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BaLLatris [955]
4 years ago
11

What happens to the saturation for when adding salt to water at room temperature

Physics
1 answer:
saw5 [17]4 years ago
5 0
Saturated and Unsaturated Solutions
Table salt (NaCl) readily dissolves in water. Suppose that you have a beaker of water to which you add some salt, stirring until it dissolves. So you add more and that dissolves. You keep adding more and more salt, eventually reaching a point that no more of the salt will dissolve no matter how long or how vigorously you stir it. Why? On the molecular level, we know that action of the water causes the individual ions to break apart from the salt crystal and enter the solution, where they remain hydrated by water molecules. What also happens is that some of the dissolved ions collide back again with the crystal and remain there. Recrystallization is the process of dissolved solute returning to the solid state. At some point the rate at which the solid salt is dissolving becomes equal to the rate at which the dissolved solute is recrystallizing. When that point is reached, the total amount of dissolved salt remains unchanged. Solution equilibrium is the physical state described by the opposing processes of dissolution and recrystallization occurring at the same rate.
While this shows the change of state back and forth between solid and aqueous solution, the preferred equation also shows the dissociation that occurs as an ionic solid dissolves.



When the solution equilibrium point is reached and no more solute will dissolve, the solution is said to be saturated. A saturated solution is a solution that contains the maximum amount of solute that is capable of being dissolved. At 20°C, the maximum amount of NaCl that will dissolve in 100. g of water is 36.0 g. If any more NaCl is added past that point, it will not dissolve because the solution is saturated. What if more water is added to the solution instead? Now more NaCl would be capable of dissolving in the additional solvent. An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved.

When 30.0 g of NaCl is added to 100 ml of water, it all dissolves, forming an unsaturated solution. When 40.0 g is added, 36.0 g dissolves and 4.0 g remains undissolved, forming a saturated solution.

How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.
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Explanation:

As the context of the data is not available, thus following data is utilized from the question as attached above

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Force on the block is given as

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F_m=0.71 N

Net Force

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F=1 N

As Ft=Δp

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Δp=1x0.07=0.07 kgm/s

Δp=p_final-p_initial

0.07=p_final-0

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v_f=(p_final)/(m)

v_f=0.07/0.073

v_f=0.95 m/s

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time is 0.07 s

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y_new=0.12+(0.95 x 0.07)

y_new=0.12+0.065

y_new=0.1865 m

So the new position is 0.1865 m

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Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.  

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Step Process

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Where V₀ = 6m/s,

We have,

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                 = 0.411 m

Where V₀ = 9.5m/s

We have,

     h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]

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Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s

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