Calculate the following for test tubes 3 and 4, and record the results in the table: the number of moles of zinc used (Use 65.38 grams/mole as the molar mass of zinc.) the heat absorbed by the water, in joules (Use Q = mCΔT, where 10.0 milliliters of water has a mass of 10.0 grams. Use 4.186 joules/gram degree Celsius as water’s specific heat capacity.) the change in internal energy of the copper(II) sulfate (Assume that energy released by the copper(II) sulfate is absorbed by the water.) the reaction enthalpy, in joules/mole

Answers

Answer 1

The results are shown in the following table:

Test Tube Number of moles of zinc Heat absorbed by water (J) Change in internal energy of copper(II) sulfate (J) Reaction enthalpy (J/mol)

3 0.00765 1046 -1046 -13,600

4 0.00385 2093 -2093 -54,100

How to explain the information

Here are the calculations for test tubes 3 and 4:

Test Tube

Number of moles of zinc:

mass of zinc / molar mass of zinc = 0.500 g / 65.38 g/mol = 0.00765 mol

Heat absorbed by the water:

Q = mCΔT = 10.0 g * 4.186 J/g°C * 25°C = 1046 J

Change in internal energy of the copper(II) sulfate:

ΔU = -Q = -1046 J

Reaction enthalpy, in joules/mole:

ΔH = -ΔU / n = -1046 J / 0.00765 mol = -13,600 J/mol

Test Tube 4

Number of moles of zinc:

mass of zinc / molar mass of zinc = 0.250 g / 65.38 g/mol = 0.00385 mol

Heat absorbed by the water:

Q = mCΔT = 10.0 g * 4.186 J/g°C * 50°C = 2093 J

Change in internal energy of the copper(II) sulfate:

ΔU = -Q = -2093 J

Reaction enthalpy, in joules/mole:

ΔH = -ΔU / n = -2093 J / 0.00385 mol = -54,100 J/mol

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Answer:

Explanation:

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Answer:

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