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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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lisov135 [29]

Answer:

0.07172 L = 7.172 mL.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT. </em>

where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 273 K, Standard T).

<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>

8 0
3 years ago
A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r
aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:

  • V₁n₂=V₂n₁

Where in this case:

  • V₁ = 6.13 L
  • n₂ = ?
  • V₂ = 11.3 L
  • n₁ = 2.51 mol

We <u>input the data</u>:

  • 6.13 L * n₂ = 11.3 L * 2.51 mol
  • n₂ = 4.63

As <em>4.63 moles is the final number of moles</em>, the number of moles added is:

  • 4.63 - 2.51 = 2.12 moles
5 0
3 years ago
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


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Answer:

Its B

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Products <=> reactants; products can go back to reactant and reactants can form products, depending on the equilibrium position.

The equilibrium position, if it is on the left, and much of the product is formed. If the equilibrium position is not the right, then its the vice versa The equilibrium position is determined by ;

  • <em><u>Temperature</u></em>

If the reaction is endothermic, it will be favoured by increase in temperature and equilibrium position will shift to the right ( reactants )

If the reaction is exothermic, its the vice versa

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