Answer to Question #193593 in Chemistry for Seb

Question

6350 J of heat is required to raise the temperature of a sample of AlF3 from 150C to 750C. What is the mass of the sample?

Answer 1

According to the heat law:

Q = cn(T₂ - T₁)

where Q - absorbed or released heat, c - heat capacity (75.1 J/mol·K), n - number of moles, T2 - final temperature (75°C = 348.15 K), T1 - initial temperature (15°C = 288.15 K).

From here:

n = Q / [c × (T₂ - T₁)] = 6350 J / [75.1 J/mol·K × (348.15 - 288.15 K)] = 1.4 mol

Finally:

m = n × Mr = 1.4 mol × 84 g/mol = 0.017 g

Answer: 0.017 g