Mothballs are composed primarily of naphthalene (C10H8). When 1.025 g of naphthalene burns in a bomb calorimeter, the temperature rises from 24.25 C to 32.33 C. (The Heat Capacity of the Calorimeter is 5.11 kJ/C.) Find DeltaE (in x 103 kJ) for the combustion of naphthalene.

-5.16 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat released by the combustion of naphthalene (Qcomb) and the heat absorbed by the bomb calorimeter (Qcal) is zero.

Qcomb + Qcal = 0

Qcomb = -Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression:

Qcal = Ccal × ΔT

where,

Qcal = Ccal × ΔT = 5.11 kJ/°C × (32.33°C-24.25°C) = 41.3 kJ

From [1],

Qcomb = -41.3 kJ

41.3 kJ are released upon the combustion of 1.025 g of naphthalene (MW 128.17). The change in the internal energy (ΔE) is:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

if you're asking what comes out, it's steam.