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igomit [66]
3 years ago
11

A solution of 3 moles of carbon disulfide in 10 moles of ethyl acetate has a vapor pressure of 170 torr at 28°c. the vapor press

ure of pure carbon disulfide at 28°c is 390 torr. what is the vapor pressue of pure ethyl acetate at 28°c
Chemistry
1 answer:
nika2105 [10]3 years ago
7 0
Total pressure of solution = (P⁰carbon sulfide * X carbon sulfide) + (P⁰ethyl acetate * X ethyl acetate)
where:
P⁰ carbon sulfide is vapor pressure of pure Carbon sulfide
X carbon sulfide is mole fraction of carbon sulfide = (number of moles of carbon sulfide / total number of moles)
P⁰ ethyl acetate is vapor pressure of pure ethyl acetate
X ethyl acetate is mole fraction of ethyl acetate = (number of moles of ethyl acetate / total number of moles)

P soln = (P⁰CS₂ * X CS2) + (P⁰ethyl acetate * X ethyl acetate)
170 torr   = (390 torr * 3 / 13) + (P⁰ ethyl acetate * 10 / 13 )
P⁰ ethyl acetate = 104 torr


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A sample of nitrogen gas has a volume of 478 cm3 and a pressure of 104.1 kPa. What volume would the gas occupy at 88.2 kPa if th
nadezda [96]

<span>1.    </span>To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

 V2 = 104.1 x 478 / 88.2

<span> V2 =564.17 cm^3</span>

6 0
3 years ago
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What are essential and non essential proteins ?
katen-ka-za [31]
Nonessential amino acids can be made by the body, while essential amino acids cannot be made by the body so you must get them from your diet.
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6 0
3 years ago
What is capillary action? Describe how capillary action shown by water is useful for plants and animals on Earth.
Shtirlitz [24]

Answer:

Look Below

Explanation:

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3 years ago
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If 12g nitrogen gas,0.40 of H2 gas and 9.0 gram of oxygen are put into 1 litre container of. 27°C what is the total pressure on
Vadim26 [7]

Answer:

Total pressure = 27.35 atm

Explanation:

Given data:

Mass of nitrogen = 12 g

Mass of H₂ = 0.40 mol

Mass of oxygen = 9.0 g

Volume of Container = 1 L

Temperature = 27 °C (27+273 = 300 K)

Total Pressure = ?

Solution:

First of all we will calculate the number of moles of individual gas.

Number of moles = mass/ molar mass

Number of moles = 12 g/ 28 g/mol

Number of moles = 0.43 mol

Pressure of N₂:

PV = nRT

P = nRT/V

P = 0.43 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 10.6 atm

Number of moles of Oxygen:

Number of moles = mass/ molar mass

Number of moles = 9 g/ 32 g/mol

Number of moles = 0.28 mol

Pressure of O₂:

PV = nRT

P = nRT/V

P = 0.28 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 6.9 atm

Pressure of H₂:

PV = nRT

P = nRT/V

P = 0.40 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 9.85 atm

Total pressure:

Total pressure = Pressure of H₂ + Pressure of O₂ + Pressure of N₂

Total pressure = 9.85 atm + 6.9 atm + 10.6 atm

Total pressure = 27.35 atm

8 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
EleoNora [17]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

4 0
3 years ago
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