What is the final temperature, in °C, in a squeezed cold pack that contains 34.2

Question

What is the final temperature, in °C, in a squeezed cold pack that contains 34.2 g of NH4NO3 dissolved in 131 g of water? (Assu me that the solution has the same specific heat as that of water, an initial temperature of 25.0 C, and no heat transfer between the cold pack and the environment. Hint: A solution is composed of a solute solvent. Number

A 90.0-mL sample of 1.000 M NaOH is mixed with 45.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 21.5°C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.3 C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g: "C), and that no heat is lost to the surroundings (a) Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. (b) Is any NaOH or H2SO4 left in the Include phases in the balanced chemical Styrofoam cup when the reaction is over? equation. Type an open parenthesis to add a phase. Phases should not be subscripted. Use the left and right arrow keys to move the cursor Yes out of a superscript or subscript in the module. O No (c) Calculate the enthalpy change per mole of H2SO4 in the reaction. Number kJ mol

Explanation / Answer

Enthalpy of dissolution of NH4NO3= 25.41 Kj/mole

moles of NH4NO2 in 34.2 gm = mass/molar mass = 34.2/64 =0.5343

enthalpy corresponding to tihs mass = 0.5343*25.41= 13.57 Kj=13.57*1000 Joules

this dissolution is endothermic as indicated by +ve sign of heat of dissolution

total mass of mixture = 131+34.2= 165.2 gm

Heat of dissolution= mass* specific heat* temperature difference ,13.57*1000= 165.2*4.184*(25-T), T= 5.4 deg,c

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