9. Si en una probeta colocamos 25 ml de aceite, 50 ml de agua y 20 g de mercurio y mezclamos las sustancias. ¿Qué sustancia queda en la

superficie?

¿Cuál en la Interfase?

y ¿Cuál en el fondo de la

probeta?

DEL ESTADO DE

BAJA CALIFORNIA

the order of liquids from bottom to top is    Mercury - Water - Oil

Explanation:

In general, when several substances are mixed we can have two cases, that the substances dissolve some enter and that the substances have not been miscible (insoluble) this is our case, the three substances do not dissolve.

Therefore they must be deposited according to the principle of Archimedes,

           B -W = 0

if the thrust (B) of the liquid is greater than the weight of the upper body, the body floats and if less, the body sinks.

           B = ρ_{liquid1}   g  V_{liquid2}

           W = m_{liquid2} g

substance densities are tabulated

ρ_{oil} = 750 to 950 kg / m3

ρ_ water = 1.00 10³ kg / m³

ρ_ Hg = 13.6 10³ kg / m #

The thrust created by the mercury is

          B_Hg = ρ_Hg g V_ Liquid2

          B_Hg = 13.6 10³  9.8 V

          B _Hg = 133.3 10³ V

The push of the water

          B_water = 1  10³ 9.8 V

          B_water = 9.8 10³ V

The oil density, let us use ρ_oil =  ​​0.75 10³ kg / m³

          B_ oil = 0.750 10³  9.8 V

          B _ oil = 7.35 10³ V

let's find the volumes of the bodies

Mercury

          ρ = m / V

         V_hg = m /ρ_Hg

          V_Hg = 20 10⁻³/ 13.6 10³

          V_Hg = 1.47 10⁻⁶ m³

Water

          V_water = 50 10⁻³ L

we reduce from liter to m³ 1000 l = 1 m³

           V_water = 50 10⁻⁶ m³

Oil

          V_oil = 25 10⁻³ l

           V_oil = 25 10⁻⁶ m³

Let's look for the push for every two substances

The mercury  in the bottom and upper  sustabce the water and oil  

           B_hg = 133.3 10³ (V_water + V_oil)

           B = Hg = 133.3 10³ (50 + 25) 10⁻⁶

           B = 10 N

the weight of substance on mercury is

          W = W_water + W_oil

          W = ρ_water g V_water + ρ_oil ​​g V_oil

          W = g (ρ_water V_water + ρ_oil ​​V_oil)

          W = 9.8 (1 10³ 50 10⁻⁶ + 0.75 10³ 25 10⁻⁶)

          W = 9.8 68.75 10⁻³

          W = 6.7 10⁻¹ N

We can be that the weight of this substance in less than the push created by the mercury therefore they float.

In resume the mercury is the substance that will be at the bottom of the cylinder.

Let's repeat the calculation for water and oil.

Suppose the water is below

            B₂ = ρ_water g V_oil

            B₂ = 1 10³ 9.8 25 10⁻⁶

            B₂ = 245 10⁻³ N

            B₂ = 2.5 10⁻¹ N

The weight of the oil

            W = ρ_oil ​​g V_oil

            W = 0.750 10³ 9.8 25 10⁻⁶

             W = 183.75 10⁻³

            W = 1.8 10⁻¹ N

we see that the thrust of the water is greater than the weight of the oil therefore the oil floats in the water

In summary the order of liquids from bottom to top is

Mercury - Water - Oil

i cant see the choices man

your answer is, orbit around the nucleus