coffee cup calorimeter calculations
coffee cup calorimeter calculations
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Problem with the molar heat capacity?
I've been running in circles trying to solve this problem and would welcome any Help! Thank you! This is part of the calculations of a laboratory report. NaOH and HCl were mixed in a calorimeter cup of coffee, and I calculated the temperature variation. Lab manual says to use "simplified expression q (sub p) = C (sub p) x (delta) T (delta) H is given as -55.81 kJ / mol Now, I have all the temperatures involved, the volumes and molarities of HCl and NaOH, and the value (delta) H = -55.81 kJ / Mol. How can I know the heat capacity of calorimeter?
Guide units always use them in your calculation to prevent errors Any First, the following equation: HCl + NaCl + NaOH = heat ---- H2O_____dHneut -55.81kJ/mole = Heat absorbed by water and heat from the heat produced = (less than moles of H + and OH-) * 55 810 J / mol =? heat absorbed = [4.18 J / gC * gsoln + (Calo Cap Heat J / C )???] * (Tf - Ti) Plug and resolved early in the heat of the calorimeter J Math Basic / C is a prerequisite to chemistry - I just try to help with the methodology solve the problem.
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