Picture of molecular structure bodies at first glance does not agree with our usual experience: we do not observe these individual particles, the bodies seem to us to be continuous. However, this objection cannot be considered convincing. M. V. Lomonosov wrote in one of his works: “It is also impossible to deny movement where the eye does not see it; who will deny that the leaves and branches of trees move in a strong wind, although from a distance he will not notice any movement. As here, due to remoteness, so in hot bodies, due to the smallness of particles of matter, movement is hidden from view. So, the reason for the apparent disagreement is that atoms and molecules are extremely small.
In the best optical microscope, which makes it possible to distinguish objects whose dimensions are not less than , it is impossible to consider individual molecules, even the largest ones. However, a number of indirect methods made it possible not only to reliably prove the existence of molecules and atoms, but even to establish their sizes. So, the size of a hydrogen atom can be considered equal; the length of a hydrogen molecule, i.e., the distance between the centers of the two atoms that make it up, is equal to. There are larger molecules, for example, protein molecules (albumin) are . In recent years, thanks to the device of a special device that makes it possible to study objects of extremely small sizes - an electron microscope - it has become possible to photograph not only large molecules, but also atoms.
The fact that the sizes of molecules are extremely small can be judged even without measurements, based on the possibility of obtaining very small amounts of different substances. By diluting ink (for example, green) in a liter of pure water, and then diluting this solution again in a liter of water, we will get a dilution of one time. Nevertheless, we will see that the latter solution has a noticeable green color and at the same time is completely homogeneous. Therefore, in the smallest volume that the eye can still discern, even at this dilution, there are a lot of dye molecules. This shows how small these molecules are.
Gold can be flattened into sheets of thickness, and by treating such sheets with an aqueous solution of potassium cyanide, sheets of gold of thickness can be obtained. Therefore, the size of a gold molecule is much less than one hundredth of a micrometer.
In the figures, we will depict molecules in the form of balls. However, molecules (and, as we will see later, atoms) have a structure that is different for different substances, often quite complex. For example, the shape and structure are known not only of such simple molecules as (Fig. 370), but also incomparably more complex ones, containing many thousands of atoms.
Fig. 370. Schemes of the structure of water molecules (a) and carbon dioxide (b)
I would like to talk about important things that are rarely explained on the websites of companies that sell cleaning systems, but it is much more pleasant to understand what is at stake when choosing a filter for your family or for work. This overview presents some important aspects to consider when choosing a filter.
What is micron and nanometer?
If you were looking for a water filter, then most likely you came across the name "micron". When it comes to mechanical cartridges, you can often see phrases such as "the unit filters coarse particles of dirt up to 10 microns or more." But how much is 10 microns? I would like to know what kind of contamination and use a cartridge designed for 10 microns will miss. Regarding membranes (be it a flow filter or reverse osmosis), another term is used - a nanometer, which is also a difficult size to represent. One micron is 0.001 millimeters, that is, if you conditionally divide one millimeter into 1000 divisions, then we just get 1 micron. A nanometer is 0.001 micron, which is essentially one millionth of a millimeter. The names "micron" and "nanometer" are coined to simplify the representation of such small numbers.
Microns are most often used to represent the depth of filtration produced by polypropylene or carbon cartridges, nanometers to represent the level of filtration produced by ultrafiltration or reverse osmosis membranes.
How are water filters different?
There are 3 main types of filters: flow filters, flow filters with an ultrafiltration membrane (membrane) and reverse osmosis filters. What is the main difference between these systems? A flow filter can be considered basic purification, since it rarely purifies water to a drinking state - that is, unlike the other two types of filters, after running water, you need to boil water before drinking (the exceptions are systems containing Aragon, Aqualen and Ecomix material). Membrane filters - filters with an ultrafiltration membrane purify water from all types of contaminants, but leave the salt balance of water intact - that is, natural calcium, magnesium and other minerals remain in the water. The reverse osmosis system purifies water completely, including minerals, bacteria, salts - at the filter outlet, the water contains, oddly enough, only water molecules.
Chlorine is the most cunning of the water pollutants.
Generally, in order to purify water from contaminants with a membrane system, the pores of the membrane must be smaller than the dimensions of the element. However, this does not work with chlorine, since the size of its molecule is equal to the size of a water molecule, and if the pores of the membrane are made smaller than the size of chlorine, then water will not be able to pass either. Here is such a paradox. Therefore, all reverse osmosis systems as part of pre-filters and as a post-filter have carbon cartridges that thoroughly purify chlorine from water. And note, since the main " headache"Ukrainian water is exactly chlorine, if you want to buy reverse osmosis, you should choose a system with two carbon cartridges in the pre-filter - this indicates the quality of cleaning.
