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fenix001 [56]
2 years ago
9

Which of the species would be expected to be the most miscible in benzene (c6h6)? 1. decane, c10h22 2. hydrogen fluoride, hf 3.

methanol, ch3oh 4. silicon dioxide, sio2 5. sugar, c6h12o6?
Chemistry
2 answers:
max2010maxim [7]2 years ago
7 0

Answer is: 1. decane, c10h22

This pair will most likely form a homogeneous solution because they are both nonpolar substances and "like dissolves like".

Other pairs will not form homogeneous solution because polar (methanol, sugar) or ionic (h<span>ydrofluoric acid)</span> substances have low solubility in nopolar (benzene) substances.

Whitepunk [10]2 years ago
3 0

Decane can be the most miscible species in benzene.

Answer: 1. decane, c10h22

Explanation

The miscibility is the measure of solubility of a solute in a solvent.

As per the rules of solubility, like substances dissolve and unlike substances get suspended.

So here benzene is a non-polar solvent, so the solute which can be most miscible in benzene should also a non-polar nature.

Thus the solute should be non-polar substance to dissolve in non-polar solvent.

Among the options given, decane C10H22 is the non-polar substance which can act as solute.

Thus decane can be the most miscible species in benzene.

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Answer:

100

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Answer:

1. C. no change

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3. E. shift to the right

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5. E. shift to the right

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Explanation:

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From the equation: C(s) + H2O(g) ⇌ CO(g) + H2(g),

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1. If the pressure of the system is increased, there would be no change to the system because there are equal number of moles of products and reactants.

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3. If H2 concentration is decreased, the equilibrium will shift to the right to annul the effects of the decrease in the concentration of a product.

4. If the concentration of H2 is increased, the equilibrium will shift to the left to annul the effects of increased concentration of a product. Hence, more H2O would be formed.

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6. If the amount of C (a reactant) is increased, the equilibrium will shift to the right. Hence, more H2 will be formed.

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9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

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