Heat of Neutralization 28 Pre-lab Questions Before beginning this experiment in

Question

Heat of Neutralization 28 Pre-lab Questions Before beginning this experiment in the laboratory, you should be able to answer the following questions. 1. Define endothermic and exothermic reactions in terms of the sign of H. 720 mL sample of water was cooled from 50.0 °C to 100 °C. How much heat was lost? 3. Define the term heat capacity. 50.0 g of water from 21.3 °C to 43.5 °C? 4. How many joules are required to change the temperature of 5. Define the term specific heat. 6. Calculate the final temperature when 50 mL of water at 65 ° C are added to 25 mL of water at 25 °c. Describe how you could determine the specific heat of a metal by using the apparatus and techniqu from this experiment. 7. Copyright © 2015 Pearson Education, Inc.

Explanation / Answer

Solution:- (8) q = m c delta T

where q, is heat energy, m is mass in g, c is specific heat in J/g.C, and delta T is change in temperature.

On addition metal to water, the temperature of water is changing from 22.1 degree C to 29.2 degree C.

So, delta T of water = 29.2 - 22.1 = 7.1 degree C

mass of water = 20.00 g (since, density of water is 1.00 g/mL, so 20.00 mL = 20.00g)

specific heat for water = 4.184 J/g.C

qwater = 20.00 g x 4.184 J/g.C x 7.1 C = 594.128 J

The heat gained by water is the heat lost by metal.

So, qmetal = -qwater = -594.128 J

delta T for metal = 29.2 - 48.6 = -19.4 C

mass of metal = 5.10 g

-594.128 J = 5.10 g x c x (-19.4 C)

c = -594.128 J/(-19.4 C x 5.10 g)

c = 6.00 J/g.C

So, the specific heat of metal is 6.00 J/g.C

(9) m = 50 g

c = 2.51 J/K.g or 2.51 J/g.C

delta T = 60 - 20 = 40 C

q = 50 g x 2.51 J/g.C x 40 C

q = 5020 J

So, the heat required to raise the temperature of ethanol is 5020 J.

(10) mass of water = 100.0 g

specific heat of the solution = 4.184 J/g.C

delta T of water = 32.0 - 23.9 = 8.1 C

q = 100.0 g x 4.184 J/g.C x 8.1 C

q = 3389.04 J

or q = 3.389 kJ

q(NaOH) + q(water = 0

so, q(NaOH) = - q(water) = -3.389 kJ

moles of NaOH = 3.25 g x 1mol/39.997 g = 0.0813 mol

delta H = q/n

where n is the number of moles of NaOH.

so, delta H = -3.389 kJ/0.0813 mol

delta H = -41.69 kJ/mol

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