Question

Ammonium nitrate is an ingredient in cold packs used by sports trainers for injured athletes. Calculate the change in temperature when 42 g of ammonium nitrate (NH&NOs, 80.1 g/mol) dissolves in 250 g water. Assume the specific heat of the solution is 4.18 J/(g °C). NH4NO3(s)-> NH4"(aq) + NO3-(aq) ΔΗ-25.7 kJ/mol

Answer 1

Answer:

See explanation

Explanation:

First, we need to use the correct expression:

Q = m*Cp*ΔT  (1)

Q = n*ΔH  (2)

These are the 2 expressions to calculate heat or energy.

Now, we want to know the change of temperature of the nitrate in water after being added, so with the innitial data of nitrate, we can calculate heat using the second expression. First, we need to calculate moles with the molecular mass:

n = m/MM

n = 42/80.1 = 0.52 moles

With these moles, we can calculate heat with the ΔH of this reaction:

Q = 0.52 * 25.7 = 13.364 kJ or 13,364 J

However, this heat as is being absorbed, the value would be negative.

Now that we have heat, we can use expression (1) and plug these values to solve for ΔT, but before, we need to know the total mass of the solution (water + nitrate)

m = 250 + 42 = 292 g

now, solving for ΔT:

-13,364 = 292 * 4.18 * ΔT

ΔT = -13,363 / (292 * 4.18)

ΔT = -10.95 °C