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Firlakuza [10]
3 years ago
10

A saturated solution of lead(II) nitrate was made at 50 °C using 100 g of water, and this was then allowed to cool to 30 °C. Pre

dict what would happen as the solution cooled?
Physics
2 answers:
olchik [2.2K]3 years ago
8 0
The solubility of the solution would decrease and some of the lead(II) nitrate would precipitate out 
Sunny_sXe [5.5K]3 years ago
4 0

Answer: The temperature of solution will decease along with the precipitation of solid lead nitrate in solution.

Explanation:

Solubility of the substance is directly effect by the temperature of the solvent.

This is because the kinetic energy of the solvent increase as temperature increase due to which solvent easily breaks apart the solute particles easily and faster which will result in solubility of more solute in solvent and vice versa.

So, when prepared saturated solution of lead nitrate at 50°C was allowed to cool down to 30°C one will observe that the precipitation of solid lead nitrate in solution. And this means that solubility of lead nitrate with decrease in temperature decreases as well.

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3 years ago
The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

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