200.0 mL of 0.200 M HNO3 is added to 200.0 mL of 0.200 M KOH in a coffee cup calorimeter. The neutralization reaction that occurs can be described by the chemical equation- HNO3(aq) + KOH(aq) → KNO3(aq) + H₂O(l) Before the solutions are mixed they are both at 23.00 °C. After the mixing of the solutions the temperature increases until it remains constant at 24.39 °C. Assuming no heat lost to the surroundings, how much heat was lost in this reaction?
200.0 mL of 0.200 M HNO3 is added to 200.0 mL of 0.200 M KOH in a coffee cup calorimeter. The neutralization reaction that occurs can be described by the chemical equation- HNO3(aq) + KOH(aq) → KNO3(aq) + H₂O(l) Before the solutions are mixed they are both at 23.00 °C. After the mixing of the solutions the temperature increases until it remains constant at 24.39 °C. Assuming no heat lost to the surroundings, how much heat was lost in this reaction?
General Chemistry - Standalone book (MindTap Course List)
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Chapter6: Thermochemisty
Section: Chapter Questions
Problem 6.109QP: A 21.3-mL sample of 0.977 M NaOH is mixed with 29.5 mL of 0.918 M HCl in a coffee-cup calorimeter...
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200.0 mL of 0.200 M HNO3 is added to 200.0 mL of 0.200 M KOH in a coffee cup calorimeter. The neutralization reaction that occurs can be described by the chemical equation-
HNO3(aq) + KOH(aq) → KNO3(aq) + H₂O(l)
Before the solutions are mixed they are both at 23.00 °C. After the mixing of the solutions the temperature increases until it remains constant at 24.39 °C. Assuming no heat lost to the surroundings, how much heat was lost in this reaction? (Assume the specific heats and densities of the solutions is the same as water.)
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