### 1) 100.0 ml of a 1.02 M acid, HA, is mixed with 100.0 ml of a 1.02 M base, BOH, in a calorimeter whose calorimeter constant is 38.7 J/ o C, making a…

1) 100.0 ml of a 1.02 M acid, HA, is mixed with 100.0 ml of a 1.02 M base, BOH, in a calorimeter whose calorimeter constant is 38.7 J/ o C, making a 200.0 ml AB salt solution. HA(aq) + BOH(aq)  AB(aq) + H2O(l) The temperature of the solution and calorimeter increased from 20.9 oC to 27.5 oC. Assume that the density and specific heat of the salt solution is the same as that of water, which is the dominant component. Water: d = 1.00 g/ml s = 4.184 J/(g o C)

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1) 100.0 ml of a 1.02 M acid, HA, is mixed with 100.0 ml of a 1.02 M base, BOH, in a calorimeter whose calorimeter constant is 38.7 J/ o C, making a…
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a) How much heat was absorbed by the solution?

b) How much heat was absorbed by the calorimeter?

The sum of these heat energies is equal to energy that was released by the neutralization reaction (so the sign is (-), negative). An equal number of moles of acid and base reacted.

c) How much heat energy was released by the neutralization reaction? ___________J

d) How many moles of acid (or base) were there? _________moles

Using this data, what is the heat released per mole of acid (or base)? ___________J/mol

This is an experimental determination of delta Hneutralization

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