An ideal solution consisting of 79 wt% benzene (C6H6) and 21 wt% toluene (C7H8) was heated in a closed vessel to 50 ºC. What is the mole fraction of benzene in the vapour phase when equilibrium was reached at 50 ºC? Round your answer to two significant figures.



Data: Vapour pressure of benzene at 50 ºC = 271 mmHg

Vapour pressure of toluene at 50 ºC = 91.5 mmHg

Answers

Answer 1

In this case, according to the Raoult's law, we can find the mole fraction of 79 wt% benzene and 21 wt% toluene at 50 °C in the vapor phase as follows:

[tex]y_iP=x_iP_i[/tex]

However, we first have to calculate the mole fractions in the solution as follows (b stands for benzene and t for toluene):

[tex]\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{(78.11*x_b+92.14*x_t)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene} \\\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{78.11*x_b+92.14*(1-x_b)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene}\\\\x_b=0.79*(x_b+1.18*(1-x_b))\\\\x_b=0.79x_b+0.932-0.932x_b\\\\x_b=\frac{0.932}{1+0.932-0.79} =0.816\\\\x_t=1-x_b=1-0.816=0.184[/tex]

Next, we calculate the total pressure as follows, according to the Dalton's law:

[tex]P=x_bP_b+x_tP_t=0.816*271mmHg+0.184*91.5mmHg=237.972mmHg[/tex]

Finally, the mole fractions of the vapor phase turn out:

[tex]y_b=\frac{0.816*271mmHg}{237.972mmHg}=0.929\\\\y_t=1- 0.929=0.071[/tex]

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five differences between true solution and false solution​

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A very long tube with a cross-sectional area of 1.00 cm2 is filled with mercury to a height of 76.0 cm. At what height would water stand in this tube if it were filled with a mass of water equal to that of the mercury? (The density of mercury is 13.60 g/cm3 and the density of water is 1.00 g/cm3.)

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The water would stand at a height of 1033.6 cm

If the mass of the mercury, m equals the mass of the water, m' at its height,h', then the weight of the mercury W equals the weight of the water, W'

So, W = W'

mg = m'g

ρV = ρ'V' where ρ = density of mercury = 13.60 g/cm³, V = volume of mercury = Ah where A = cross-sectional area of tube = 1.00 cm and h = height of mercury = 76.0 cm, ρ' = density of water = 1.00 g/cm³, V = volume of water = Ah' where A = cross-sectional area of tube = 1.00 cm and h' = height of water.

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What is the concentration of chloride ions in solution after the reaction of 5.8 mL 0.12 M ammonium chloride with 3.2 mL 0.21 M silver nitrate

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Answer:

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A piece of potato had an original mass of 3.2g and after being left in a sugar solution for 24 hours it then had a mass of 2.8g. Calculate the percentage mass decrease of the potato.

Answers

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Explanation:

A piece of potato had an original mass of 3.2g and after being left in a sugar solution for 24 hours it then had a mass of 2.8g. Calculate the percentage mass decrease of the potato.

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or  -13% to 3 sig figs

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Answer:

213 kJ-1704 kJ852 kJ (ignore what's on my paper, accidentally added a negative)

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These problems are pretty easy, you just need to do to the enthalpy what is done to the equation itself. Numbers one and 3 are reversed, so they become positive values. When a coefficient changes in the equation, you multiply the enthalpy by that number. I have attached my work below. Let me know if there is any confusion left.

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Answer:

A student performed titration to find the concentration of acetic acid. Titrant was sodium hydroxide. He went beyond the indicator color change by adding too much titrant.

Explanation:

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Answer:

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Answer:

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Answer:

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Answer :

[tex] \: [/tex]

Concentration of Solutions

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Answers

Answer:

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63

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Explanation:

Our Direct Nanofiltration (dNF) membranes offer the unique combination of a low fouling hollow fiber configuration with the ability to remove organics and salinity (hardness) from water in one simple step. Other than a strainer, no further pre-treatment is required. We manufacture our membranes with a patented layer-by-layer process, where multiple nano-scale layers are deposited on a membrane support. This method enables very precise and controlled rejection and flux properties of the membrane. A unique solution for troubled water.

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Answers

Answer:

Rb, Zn, P, S, F, Ca, Co, Cr

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Answers

Answer:

-106°C

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1558 mmHg / 1 × 1 atm / 760. mmHg = 2.05 atm

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Answers

Answer:

Explanation:

Determine the treatments you’ll be using on your eggs, and prepare the substances you’ll need. You can make salt-water solutions by dissolving different amounts of table salt in containers of water (e.g. 100g, 200g, 300g of salt (NaCl) per liter). You can make solutions of food coloring by adding a few drops of each color into containers of  

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Answers

Answer:

Use the Rydberg Equation 1/λ = R(1/n²(final) - 1/n²(initial) where R = 109,678 cm⁻¹. Substitute initial orbital number and final orbital number and solve for wavelength (λ).

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magnesium salt is the type of substance

Answer:

MgCO 3 is an inorganic salt with chemical name Magnesium Carbonate. It is also called Magnesite or Hydromagnesite or Barringtonite. Hydrated forms of magnesite such as di, tri, tetrahydrates are present as minerals. It acts as a fertilizer and as an antacid.

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A: Heat Energy
B: Potential Energy
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D: Kinetic Energy​

Answers

Answer:

D: Kinetic Energy​

Explanation:

Temperature is a measure of kinetic energy. Temperature is a measure of the average kinetic energy of the particles in an object. When temperature increases, the motion of these particles also increases.

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perdon no se es que es para que me dé ingreso

A 1.8 mole sample of a compound weighs 195 g and is found to be 11.18% H and 88.82% C. What is the molecular formula for the compound?

Answers

Answer:

C8H12

Explanation:

First find the molar mass

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Then the molecular formula

Mass of C in 1 mole of compound = 88.82×108.3/100

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Mass of H in 1 mol of.the compound = 108.3×11.18/100 = 12.11 g

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Answers

The alcohol content is 450 proof

In the given example, the measurement of the alcohol content is said to be measured in alcohol proof.

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Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of
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Answers

Answer:

[tex]2al + 3h2so4 = > al2(so4)3 + 3h2 \\ eqvt \\ = > \frac{5.4}{27} = > \frac{5 \times 50}{1000} \\ .2 \: mol = > .25 \: mol \\ \\ so \: .25 \: is \: limiting \: reagent \\ according \: to \: ideal \: gas \: equation \\ pv = nrt \\ v = \frac{nrt}{p} \\ v = \frac{.25 \times .082 \times 300}{1} \\ = \frac{1}{4} \times 8.2 \times 3 \\ = 2.05 \times 3 \\ = 6.15litre \\ thank \: you[/tex]

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