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LuckyWell [14K]
3 years ago
7

What happens if more solute is added to a saturated solution?

Chemistry
2 answers:
uranmaximum [27]3 years ago
6 0

Answer:

will not change

Explanation:

A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.

Alexxx [7]3 years ago
5 0

Answer:

A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.

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True the first step is nucleation
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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
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Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

  10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

                  T_{f} - 100^{o}C = 0^{o}C - T_{f}    

                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

                         = 10000 g \times 0.385 J/K g \times 323 K

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

            = 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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Answer:

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Explanation:

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