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svetoff [14.1K]
2 years ago
10

Which element has fewer than 4 dots In its Electron dot diagrams

Chemistry
1 answer:
nadya68 [22]2 years ago
4 0

Answer:Strontium (Sr) and Gallium (Ga)

Explanation:

Strontium (Sr) and Gallium (Ga) are the elements which have fewer than four dots in the electron dot diagrams. Explanation: Electron dot diagrams are the diagrams which represent the valence electrons in an element. The electrons are represented by the dots in these diagrams.

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  • Boiling point of the sugar solution will be higher than that of water's boling point.
  • Freezing point of the sugar solution will be lower than that of water's freezing point.

Explanation:

  • Boiling point of a liquid is defined as temperature at which vapor pressure of liquid becomes equal to the atmospheric pressure.

Boiling point of solution is always higher than that of the pure solvent

Vapor pressure increases with increase in temperature which means sugar solution will be heated more to make vapor pressure equal to atmospheric pressure.

  • Freezing point is defined as temperature at which solid and liquid phase are at equilibrium or temperature at which vapor pressure of liquid becomes equal to the vapor pressure in its solid phase.

Freezing point of solution is always lower than that of the pure solvent.

Lower the temperature, lower will be the vapor pressure which sugar solution solution will get freeze at lower temperature than that of the water.

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Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
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Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

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