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GuDViN [60]
3 years ago
12

In trying to control fall armyworms in crops, an Agriculture extension officer applied cypermethrin which was prepared by dissol

ving 200g of the cypermethrin , C22H19Cl2NO3 in 1000g of water H2O . Calculate the mole fraction of cypermethrin in the solution.
Chemistry
1 answer:
sveta [45]3 years ago
8 0

Answer:

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.0086

Explanation:

Mole fraction remains a sort of concentration. It indicates:

moles of solute / (moles of solute + moles of solvent)

Moles of solute / Total moles.

Solute: Cypermethrin → C₂₂H₁₉Cl₂NO₃

Solvent: Water (PM = 18g/mol)

We calculate moles from solvent: 1000g /18 g/mol = 55.5 moles

We calculate PM for C₂₂H₁₉Cl₂NO₃

12g/mol . 22 + 1g/mol . 19 + 35.45 g/mol . 2+ 14g/mol + 16g/mol . 3 = 416 g/m

Moles of solute: 200 g / 416g/mol = 0.481 moles

Total moles: 0.481 + 55.5 = 55.98 moles

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.481 moles / 55.98 moles = 0.0086

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True

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[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

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Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

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Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

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In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

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The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

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