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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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Answer:

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Explanation:

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Let's determine the moles of our reactants:

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1 mol of phosphate reacts with 3 mol of sulfuric so

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I have 0.554 of sulfuric, so this is the reactant in excess.

The limiting reagent is the Phosphate.

1 mol of PO₄⁻³ produces 2 mol of phosphoric

0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles

Mol . molar mass = Mass

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Percent yield = (Produced / Theoretical) .100

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Explanation:

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In this isotope, we can deduce that the mass number is the superscript and the atomic number is the subscript;

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Mass number is the number of protons and neutrons in an atom;

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Atomic number is the number of protons

   

So,  Number of protons  = 46

Number of neutrons  = Mass number  - Atomic number

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