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

Substances that cannot be separated and found on a periodic table

GaryK [48]

Substances that cannot be separated and found on a periodic table are elements.

4 0

3 years ago

Read 2 more answers

What is the H+ in solution with pH of 10.9

Soloha48 [4]

Answer:

Explanation:

has a pH of 6.6, then what is the H3O+ in solution X? View Answer · What is the pOH of a solution in which (H+)

8 0

3 years ago

Which are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3C―CH3 sp sp2 sp3 (b) H3C―CH═CH2 sp sp2 sp3 (

kherson [118]

Answer:

(a) sp³ sp³

H₃C - CH₃

(b) sp³ sp²

H₃C - CH = CH₂

sp²

H₃C - C ≡ C - CH₂OH

sp sp³

(d) sp³ sp²

H₃C - CH=O

Explanation:

Alkanes or the carbons with all the single bonds are sp³ hybridized.

Alkenes or the carbons with double bond(s) are sp² hybridized.

Alkynes or the carbons with triple bond are sp hybridized.

Considering:

(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.

Hence,

sp³ sp³

H₃C - CH₃

(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.

Hence,

sp³ sp²

H₃C - CH = CH₂

sp²

(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.

Hence,

sp³ sp

H₃C - C ≡ C - CH₂OH

sp sp³

(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.

Hence,

sp³ sp²

H₃C - CH=O

8 0

3 years ago

An analytical chemist has determined by measurements that there are 96.5 moles of carbon in a sample of acetic acid. how many mo

Nadya [2.5K]

The formula of acetic acid is CH3COOH => C2H4O2.

So, the acetic acid has the same number of atoms of carbon (C) than of oxygen (O).

Therefore, the sample that contains 96.5 moles of carbon, will contain also 96.5 moles of O.

Answer: 96.5 moles of oxygen.

3 0

3 years ago

The standard heats of formation for CO2(g), C2H6(g), and H2O(l) are -394.0 kJ/mol, -84.00 kJ/mol, and -286.0 kJ, respectively. W

Ivanshal [37]

Answer:

ΔH°r = -1562 kJ

Explanation:

Let's consider the following combustion.

C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)

We can calculate the standard heat of reaction (ΔH°r) using the following expression:

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(i) are the standard heats of formation of reactants and products

The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]

ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]

ΔH°r = -1562 kJ

6 0

3 years ago