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owlet
 one year ago
Please, I need a stepbystep guidance on how to solve ths:
The temperature rises from 25°C to 29° in a bomb calorimeter when 3.50g of sucrose undergoes combustion in a bomb calorimteter. Calculate the heat of combustion of sucrose in kJ/mol sucrose. The heat capacity of the calorimeter is 4.90 kJ/°C. The molar mass of sucrose is 342.3g/mol.
I'm confused with the units.
owlet
 one year ago
Please, I need a stepbystep guidance on how to solve ths: The temperature rises from 25°C to 29° in a bomb calorimeter when 3.50g of sucrose undergoes combustion in a bomb calorimteter. Calculate the heat of combustion of sucrose in kJ/mol sucrose. The heat capacity of the calorimeter is 4.90 kJ/°C. The molar mass of sucrose is 342.3g/mol. I'm confused with the units.

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owlet
 one year ago
Best ResponseYou've already chosen the best response.0\(q=mc \Delta T\\q=(3.5g)(4.90\ kJ/°C)(4°C)\\q=68.6\ kJ•g\) then if I convert it to kj/mol \(68.6\ kJ•g(1\ mol/342.3g)\) the unit that I will get is kj•mol not kj/mol

owlet
 one year ago
Best ResponseYou've already chosen the best response.0but if i will multiply it by the molar mass, i will get a unit of kj•g^2/mol

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2The set up looks like this: dw:1444263286979:dw The reaction gives off thermal energy which is absorbed by the water (not mentioned in the question) and the temperature rises. The specific heat capcity of the calorimeter here includes the specific heat capacity of the water, so the formula for this is reduced to: \(\sf \large q_{absorbed}=C_{calorimeter}*\Delta T\) The mass of the sugar and it's heat of combustion is used to find exactly how much thermal energy was given to the water. The formula for this is: \(\sf \large q_{released}=moles*\Delta H_{comb}\) Solving for \(\sf \Delta H_{comb}\) gives: \(\sf \large \Delta H=\dfrac{q_{released}}{moles }\) (NOTE: sucrose needs to be in moles not grams look at the units of the heat of combustion) The energy absorbed by the water and energy given off are related by: \(\sf \large q_{absorbed}=q_{released}\) So the combined formulas are: lving for \(\sf \Delta H_{comb}\) gives: \(\sf \large \Delta H=\dfrac{q_{released}}{moles }=\dfrac{C_{calorimeter}*\Delta T}{moles_{sucrose}}\)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2sorry, i messed up the eiditing (ignore that last "lving fro Hcomb gives" the formula is correct though

owlet
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much. I'll try to absorb everything you said, it is really helpful! :)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2no problem! ask if you need clarification on anything

owlet
 one year ago
Best ResponseYou've already chosen the best response.0Actually, I'm confused with exothermic and endothermic reaction. In exothermic, heat is released, so the object should get colder and the surrounding system will get hotter right? In endothermic, heat is absorbed, so the object should get hotter while the surrounding system get colder... am I understanding this right or it's the other way around?

owlet
 one year ago
Best ResponseYou've already chosen the best response.0So if I'm right, it means the combustion reaction in the problem would be exothermic? Is that the reason why the heat of combustion is negative?
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