15.3 g of nano3 were dissolved in 100g of water in a calorimeter. the temperature of the water drops from 25.00°c to 21.56°c. calculate delta h for the solution process. (kj/mol)

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 100.0 g).

c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

∵ Q = m.c.ΔT

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

∵ ΔH = Q/n

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

i think the answer is terrae but not 100 percent sure. ▓▬▬ §