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Kitty [74]
3 years ago
10

The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What

is the heat of vaporization at 75oC?
Chemistry
1 answer:
pentagon [3]3 years ago
8 0

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

=  Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

or

The heat of vaporization at 75° C = 900 J/mol

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Total number of valence electrons = 5 + 3 = 8 electrons

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A single hydrogen atom has a mass of 1.67 × 10−24 g. A sodium atom has an atomic mass of 23. How many sodium atoms are required
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<h3>Answer:</h3>

                      2.55 × 10²² Na Atoms

<h3>Solution:</h3>

Data Given:

                 M.Mass of Na  =   23 g.mol⁻¹

                 Mass of Na  =  973 mg  =  0.973 g

                 # of Na Atoms  =  ??

Step 1: Calculate Moles of Na as:

               Moles  =  Mass ÷ M.Mass

               Moles  =  0.973 g ÷ 23 g.mol⁻¹

               Moles  =  0.0423 mol

Step 2: Calculate No, of Na Atoms as:

As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,

               Moles  =  No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹

Solving for No. of Na Atoms,

               No. of Na Atoms  =   Moles × 6.022 × 10²³ Na Atoms.mol⁻¹

               No. of Na Atoms  =   0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹

               No. of Na Atoms  =  2.55 × 10²² Na Atoms

<h3>Conclusion: </h3>

                          2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.


5 0
3 years ago
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