A15.7 g aluminum block is warmed to 53.6 ∘c and plunged into an insulated beaker containing 32.1 g of water initially at 24.2 ∘c. the aluminum and the water are allowed to come to thermal equilibrium. assuming that no heat is lost, what is the final temperature of the water and aluminum?

27.055 (°C)

Explanation:

the final temperature of the water and the aluminium has the same value: the aluminium block lost the heat, the water 'takes' the heat.

All the details are in the attached picture, the answer is marked with green.

"the periodic table is arranged in order of atomic number."

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 :