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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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Firlakuza [10]

Answer:

1807.24L

Explanation:

Using combined gas law equation:

P1V1/T1 = P2V2/T2

Where;

P1 = pressure on Earth

P2 = Pressure on Mars

V1 = volume on Earth

V2 = volume on Mars

T1 = temperature on Earth

T2 = temperature on Mars

According to the information provided of the balloon in this question;

P1 = 1 atm

P2 = 4.55 torr = 4.55/760 = 0.00599atm

V1 = 14.5L

V2 = ?

T1 = 19°C = 19 + 273 = 292K

T2 = -55°C = -55 + 273 = 218K

Using P1V1/T1 = P2V2/T2

1 × 14.5/292 = 0.00599 × V2/218

14.5/292 = 0.00599V2/218

Cross multiply

14.5 × 218 = 292 × 0.00599V2

3161 = 1.74908V2

V2 = 3161 ÷ 1.74908

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3 0
2 years ago
Help me please I need help
yanalaym [24]
37. Is D
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The half-life of 131I is 8.021 days.What fraction of a sample of 131I remains after 24.063 days
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24.063/8.021 = 3 half lives
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3 years ago
CORRECT ANSWER GETS BRAINLIEST! PLEASE HELP AND EXPLAIN!!
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Answer:

A car stopped at the top of the hill

Explanation:

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3 years ago
A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the
Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}

R is Gas constant having value = 0.008314 kJ / K mol  

Given temperature, T = 293 K

Given, \Delta{G^0}=15\ kJ/mol

So,  Applying in the equation as:-

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

Thus,

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}

\frac{[His]}{[His+]}=0.00211

Also, considering:-

pH=pKa+log\frac{[His]}{[His+]}

Given that:- pH = 7.0

So, 7.0=pKa+log0.00211

<u>pKa of the histidine = 9.67</u>

8 0
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