All categories

3 years ago

Which of these processes involve the weakening of the attraction between particles

Chemistry

1 answer:

AveGali [126]3 years ago

3 0

Answer : The correct option is (3) Evaporation.

Explanation :

Evaporation : It is a process in which the liquid particles are converted to gaseous particles. The inter-molecular forces of attraction between the liquid particles will be weakened. As a result the phase change occurs.

Freezing : It is a process in which a phase changes from liquid state to solid state.

Condensation : It is a process in which a phase changes from gaseous state to liquid state.

Crystallization : It is a process in which a solute particles get separated from the solvent in the form of crystals ( in solid phase).

Whereas in freezing, condensation and crystallization we are moving from one phase having weaker inter-molecular forces of attraction to another phase having stronger inter-molecular forces of attraction. So, in these processes strengthening of attraction between the particles takes place.

Hence, the correct answer is Evaporation.

You might be interested in

Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)

miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

Adding solute to water elevates the boiling point.

The elevation in boiling point (ΔTb) can be calculated using the relation:

ΔTb = i.Kb.m,

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

(1) 1.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

(2) 2.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

(3) 1.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

(4) 2.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).

6 0

3 years ago

A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0g sample of a pure substance from to −

astraxan [27]

Answer:

C = 0.2349 J/ (g °C)

Explanation:

Mass, m = 894.0g

Initial Temperature = −5.8°C

Final Temperature = 17.5°C

Temperature change = 17.5°C - (−5.8°C) = 23.3

Heat, H = 4.90kJ = 4900 J

Specific heat capacit, C = ?

The relationship between these quantities is given by the equation;

H = mCΔT

C = H / mΔT

C = 4900 / (894)(23.3)

C = 0.2349 J/ (g °C)

4 0

3 years ago

A sample of the amino acid tyrosine is barely soluble in water. Would a peptide containing only a string of Tyr residues be more

Misha Larkins [42]

Answer:

Tyrosine is a polar and aromatic compound. its side chain acidity and basicity is neutral

if a peptide contain only a string of tyrosine residue especially l tyrosine the solubility increases more

Explanation:

even tyrosine number remains constant, tyrosine containing peptide will be more soluble. This peptide is soluble in 1 M HCl (100 mg/ml), with heating. The solubility in water (25 °C) is 0.45 mg/ml in the pH range 3.2 - 7.5.

2.0 mg/ml; at pH 9.5, the solubility is 1.4 mg/ml; and at pH 10, the solubility is 3.8 mg/ml.

5 0

3 years ago

A sample of monochloro pentaborane (9), b5h8cl, has a mass of 20.1 g. what amount, in moles, does this mass represent?

dolphi86 [110]

The number of moles in monochloro pentaborane (9) $B_5H_8Cl$ is 0.37056.