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quester [9]
3 years ago
10

Are the strengths of the interactions between the particles in the solute and between the particles in the solvent before the so

lute and solvent are combined greater than, less than, or equal to the strengths of the interactions between solute particles and solvent particles after dissolution? Explain.
Chemistry
1 answer:
Colt1911 [192]3 years ago
5 0

Answer:

Less than

Explanation:

The process of dissolution occurs as a kind of "tug of war". On one side are the solute-solute and solvent-solvent interaction forces, while on the other side are the solute-solvent forces.

Only when the solute-solvent forces are strong enough to overcome the pre-mixing forces do they overcome the "tug of war", and thus dissolution occurs.

Thus, it is concluded that the interaction forces between solute particles and solvent particles before they are combined are less than the interaction forces after dissolution.

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pishuonlain [190]
Among the choices provided, the statement that correctly describes the rusting process below is that "Oxygen was reduced over the course of this reaction" as <span> iron can't be the oxidizing agent, because it is the one being oxidized.

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1. Land-dwelling creatures depend on the the water cycle to ___________ fresh water from its salt solution in the Earth's oceans
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1 is evaporate and 2 is precipitate
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At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0
3 years ago
Describe how the molecular structures of alkenes and alkynes differ from the structure of alkanes
evablogger [386]

Alkanes are saturated hydrocarbon that contains only single bonds, whereas Alkenes and Alkynes are unsaturated hydrocarbons which contain one or more double bond and triple bonds.

<u>Explanation:</u>

  • A saturated hydrocarbon with an only single bond is called alkanes. Ethane consisting of two carbon atoms that are bonded with a single bond and six hydrogen atoms sharing the other valence electron of carbon atoms. The molecular structure of alkane is CnH2n+2.
  • An unsaturated hydrocarbon with a two bond is called alkenes. Ethene consisting of two carbon atoms double-bonded to each other. The molecular structure of alkene is CnH2n.
  • An unsaturated hydrocarbon with a triple bond is called as alkynes. It involves sharing three pairs of electrons. The molecular structure of alkyne is CnH2n-2.
4 0
3 years ago
A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
3 years ago
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