Calculate qsoln (in J) for a trial whose qsample was calculated to be 553 J and

Question

Calculate qsoln (in J) for a trial whose qsample was calculated to be 553 J and whose qsample was calculated to be 49 J.
Enter your answer to zero decimal places. Include a negative sign before your digit if you believe it is required. Calculate qsoln (in J) for a trial whose qsample was calculated to be 553 J and whose qsample was calculated to be 49 J.
Enter your answer to zero decimal places. Include a negative sign before your digit if you believe it is required.
The question says to use the photo. But that’s all the information she gives us. Sample Calculation: Determining the Enthalpy of Solution of Calcium Acetate A student is asked to add 5-6 g of Ca(C2H302)2 to 50 mL of water and to use the change in temperature to calculate AHson. The reaction below corresponds to the enthalpy of solution of calcium acetate. Hsolni Ca(C2H3O2)2(s) H2O(o Ca"(aq) + 2 C2H3O2-(aq) The student first checked the solublity of Ca(CaH,02)h to make sure that a 6 g sample would dissolve in 50 mL ter. She reasoned that because the solubility of this salt is 18.7 g in 50 mL of water at 20 °C, a 6 g sample should readily dissolve in 50 mL of water at 20 °C (-room tenm 50.014 g of distilled H20 and observed that the temperature changed from 24.2 C to 17.6 C. The calorimeter constant (Coal) for the calorimeter used was assumed to be 12.0 J/°C, and the specific heat (Csample) for Ca(C2HsO2)2(aq) is 3.98 JIg C. If qson is the heat transfer associated with the dissolving of the salt sample and Qsample is the heat lost or gained by the solution of water and salt, then: p). She then stirred 5.501 g of Ca(C2H302)2 into Geon + [(mass of salt + mass of water)(Csample)(ATI] + [(Ca)(AT)] = 0 qson + [(5.501 g + 50.014 g)(3.98 J/g·°C)( 17.6 -24.2°C)] + [(12.0 JC)(17.6 °C-24.2 °C)] = 0 qsoln-1458 J-79.2 J = 0 qsoln = +1537 J +279 J/g 279dx_x 158.168 g = +44.2 kJ/mol Hson (in J/g deicer) = +1537 J = + 5.501 g aHsoln (in kJ/mol deicer) = mux_x 158.168 g = t g 1000J mol

Explanation / Answer

qsample = 553 J

qcal = 49 J

qsoln + qsample + qcal = 0

qsoln + 553 + 49 = 0

qsoln = -(553+49) = -602 J

So, qsoln = -602 J

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