in part A to attain, to obtain an approximate value for the atomic mass of cadmium. So for cadmium, we're told that the specific heat is equal to point 23 jewels per gram. degrees Celsius. So let's plug that in here. We'll leave out the units now. Let's solve this for cadmium minus 0.131 from both sides, please. Smooth. 12169 is equal to 24.991 over a W divided by 24.909 And this is 0.8708 is equal to one over a w will take the inverse of this. The atomic weight is equal to Oh, I'm sorry. I think I got this wrong here. Point Hey, did. Let's fix it. 24 points. Choose 09 which will make a difference. 24 point. Let's make the change all throughout here. 24 points. It's gonna change this number. Here, make a change. Quickly. 24 to 09 Yeah, this is right. 0.8959 Now we take the reciprocal of this to give me a mass of 111.61 I a m u. Okay, for part C Oh, We are, um, to raise the temperature 75 of particular metal by 15 degrees. Requires 450 jewels of heat What metal might this be? So we're going Thio find the heat required is equal to m times the specific heat times the delta t uhh! The energy required is 450 jewels. The mass were given this 75 0 g specific heat. Is there unknown? Delta T We're raising the temperature of 15 degrees. So solving for the specific heat here we get a 150.40 jewels program degrees Celsius now bringing this into our formula. Thes specific heat is equal to 24.2091 over a W plus 0.131 We know the specific heat is 0.40 will devote the units 24.2091 over a W plus 0.131 Andi doing the math here and solving let you go ahead and solve that. Verify s something for the atomic. We we get 62.578 a. M u. Which corresponds to the element. Mhm 62.5. Closest. Okay, two, I would see the element. Copper

So we want an experiment to monitor the rate of dissolving of a metal carbonate. We could use either of the three carbonates or any mixture. Ah, but I would probably start with the picking one and then if I needed to, I would use the others each individually. And I would take, uh, and and way out identical samples of the metal carbonate s O that I had. Ah, a certain number of them. Maybe half a dozen Different. Uh, I should say half a dozen identical samples of the metal carbonate and then ah, add ah hydrochloric acid solution of known concentration and mix. And each time you mix at a certain period of time, take out ah sample of the, uh ah liquid. Either rapidly filter it or spend out the solid and pour off the liquid and then see what the concentration of the metal ion is in the solution. There lots of ways Teoh metal measure metal ion concentrations and then do it again with a different, uh, concentration of the hydrochloric acid and again sample at various times and see what concentration of the metal ion. So I would measure the metal ion in solution Ah, at various times with different identical samples of the solid in different concentrations of the H CEO

In each part of this problem were given an overall chemical reaction, and we want to determine with the overall change in Temple P or Delta. Each of reaction is we want to calculate this in two different ways. In the first way. We're going to list out the formacion reactions of each chemical species involved in the overall reaction from their respective elements. We're then going to determine the belt to each of formacion values for each one of these, and then combined the thermal, the thermo chemical equations using Hess's law in order to solve for the overall change in entropy of the given reaction. And the second way that we're going to sell for this to check our work and verify our solution is to use the overall Delta each of reaction equation that involves the entropy of formation values. So we start with part A, and we first read out the thermo chemical equations that that correspond to the overall reaction given import. A. We write out the formation reaction for each of the chemical species involved in the overall reaction, and we start with C A c 03 and we see that we need calcium, carbon and oxygen to form C A C 03 So for the formation reactions, the reactant site is going to be elements that occur naturally and do not require any energy in order to form. So that would be solid metals like calcium and carbon and die atomic gases like oxygen. For all of these thermal chemical entropy of formation, reactions were producing one mole of the desired compound. So we therefore have to adjust the with the metric coefficients of the elements on the left side so that we can produce one mole of the desire product when we balance the equation. That is why, for example, we used fractions or three halves oxygen so that there are three oxygen's on each side of this reaction. We move on to the next reaction calcium oxide, which is foreign from calcium and oxygen. And so we have solid calcium and die atomic oxygen, forming one mole of calcium oxide, and we balance it appropriately and finally for carbon dioxide. This is formed from the elements of carbon, an oxygen, so that's solid. Kurban die atomic oxygen to form one mould carbon dioxide. Next we can look up what the