When 25.0 mL of 0.700 mol/L NaOH was mixed in a calorimeter with 25.0 mL of...

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When 25.0 mL of 0.700 mol/L NaOH was mixed in a calorimeter with 25.0 mL of...

When 25.0 mL of 0.700 mol/L NaOH was mixed in a calorimeter with 25.0 mL of 0.700 mol/L HCl, both initially at 20.0 °C, the temperature increased to 22.1 °C. The heat capacity of the calorimeter is 279 J/°C. What is the enthalpy of neutralization in kJ/ mole of HCl?

Since the solutions are mostly water, the solutions are assumed to have a density of 1.0 g/mL and a specific heat of 4.18 J/g°C.

Select one:

a. −1020

b. -58.6

c. −5856

d. 58.6

science chemistry

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Answer 1

Answer #1

no of moles of HCl = molarity * volume in L

= 0.7*0.025 = 0.0175 moles

total volume of solution = 25 +25 = 50ml

mass of solution = volume * desnity

= 50*1 = 50g

The heat involves are

heat from neutralization +heat to warm solution + heat to warm calorimeter = 0

q1 + q2 + q3 = 0

n $\Delta$ H = mC $\Delta$ T + C $\Delta$ T = 0

0.0175mole* $\Delta$ H = 50*4.18*(22.1-20) + 279*(22.1-20) = 0

$\Delta$ H = -58560J/mole

= -58.56KJ/mole

round off -58.6KJ/mole

b. -58.6 >>>>answer

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Answer 2

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