All categories

3 years ago

How physical properties are important for identifying unknown substances

Physics

1 answer:

dmitriy555 [2]3 years ago

6 0

Answer:

Every characteristic property is unique to one given substance. Scientists use characteristic properties to identify an unknown substance. Characteristic properties are used because the sample size and the shape of the substance does not matter.

Explanation:

A characteristic property is a chemical or physical property that helps identify and classify substances. The characteristic properties of a substance are always the same whether the sample being observed is large or small. Examples of characteristic properties include freezing/melting point, boiling/condensing point, density, viscosity and solubility.

You might be interested in

A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i

Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0

3 years ago

Gravitational Potential Energy = mass x gravity x height

Blizzard [7]

M= gpe / gh
G= gpe / mh
H=gpe / mg

7 0

3 years ago

Read 2 more answers

A vessel that contains a gas has two pressure gauges attached to it. One contains liquid mercury, and the other an oil such as d

castortr0y [4]

Answer:

Pressure of the gas = 12669 (Pa) and height of the oil is 1,24 meters

Explanation:

First, we can use the following sketch for an easy understanding, in the attached image we can see the two pressure gauges the one with mercury to the right and the other one with oil to left. We have all the information needed in the mercury pressure gauge, so we can determine the pressure inside the vessel because the fluid is a gas it will have the same pressure distributed inside the vessel (P1).

Since P1 = Pgas, we can use the same formula, but this time we need to determine the height of the column of oil in the pressure gauge.

The result is that the height of the oil column is higher than the height of the one that uses mercury, this is due to the higher density of mercury compared to oil.

Note: the information given in the units of the fluids is not correct because the density is always expressed in units of (mass /volume)

4 0

3 years ago

Pyramid math (algebra 1)

mixas84 [53]

5436 7699 2360 3488 2113

3 0

3 years ago

Suppose Mary would like to design an emergency overheating alarm using one of these materials. In her design, at room temperatur

Reptile [31]

Answer:

A. 92.88 °C

B. 401.535 °C

Explanation:

So, the overheating system is going to be based on the principle of thermal linear expansion. The behavior of this principle is ruled by the equation:

∆L= L_i * ∆T * α

Where α is a coefficient of linear expansion. For a cylinder made from polycarbonate α = 70,2*10^(-6) °C^(-1) and for a cylinder made from cast iron α = 12*10^(-6) °C^(-1). If we isolate the term of the temperature’s difference, we have:

∆L/(L_i * α) = ∆T → T_f = T_i + ∆L/(L_i * α)

Replacing the values, for the case of the Polycarbonate we have:

T_f = T_i + ∆L/(L_i * α) = 23°C+0,0273cm/(6,01cm *70,2 * 10^(-6)°C^(-1) ) = 92,88 °C

Replacing the values, for the case of the Cast Iron we have:

T_f = T_i +∆L/(L_i * α) = 23°C + 0,0273cm/(6,01cm * 12 * 10^(-6) °C^(-1) ) = 401,535 °C

As we see, is way better to use the polycarbonate in this application.

Have a nice day. Let me know if I can help you with anything else. :D

4 0

3 years ago