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Feliz [49]
2 years ago
8

A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper

ature is 74.61 mmHg and its density is 0.87865 g/cm3; the vapor pressure of pure carbon tetrachloride is 91.32 mmHg and its density is 1.5940 g/cm3. If this solution is ideal, its total vapor pressure at 20°C is
Chemistry
1 answer:
finlep [7]2 years ago
7 0

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

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Answer:

See explanation

Explanation:

A <u>trans</u> alkene is more stable than a <u>cis alkene</u> because they have fewer steric interactions.

<em>⇒ In a cis alkene there is steric hindrance, because the methyl groups are on the same side of the double bond. </em>

<em>Because of this steric crowding, there are van der Waals repulsive forces between the electron clouds of the groups. </em>

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In an elimination reaction, a geometry where the β hydrogen and the leaving group are on opposite sides of the molecule is called <u>anti</u> periplanar.

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In an <u>SN1 </u>mechanism, a nucleophile attacks the carbocation, forming a substitution product,

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6 0
2 years ago
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Answer: 4.86 x 10⁻³ mole.

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8 0
3 years ago
How many atoms are in 73.9g of potassium oxide
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Answer: 4.69(10)^{23} atoms

Explanation:

Firstly, we have to find the Molecular mass of potassium oxide (K_{2}0):

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This means that in 1 mole of K_{2}O there are 94 g and we need to find how many moles there are in 73.9 g K_{2}O:

1 mole of K_{2}O-----94 g of K_{2}O

X-----73.9 g of K_{2}O

X=\frac{(73.9 g)(1 mole)}{94 g}

X=0.78 mole This is the quantity of moles in 73.9 g of potassium oxide

Now we can calculate the number of atoms in 73.9 g of potassium oxide by the following relation:

N_{atoms}=(X)(N_{A})

Where:

N_{atoms} is the number of atoms in 73.9g of potassium oxide

N_{A}=6.0221(10)^{23}/mol is the Avogadro's number, which is determined by the number of particles (or atoms) in a mole.

Then:

N_{atoms}=(0.78 mole)(6.0221(10)^{23}/mol)

N_{atoms}=4.69(10)^{23} atoms This is the quantity of atoms in 73.9g of potassium oxide

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3 years ago
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