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Andreas93 [3]
3 years ago
11

A partial molar property of a component in a mixture may be either greater than or less than the corresponding pure-component mo

lar property. Furthermore, the partial molar property may vary with composition in a complicated way. Show this to be the case by computing (a) the partial molar volumes and (b) the partial molar enthalpies of ethanol and water in an ethanol-water mixture. (The data that follow are from Volumes 3 and 5 of the International Critical Tables, McGraw-Hill, New York, 1929.)
Chemistry
1 answer:
ioda3 years ago
4 0

Answer:

Partial molar volumes help to assess the influence of pressure on phase equilibria or reaction equilibria. The prediction of partial molar volumes or excess volumes is a sharp test for theories of the fluid state.

On the other hand, the interpretation of partial molar volumes or excess volumes is usually difficult, if not impossible. Many mixtures exhibit positive and negative excess volumes, depending on composition, temperature, and pressure. Speculations why some mixture exhibits positive or negative excess volumina are futile, particularly if they are based on measurements at ambient pressure and temperature only.

Explanation:

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44.90 g of reactants are placed in a beaker
sergiy2304 [10]

Answer:

D. -1882J

Explanation:

We can solve the energy released in a chemical reaction in an aqueous medium using the equation:

Q = -m*C*ΔT

<em>Where Q is energy (In J),</em>

<em>m is mass of water (45.00g)</em>

<em>C is specific heat of water (4.184J/g°C)</em>

<em>And ΔT is change in temperature (25.00°C - 15.00°C = 10.00°C)</em>

<em />

Replacing:

Q = -45.00*4.184J/g°C*10.00°C

Q = -1882J

Right answer is:

<h3>D. -1882J</h3>

<em />

7 0
3 years ago
Calculate the freezing point of a solution containing 15 grams of kcl and 1650.0 grams of water. the molal-freezing-point-depres
Soloha48 [4]
Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.
4 0
3 years ago
Read 2 more answers
A gas initial volume and pressure is 5m^3 and 101320 Pa and allows to expand up to 12m^3 then what is the final pressure?
Anton [14]

Answer:

New pressure = 42216.66 Pa

Explanation:

Given that,

Initial volume, V₁ = 5 m³

Final pressure, P₁ = 101320 Pa

Final volume, V₂ = 12 m³

We need to find the final pressure of the gas. We know that the relation between pressure and volume is given by :

P\propto \dfrac{1}{V}\\\\P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{101320 \times 5}{12}\\\\P_2=42216.67\ Pa

So, the new pressure is equal to 42216.66 Pa.

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True would be the answer
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I just need the parallel, please help:)
Lynna [10]

Answer:

1. seperate paths

Explanation:

5 0
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