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True [87]
3 years ago
12

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

aw. For Example, consider two components A and B to form non-ideal solutions. Let the vapour pressure, pure vapour pressure and mole fraction of component A be PA, PAo and XA respectively and that of component B be Ps, PB° and xB respectively. These liquids will show positive deviation when Raoult's Law when: a. PIA]> P[AO] times x[A] and PIB] > PIBO] times xIB], as the total vapour pressure (PIAO] XIA] + P[B0] x[B]) is greater than what it should be according to Raoult's Law. O b. The solute-solvent forces of attraction is weaker than solute-solute and solvent-solvent interaction O c. The enthalpy of mixing is positive because weaker binding forces or even repulsion are resulted O d. The volume of mixing is positive as weaker binding forces have led to an expansion in volume O a and b only O a, b and c only O All of the above
Chemistry
1 answer:
8090 [49]3 years ago
6 0

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's  Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure  of component A is P_A and vapour pressure of component B is P_B.

P^0_A= Vapour pressure of component A in pure form

P^0_B= Vapour pressure of component B in pure form

x_A=Mole fraction of component A

x_B==Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

P_A >P^0_A\cdot x_A,P_B>P^0_B\cdot x_B

Therefore, P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

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If the energy of photon emitted from the hydrogen atom is 4.09 x 10-19 J, what is
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Answer:

486 nm

Explanation:

From the question given above, the following data were obtained:

Energy (E) = 4.09×10¯¹⁹ J

Wavelength (λ) =?

Next, we shall determine the frequency of the photon. This can be obtained as follow:

Energy (E) = 4.09×10¯¹⁹ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

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4.09×10¯¹⁹ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 4.09×10¯¹⁹ / 6.63×10¯³⁴

f = 6.17×10¹⁴ Hz

Next, we shall determine the wavelength of the photon. This can be obtained as follow:

Frequency (f) = 6.17×10¹⁴ Hz

Velocity of photon (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 6.17×10¹⁴

Divide both side by 6.17×10¹⁴

λ = 3×10⁸ / 6.17×10¹⁴

λ = 4.86×10¯⁷ m

Finally, we shall convert 4.86×10¯⁷ m to nm. This can be obtained as follow:

1 m = 1×10⁹ nm

Therefore,

4.86×10¯⁷ m = 4.86×10¯⁷ m × 1×10⁹ nm / 1 m

4.86×10¯⁷ m = 486 nm

Therefore, the wavelength of the photon is 486 nm

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