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True [87]
4 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]4 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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4 years ago
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Why are the components of the mixture separated ? Give any 6 reasons........​
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We need to separate different components of a mixture to separate the useful components from the non-useful or some harmful components.

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(a) Tea leaves are separated from tea.

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3 years ago
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9. Let's say, you need to make 2.00 L of 0.05 M copper (II) nitrate solution. How many grams of
Vlad [161]

Answer:

18.76 g of copper II nitrate

Explanation:

Now recall that we must use the formula;

n= CV

Where;

n= number of moles of copper II nitrate solid

C= concentration of copper II nitrate solution

V= volume of copper II nitrate solution

Note that;

n= m/M

Where;

m= mass of solid copper II nitrate

M= molar mass of copper II nitrate

Thus;

m/M= CV

C= 0.05 M

V= 2.00 L

M= 187.56 g/mol

m= the unknown

Substituting values;

m/ 187.56 g/mol = 0.05 M × 2.00 L

m= 0.05 M × 2.00 L × 187.56 g/mol

m= 18.76 g of copper II nitrate

Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.

6 0
3 years ago
When two atoms with different electronegativities are bonded together, a bond dipole exists. These are displayed with an arrow o
____ [38]

Answer:

Dipole, less electronegativity,  higher electronegativity

Explanation:

!!! Incomplete Question

When two atoms with different electronegativities are bonded together, a bond dipole exists.

A bond dipole is a partial charge assigned to bonded atoms due to difference in electron density, difference in electronegativity is a factor to this.

These are displayed with an arrow originating at the atom with the less electronegativity and pointing toward the atom with the higher electronegativity value

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3 years ago
Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the den
NISA [10]

Answer:

\rho=1.54\ g/cm^3

Explanation:

The expression for density is:

\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}

N_a=6.023\times 10^{23}\ {mol}^{-1}

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

\rho is the density

Radius = 197 pm = 1.97\times 10^{-8}\ cm

Also, for fcc, Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm

Thus,  

\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3

\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3

\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3

\rho=\frac{160.312}{104.19483}\ g/cm^3

\rho=1.54\ g/cm^3

7 0
3 years ago
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