When 4.37 g of potassium bromide
(KBr) is dissolved in 109 g of
water in a styrofoam calorimeter of negligible heat capacity, the
temperature drops from 25.00 to
23.34 °C. Based on this observation, calculate q
for the water and ΔH° for the process, assuming that the heat
absorbed by the salt is negligible.
The specific heat of water is 4.184 J °C-1 g-1.
4.37 g of potassium bromide (KBr) = mass / molar mass = 4.37 g / 119.002 g/mole = 0.0367 mole.
heat gain by water (q) = m * s * dT = 109 * 4.184 * (23.34 - 25.00) = - 757.1 J
ΔH° for the process = 757.1 J / 0.0367 mole = 20628 J / mole = 20.6 KJ / mole.
When 4.37 g of potassium bromide (KBr) is dissolved in 109 g of water in a...
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