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vova2212 [387]
3 years ago
13

You have a cup of hot water and you dissolve as much salt in it as possible. You then put the cup into the refrigerator and you

check it again after 24 hours. What would you notice when you checked the cup after 24 hours in the refrigerator? None of the salt would settle out of the water. Some of the salt would settle out of the water. All of the salt would settle out of the water. The salt water would have changed colors.
Chemistry
2 answers:
n200080 [17]3 years ago
7 0

Answer:

Some of the salt would settle out of the water.

Explanation:

This is an example of a supersaturated solution.

In this example, the salt is the analyte, and the water is the solvent.

When you mix the salt into the water in an ambient temperature there is a point where you can't solve more salt, because the amount of water isn't able to solubilize any more quantity of analyte. This is named supersaturation point.

If you expose the mix to the heat, the energy supply to the solution make that the excess of salt can be solubilized, but this effect only last until the heat still going on.

So when you put the solution into the refrigerator, the quick down of the heat makes that the excess of salt settles out of the water.

And all this process occurs because the solution always wants to reach the equilibrium between the analyte and the solvent.

hoa [83]3 years ago
6 0
Some of the salt would settle out. When the water was heated, it was able to absorb more salt than usual. This is known as super saturation. When the water is frozen it cannot hold as much salt, so some of it has to come out.
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Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

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Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

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\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

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\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

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\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

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\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

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