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ExtremeBDS [4]
1 year ago
5

The mole fraction of A (XA ) in the vapor phase of a mixture of two liquids is 0.24 and the sum of the partial pressures is 740

torr. Calculate the mol fraction of B (report to 2 decimal places) in liquid phase if the vapor pressure of pure B (PB0) 800 Torr. Show calculations.
Chemistry
1 answer:
Yakvenalex [24]1 year ago
4 0

Answer:

Mole fraction of B is 0.76

Explanation:

The pressure of a mixture of gases using vapor pressure and mole fraction of each gas is:

P = P°aₓXa + P°bₓXb + ... + P°nₓXn

<em>Where P is pressure of the system, P° is vapor pressure of pure gas and X is its mole fraction.</em>

In the problem, the pressure of the mixture of two gases is 740torr, mol fraction of A is 0.24 and vapor pressure of B is 800torr, that is:

740torr =  P°aₓ0.24 + 800torrₓXb

Also, the sum of mole fractions for each of the compounds in the mixture is 1, that is:

1 = Xa + Xb

As Xa = 0.24

1 = 0.24 + Xb

1-0.24 = Xb

0.76 = Xb

<em>Mole fraction of B is 0.76</em>

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BARSIC [14]

Answer:

Dust and smoke.

Explanation:

Dust and smoke are two different particles present in the air. Dust and smoke are different from one another due to their origin. Smoke formed from burning of materials while dust refers to the soil particles lifted by the wind due to their light weight. Dust and smoke are similar to each other due to their small in size, infinite number means uncountable and light weight.

6 0
2 years ago
NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including f
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First, let's write down the balanced chemical reaction between the given reactants:

NO₂ + NO → N₂O + O₂

The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:

Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)

For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1

6 0
1 year ago
Procaine hydrochloride ( = 272.77 g/mol) is used as a local anesthetic. Calculate the molarity of a 4.666 m solution which has a
Mkey [24]

Procaine hydrochloride ( = 272.77 g/mol) is used as a local anesthetic. Calculate the molarity of a 4.666 m solution which has a density of 1.1066 g/ml.

molarity = Moles of solute /  volume of solution

          Molarity = m d / [ 1 + (mW / 1000)]

Molarity = 4.666 X 1.1066 / [ 1 + (4.666 X 272.77 / 1000)]

Molarity =  5.16 / 2.272= 2.271 M

5 0
2 years ago
Read 2 more answers
A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of th
valkas [14]

Answer:

a. Ca₃(PO₄)₂.

b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. 3.1g of Ca₃(PO₄)₂

d. Percent yield = 93.5%

Explanation:

a. Based on the reaction:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)

<em>3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.</em>

<em />

As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,

<h3>Ca₃(PO₄)₂</h3><h3 />

b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate

0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =

<h3>0.010 moles of Ca₃(PO₄)₂ can we expect to be produced</h3><h3 />

c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;

0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =

<h3>3.1g of Ca₃(PO₄)₂</h3><h3 />

d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

\frac{2.9g}{3.1g} *100

<h3>Percent yield = 93.5%</h3>
7 0
2 years ago
COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at
mestny [16]

<u>Answer:</u> The value of K_p for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of COCl_2 = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

                  COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial:            1                    -         -

At eqllm:       1-x                 x        x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

[(1 - x) + x+ x]=1.2\\\\x=0.2atm

The expression for K_p for above equation follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm

Putting values in above equation, we get:

K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05

Hence, the value of K_p for the reaction at 690 K is 0.05

3 0
1 year ago
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