When a 3.25 g sample of solid NaOH was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 degrees C to 32.0 degrees C, calculate the heat of neutralization (deltaH) in kJ/mol NaOH for the solution process:
NaOH(s) --> Na+ (aq) + OH(aq)
assume it's a perfect calorimeter and that the specific heat of the solution is the same as that of pure water.
2007-08-01 02:48:39 · 2 answers · asked by calc25 1 in Science & Mathematics ➔ Other - Science
H = (103.25 g)(4.186 J/g-°C)(32.0°C - 23.9°C)(40 g/mol.)(1 kJ/1,000 J) / (.3.25 g)
H ≈ 43.067 kJ/mol.
2007-08-03 10:37:36 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
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2016-10-13 08:27:34 · answer #2 · answered by ? 4 · 0⤊ 0⤋