entropy of formation values are each one of these thermo chemical equations in the appendix or through some other appropriate source. And now we need to use Hess's law to combine these three formation reactions in order to determine what the Delta H is of the given reaction. If we compare the given reaction to the formation reactions, we see that one mole of calcium carbonate is a reactant in the desired reaction in a product in the formation reaction. So we therefore need to flip the first reaction by multiplying it by negative one so that we can get C A c 03 to be a reacted instead of a product. We go through the other chemical species. We see that one mole of CEO and one mole of CO two are produced, and that is also the case for the formation of the act reactions. We have one mole of each one of those species being produced, so now we manipulate those equations in that way where we flip the first reaction and keep the second and third reactions as they are. And when we flip that first reaction, we multiply you delta, each value by negative one so that it becomes positive delta each of formation values are given in units of killing joules per mole. But for all these formation, reactions were forming one mole of each desire product. And so therefore we can say that this corresponds to units of killing jewels were overall energy or each one of these. Now we have to cancel out like terms. We see that we have 1/2 plus one. So 1.5 moles of oxygen on this side to cancel out with 1.5 moles of oxygen on the products, we also have one mole of solid carbon on each side in one mole of solid calcium on each side. And so when we cross that all out and we combine these three reactions, we see that we are left with the overall desired reaction. So now though, all we have to do due to find the well to each of the reaction, is to add these three terms together and we should get an answer of Delta H of reaction to be 179.2 village ALS The second we that we are going to determine this is using the Yeah, he the entropy of reaction equation from the until piece of formation. So we subtract the reactant from the products that is the info peas of the overall entropy. Change of the products minus the overall entropy. Change of the reactant. We need to multiply the number of moles of each species by their respective mental piece of information, so that when we cancel out units, the most cancel out and were left in units of energy killer JAL's. So just looking at the reaction, we see that we have one mole of each chemical species and we use the built to each of formacion values and multiply them together. And to get for the product side where we have CEO in co two to be a value of negative 1028.4. For the reactant side, we have one mole of CEO in multiplying that by its delta each information value, we get that same value and now we we just subtract that delta. Each of the reactant is from the delta each of the products to get the delta h of the overall reaction we noticed said that corresponds to the same value that we got the other way 179 point to kill it, Jules. And so that is why that appropriately verifies our work and that that means that we calculated the correct Delta. Each value both ways since we got the same answer. That's the same process that we used for Parts B and C of this problem. So again we're forming calcium hydroxide from calcium, oxygen and hydrogen, balancing out appropriately to produce one mole. And we do that for the other two reactions for CEO and H two of those formation reactions. And we can look up the values for Delta H again. And then we use Hess's law to combine those thermo chemical reactions is we see that we need one mole of CEO in the products in that this president in reaction be one mole of H 20 in the products which is president reacted. See And we need one mole c a o. H. Two in the reactant. But we need to flip the first reaction in order to move that from the product side to the reactant side. So we manipulate those equations in this way. We see, have we see we have 1 1/2 plus 1/2 so one oxygen on the reactive side and one on the product side. We have one mole of each two on each side, and we have one mole of solid calcium on each side. We multiply that delta each of formation by negative one for the first reaction since we flipped it. And we added all together and get the final answer of 106.5 killer jewels for the overall entropy change or this desired reaction that we conform. After using Hess's law, you didn't use the until p of reaction from entropy of formation equation. Do you see that we again have one mole of each chemical species and we multiply it by the delta h of formacion value for each one of those species and do the total entropy change of the products minus the total entropy change of the reactant. When we do that, we see that we get the same value for Delta H to be 106.5 village ALS for part B. Finally, in part C, we have our given chemical reaction and we form one mole of each chemical species in this way, But we do not have a reaction for solid iron, since it occurs naturally and has a Delta