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dezoksy [38]
3 years ago
14

Consider two flasks at 25 degree Celsius, one contains an ideal gas and the other contains the real gas SO3. Which statement reg

arding these gases is true?
A. As the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of SO3 because ideal gases lack inter-molecular forces.
B. As the temperature is decreased, the ideal gas will liquefy first because ideal gases have larger values of the van der Waals coefficient b.
C. As the temperature approaches 0 K, the volume of the ideal gas will be smaller than the volume of SO3 because ideal gases have larger values of the van der Waals coefficient a.
D. After the temperature has been lowered about 100 K, the pressure of the ideal gas will be smaller than the pressure of SO3 because the van der Waals coefficient a for SO3 is large.
E. After the temperature is increased about 100 K, the pressure of the ideal gas will be smaller than the pressure of SO3 because the van der Waals coefficient a for SO3 is lar
Chemistry
1 answer:
Strike441 [17]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

In real gases, there exists force of attraction between the molecules at low temperature and high pressure. This is because at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.

As a result, forces of attraction increases as molecules come closer to each other and therefore, gases deviate from an ideal gas behavior.

And, at low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas because they have high kinetic energy. Hence, gases behave ideally at these conditions.

Thus, we can conclude that the statement as the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of SO_{3} because ideal gases lack inter-molecular forces, is true.

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In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

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A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. how much calcium is contained in 45.0 g of c
kolbaska11 [484]

In a sample of pure calcium fluoride of mass 15.0 g, 7.70 g of calcium is present. First convert the mass into number of moles as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Molar mass of Ca is 40 g/mol, putting the values,

n=\frac{7.70 g}{40 g/mol}=0.1925 mol

Similarly, molar mass of CaF_{2} is 78.07 g/mol thus, number of moles will be:

n=\frac{15.0 g}{78.07 g/mol}=0.1921 mol.

Thus, 0.1921 mol of CaF_{2} have 0.1925 mol of Ca, or 1 mole of CaF_{2} will have approximately 1 mole of Ca.

Now, mass of Ca needs to be calculated in 45.0 g of CaF_{2}. Converting mass into number of moles first,

n=\frac{45.0 g}{78.07 g/mol}=0.5764 mol

Thus, number of moles of Ca will also be 0.5764 mol, converting number of moles into mass,

m=n\times M=0.5764 mol\times 40 g/mol=23.06 g

Therefore, mass of Ca will be 23.06 g.

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