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Section 7In an experiment; 00 g of sucrose (C12Hn0,) is burned and} caue: the temperature of the bomb calorimeter t0 rise from 25.0 "€ to 30.0 "C. I...

Question

Section 7In an experiment; 00 g of sucrose (C12Hn0,) is burned and} caue: the temperature of the bomb calorimeter t0 rise from 25.0 "€ to 30.0 "C. If the heat capacity of the H,Abomb calorimeter Ceal is 8.57 kJPPC , calculate the enthalpy of combustion expressed in kJol4-15.86 10*}K -1465 10*3C -3.35*10*}D - 1.50*10*}

Section 7 In an experiment; 00 g of sucrose (C12Hn0,) is burned and} caue: the temperature of the bomb calorimeter t0 rise from 25.0 "€ to 30.0 "C. If the heat capacity of the H,Abomb calorimeter Ceal is 8.57 kJPPC , calculate the enthalpy of combustion expressed in kJol 4-15.86 10*} K -1465 10*3 C -3.35*10*} D - 1.50*10*}



Answers

When 0.514 g of biphenyl $\left( \mathrm { C } _ { 12 } \mathrm { H } _ { 10 } \right)$ undergoes combustion in a bomb cal-
orimeter, the temperature rises from $25.8 ^ { \circ } \mathrm { C }$ to $29.4 ^ { \circ } \mathrm { C } .$ Find $\Delta E _ { \mathrm { man } }$ for the combustion of biphenyl in $\mathrm { k } ] / \mathrm { mol }$ biphenyl. The heat capacity of the
bomb calorimeter, determined in a separate experiment, is 5.86$\mathrm { kJ } / ^ { \circ } \mathrm { C }$ .

Okay for this question, daughter e equals to minus on sea daughter tea. So in this question, army will strew zero point to fire 1514 grams C postal 5.86 on Dr T is 33.33 minus 24 pointed to fire themselves this degree we have easily calculated out E on DDE. The most off by Fino equals to, um, divided by the molar mass. And so the daughter ye reaction equal strode out on e miners minus the total more off the reactor and equals two minus 6.3 times 10 to the power off three killjoys hormone.

So firstly we need to convert temperature into Kelvin, where we get 3.52 Kelvin and then we have Q. It's equal to negative C. Multiplied by adults at T add dot S delta T. This is equal to negative 100 756 jewels. So we can calculate the molar mass of glucose, which is 180 g per mole. Where delta E. Is equal to -10700 56 jewels, divided by 9692 g. Multiplied by 180 g per mole. We got negative 2797.8 kg joules per mole.

To calculate the delta E. For this reaction occurring in a bomb calorie emitter, Q will be equal to DELTA E. Because in a bomb calorie emitter, that's the case because volume stays constant. The balanced chemical reaction is one more glucose reacts with six moles, oxygen producing six moles carbon dioxide and six miles water Q. Of the reaction will be equal to the negative Q. Of the calorie emitter. As heat is transferred Q. Of the calorie meter will be the calorie meter constant. 6.317 killer jewels per degree Celsius or per kelvin multiplied by the change in temperature. Given us negative 33 474 killer jewels. This should be killing jewels here. So DELTA E. For the reaction is going to be to kill the jewels divided by the moles of glucose that combusted. We can calculate the molds of glucose by taking the massive glucose, dividing by its smaller mass. Now we have killer jewels per mole of glucose, which gives us the value of negative 2805 kill jules per mole

So we were told that a certain amount of glucose is burned in a constant volume calorie meter, and we're supposed to find how much heat is evolved per mole, given the temperature rise of the water and the bomb of the Kalorama. So we know that the heat that is released or evolved by the glucose must be equal to the heat of the water as well, or in addition to the heat of the bomb. And so first, let's find the heat that is absorbed by the bomb. So are changing. Temperature is 25 0.2 to minus are 21.7, and we multiply that by the given heat capacity, which is 650. Google's per Kelvin. And so, using this, we can find that the heat absorbed by the bomb is 2000 288. Jules and we can do the same thing for the water. We know that queues equal to emcee times, Delta T mass specific eat and change in temperature. And so if we're told that there's 575 grams of water, of course, times are specific. It of water, which is 4.184 And that is Jules Per Gram Kelvin and our change in temperature, which is the same as the last, just 3.52 degrees Celsius. And that gives us times 4.4 in times 575 8000 468 jewels making our total of this plus this 10,756 jewels. And using this, we just have to divide by the the most of glucose. So we confined how much heat Permal of glucose is involved. So to do that, we just have to divide our sample of 0.6 92 grams of glucose over the molar mass of glucose. It just happens to be 180 0.2 grams per mole. Adding up all the individual more masses of glucose, which is C six h 12 06 Our grams are going to cancel here, and we will be left with our answer of 0.0 three 84 moles, you know we have to do is divide our heat 10,756 duels. Or we can equate this to kill a jewels by dividing it by 1000 which I will do it quickly. 10.756 killer jewels. We can divide this by our most his 0.38 for Morse. And we can get our final answer of 2000 800 and one killer jewels. Permal, let me fix it. That kill. Oh, really quickly. And there you have.


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