H information value of zero. We look up the Delta H information values for the other thermal chemical equations, and then we combined them as we did before. Using Hess's law. You see that we have three moles of oxygen on each side when we add. Three have twice on the product side and three moles of solid carbon cancel out on each side. And when we combine all of those together, we have to multiply each of the given Delta H values by the factor by which we slipped and manipulated those equations. So you multiply the 1st 1 by negative one second, one by negative three and 1/3 1 by three. The reason that we use those coefficients. It's because we know that we need one mole of F E 203 as a reactant, but it is formed as one mole of a product, so we have to flip the first reaction by negative one and forth three CEOs in the reactant side. You see that we need to flip this reaction and multiply it by three. So we have three CEOs and reactive side and then for three. Co two we have co two is a product that we need to multiply it by three for that stick you metric coefficient. So when we do all of that and we we adjust the values for Delta H When we add them all together we get a final answer of negative 24.8 village ALS for the overall entropy change for the reaction in part C and then again we just need to verify that using the equation again for the little entropy change of the products minus a total in Philippi Change of the reactors and we use this doohickey metric coefficients to see we have three moles of CO two in the products Multiply it by its Delta h of formacion to get that value than for the reactant one mole of F E 203 and three moles of CEO. You want to play them by their respective delta h of formacion values and add them together again. We do not pay attention to solid iron since it has a Delta H of formation value of zero when we do the total entropy change of the products minus the reactant, we get an answer of negative 24.8, killing JAL's, which again matches what we calculated before by combining the thermal chemical equations.

For this question, a chemical reaction is occurring, generating the heat in the calorie meter. We had 1.22 g of magnesium, 200 mL of six point oh two molar hcl. Then the reaction of magnesium, one mole with two moles. Hcl produces magnesium chloride and hydrogen gas. The queue for this process is going to be equal to the heat gained by the water and the calorie meter plus the heat gained by the calorie meter. The water in the calorie meter is the solution of hydrochloric acid, to which the 1.22 g of magnesium was added. Knowing the density or specific gravity of the 6 to Mueller Hcl solution, we can then calculate the heat associated or the heat absorbed by the solution. We'll take the heat capacity of the specific heat capacity of the solution. The problem tells us to assume that it is the same as water. 4.184 will take the volume of the solution and converted into grams of solution by multiplying by the specific gravity or density of the solution. Now we have the specific heat capacity, multiplied by a mass. We need to finish that by multiplying by the change in temperature. The final temperature was 45.5°C with an initial temperature of 23°C. This then corresponds to the heat absorbed by the solution. He also is absorbed by the calorie emitter. So to get the total heat we need to add on the heat absorbed by the calorie emitter, which will be equal to the uh calorie meter constant provided at 562 jewels per degree Celsius multiplied by the change in temperature of the calorie emitter, which is going to be the same as the change in temperature of the solution in the calorie emitter. and we get a total amount of 23,000 jewels or 23 Killer Jewels. This heat is released during the chemical reaction so it's represented with a negative sign. Now, knowing the heat released during the reaction, we need to identify which is the limiting reactant. This will then tell us the molds that reacted Well. 1st calculate the moles of magnesium. We have .0502 moles of magnesium. Then we'll calculate the moles of hcl melissa. Magnesium was calculated by taking the mass divided by the molar mass of magnesium moles of hcl is calculated by taking the volume of hcl in leaders 100 mL. His 1000.1 liters multiplied by the polarity of the hcl solution, We see that we have significantly more hcl more moles hcl and although we need to molds hcl for everyone, more magnesium, we definitely have more than twice the moles of magnesium. So magnesium ends up being the limiting reactant where hcl is an excess. So this amount of heat corresponds to the complete consumption of .5 02 moles magnesium. So to calculate delta H. For the reaction because one mole magnesium is part of the balanced chemical reaction, calculating the heat released per mole of magnesium gives us delta H. For the chemical reaction, we take the heat calculated up here, Divided by the moles that we determined from the mass of magnesium. And this gives us a delta h value of -460 kill a jules per mole.


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