In a constant pressure calorimeter experiment a student measures a temperature change of 2.5 degrees c. the heat measured was 2.200 kj. what was the volume of the solution? assume that csol=4.18 j/gc

The volume of the solution was 210 mL.

q = mCₛₒₗΔT

m = q/(CₛₒₗΔT) = 2200 J/(4.18 J·°C⁻¹g⁻¹ × 2.5 °C) = 210 g  

If the solution has the same density as water (1.00 g·mL⁻¹) ,

Volume = 210 g × (1 mL/1.00 g) = 210 mL

Note: The answer can have only two significant figures because that is all you gave for the temperature change.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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