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Anni [7]
2 years ago
14

Explain the process of filtration and crystallization ​

Chemistry
1 answer:
Akimi4 [234]2 years ago
8 0

Answer:

Filtration, the process in which solid particles in a liquid or gaseous fluid are removed by the use of a filter medium that permits the fluid to pass through but retains the solid particles. Either the clarified fluid or the solid particles removed from the fluid may be the desired product.

Crystallization, or crystallisation, is the process of atoms or molecules arranging into a well-defined, rigid crystal lattice in order to minimize their energetic state. The smallest entity of crystal lattice is called a unit cell, which can accept atoms or molecules to grow a macroscopic

crystal.

BRAINLIEST PLEASE, I NEED IT

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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
2 years ago
.The arrangement of constituent particles is most ordered in *
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Answer:

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Elements with similar properties are located
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Those elements with similar properties are in the same column.
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Evidence shows that lakes are being affected by acid rain. Acid rain results in
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c

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