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3 years ago

Which of these could be separated by distillation or filtration

2 answers:

6 0

Its B- a mixture.
Because I don't think you could separate and atom by distillation or filtration.
Actually I don't think you could separate an element by them.
I'm pretty sure compound can't be separated by them either.

Shkiper50 [21]3 years ago

5 0

Answer: Option (B) is the correct answer.

Explanation:

A process where solute or solid particles which are present in a liquid or gas are separated out with the help of a filter medium like filter paper is known as filtration process.

For example, sand can be separated out from water by using a filter paper.

Components of a mixture can never combine chemically with each other.

Whereas, an element is a substance which contains only one type or kind of atoms.

For example, a pure zinc will contain only atoms of zinc.

A compound is defined as the substance in which different elements are chemically combined together in a fixed ratio by mass.

For example, $MgSO_{4}$ is a compound and elements are present in 1:4 ratio.

A compound can be divided into its constituent or simpler substances.

Thus, we can conclude that a mixture could be separated by distillation or filtration.

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3 years ago

The value of ΔG° at 25 °C for the decomposition of gaseous sulfur trioxide to solid elemental sulfur and gaseous oxygen, 2SO3 (g

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Answer:

$\begin{array}{l}{\text { The value of } \Delta \mathrm{G}^{\circ} \text { at } 25^{\circ} \mathrm{C} \text { for the decomposition of gaseous sulfur trioxide to solid }} \\ {\text { elemental sulfur and gaseous oxygen is }+740.8 \mathrm{kJ} / \mathrm{mol}}\end{array}$

Option: A

<u>Explanation</u>:

$\begin{array}{l}{\text { The value of } \Delta \mathrm{G}^{\circ} \text { at } 25^{\circ} \mathrm{C} \text { in the following reaction can be calculated as follows: }} \\ {2 \mathrm{SO}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{S}(\mathrm{s}, \text { rhombic })+3 \mathrm{O}_{2}(\mathrm{g})}\end{array}$

$\begin{array}{l}{\Delta \mathrm{G}^{\circ} \text { is Standard Gibbs free energy change which can be calculated from the standard free }} \\ {\text { energies of formation of the products and the reactants from the following equation: }}\end{array}$

$\begin{array}{l}{\Delta \mathrm{G}^{\circ}=\Sigma \mathrm{G}_{\mathrm{f}(\text { products })}^{\circ}-\Sigma \mathrm{G}_{\text {creatants }}^{\circ}} \\ {\Delta \mathrm{G}^{\circ}=[\mathrm{Sum} \text { of standard free energies of formation of products }]-[\mathrm{Sum}\text { of standard } } \\ {\text { free energies bf formation of reactants] }}\end{array}$

$\begin{array}{l}{\text { Now here standard values of } \Delta G^{\circ} f(k J / m o l) \text { for } S=0, O_{2}=0 \& S O_{3}=-370.4} \\ {\text { Hence these values can be substituted in above equation: }} \\ {\Delta G^{\circ}=\left[2 G_{f}^{\circ}(0)+3 G_{f}(0)\right]-[2(-370.4)]} \\ {\Delta G^{\circ}=[-0+0]-[-740.8]} \\ {\Delta G^{\circ}=+740.8 \mathrm{kJ} / \mathrm{mol}}\end{array}$

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