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4 years ago

The purging of impurities from a compound by crystallization:

Chemistry

2 answers:

Nuetrik [128]4 years ago

5 0

C.

This is also the only change by which crystals may be produced.

Ne4ueva [31]4 years ago

3 0

<h3><u>Answer;</u></h3>

occurs after the liquid melt cools back to a solid

The purging of impurities from a compound by crystallization <u>occurs after the liquid melt cools back to a solid.</u>

<h3><u>Explanation;</u></h3>

<em><u>Purging of impurities from a compound by crystallization is done through a method known as melt crystallization. Melt crystallization is a type of cooling crystallization process which is operated close to the melting point of the pure compound.</u></em>
The sample for melt crystallization id normally an impure compound. When the impure sample is cooled below the equilibrium temperature the result is formation of a solid phase which is more pure that the initial compound, while the impurities remain in the impure sample.

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3 years ago

A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the

iren2701 [21]

Answer:

1) Endothermic.

2) $Q_{rxn}=4435.04J$

3) $\Delta _rH=15.8kJ/mol$

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

$Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J$

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

$\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol$

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3 years ago

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