4. You mix 50.0 mL. of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially at21.4°C. The final temperature of the neutralization reaction was determined to be 41.2°C. The calorimeter constant was known to be 109.6 J/°C.DensityAqueous Solution = 1.00 g/mLSpecific Heat of water 4.184 J/g.°Ca. What is the total amount of heat evolved in this reaction?b. If 145 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (kJ/mole)?

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Asked Nov 6, 2019
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Please solve for a and b

4. You mix 50.0 mL. of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially at
21.4°C. The final temperature of the neutralization reaction was determined to be 41.2°C. The calorimeter constant was known to be 109.6 J/°C.
DensityAqueous Solution = 1.00 g/mL
Specific Heat of water 4.184 J/g.°C
a. What is the total amount of heat evolved in this reaction?
b. If 145 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (kJ/mole)?
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4. You mix 50.0 mL. of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially at 21.4°C. The final temperature of the neutralization reaction was determined to be 41.2°C. The calorimeter constant was known to be 109.6 J/°C. DensityAqueous Solution = 1.00 g/mL Specific Heat of water 4.184 J/g.°C a. What is the total amount of heat evolved in this reaction? b. If 145 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (kJ/mole)?

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Expert Answer

Step 1

(a)

The total amount of heat evolved is equal to addition of heat ab...

50mL50mL = 100mL
Total volume
Density 1.00g/mL
Mass 100g/mL x100mL
=100 g
Initial Temperature = 21.4°C
Final Temperature 41.2°C
AT 41.2°C21.4°C
= 19.8°C
= mcAT + caAT
Qtotal
= |100 g x 4.184 J/g°Cx 19.8°C109.6J/C x 19.8C
- 10454.4J
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50mL50mL = 100mL Total volume Density 1.00g/mL Mass 100g/mL x100mL =100 g Initial Temperature = 21.4°C Final Temperature 41.2°C AT 41.2°C21.4°C = 19.8°C = mcAT + caAT Qtotal = |100 g x 4.184 J/g°Cx 19.8°C109.6J/C x 19.8C - 10454.4J

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