We hope the information provided has been useful to you. More information can be found on the website
Molecules have sizes and various shapes. For clarity, we will depict a molecule in the form of a ball, imagining that it is covered by a spherical surface, inside which are the electron shells of its atoms (Fig. 4, a). According to modern concepts, molecules do not have a geometrically defined diameter. Therefore, it was agreed to take the distance between the centers of two molecules (Fig. 4b) as the diameter d of a molecule, so close that the forces of attraction between them are balanced by the forces of repulsion.
From the course of chemistry "it is known that a kilogram-molecule (kilomole) of any substance, regardless of its state of aggregation, contains the same number of molecules, called the Avogadro number, namely N A \u003d 6.02 * 10 26 molecules.
Now let's estimate the diameter of a molecule, for example water. To do this, we divide the volume of a kilomole of water by the Avogadro number. A kilomole of water has a mass 18 kg. Assuming that water molecules are located close to each other and its density 1000 kg / m 3, we can say that 1 kmol water occupies a volume V \u003d 0.018 m 3. Volume per molecule of water
Taking the molecule as a ball and using the ball volume formula, we calculate the approximate diameter, otherwise the linear size of the water molecule:
Copper molecule diameter 2.25*10 -10 m. The diameters of gas molecules are of the same order. For example, the diameter of a hydrogen molecule 2.47 * 10 -10 m, carbon dioxide - 3.32*10 -10 m. So the molecule has a diameter of the order 10 -10 m. On length 1 cm 100 million molecules can be located nearby.
Let's estimate the mass of a molecule, for example sugar (C 12 H 22 O 11). To do this, you need a mass of kilomoles of sugar (μ = 342.31 kg/kmol) divided by the Avogadro number, i.e., by the number of molecules in
« Physics - Grade 10 "
What physical objects (systems) does molecular physics study?
How to distinguish between mechanical and thermal phenomena?
The molecular-kinetic theory of the structure of matter is based on three statements:
1) the substance consists of particles;
2) these particles move randomly;
3) particles interact with each other.
Each assertion is rigorously proven by experiments.
The properties and behavior of all bodies without exception are determined by the movement of particles interacting with each other: molecules, atoms or even smaller formations - elementary particles.
Estimation of the sizes of molecules. To be completely sure of the existence of molecules, it is necessary to determine their sizes. The easiest way to do this is to observe the spreading of a drop of oil, such as olive oil, on the surface of the water. The oil will never occupy the entire surface if we take a sufficiently wide vessel (Fig. 8.1). It is impossible to force a droplet of 1 mm 2 to spread out so that it occupies a surface area of more than 0.6 m 2 . Suppose that when the oil spreads over the maximum area, it forms a layer with a thickness of only one molecule - a "monomolecular layer". It is easy to determine the thickness of this layer and thus estimate the size of the olive oil molecule.
The volume V of the oil layer is equal to the product of its surface area S and the thickness d of the layer, i.e. V = Sd. Therefore, the linear size of an olive oil molecule is:
Modern appliances allow you to see and even measure individual atoms and molecules. Figure 8.2 shows a micrograph of the surface of a silicon wafer, where the bumps are individual silicon atoms. Such images were first learned to be obtained in 1981 using complex tunneling microscopes.
Molecules, including olive oil, are larger than atoms. The diameter of any atom is approximately equal to 10 -8 cm. These dimensions are so small that it is difficult to imagine them. In such cases, comparisons are used.
Here is one of them. If the fingers are clenched into a fist and enlarged to the size of the globe, then the atom, at the same magnification, will become the size of a fist.
Number of molecules.
With very small sizes of molecules, the number of them in any macroscopic body is enormous. Let us calculate the approximate number of molecules in a drop of water with a mass of 1 g and, therefore, a volume of 1 cm3.
The diameter of a water molecule is approximately 3 10 -8 cm. Assuming that each water molecule with dense packing of molecules occupies a volume (3 10 -8 cm) 3, you can find the number of molecules in a drop by dividing the drop volume (1 cm 3) by the volume, per molecule:
Mass of molecules.
The masses of individual molecules and atoms are very small. We calculated that 1 g of water contains 3.7 10 22 molecules. Therefore, the mass of one water molecule (H 2 0) is equal to:
Molecules of other substances have a mass of the same order, excluding huge molecules organic matter; for example, proteins have a mass hundreds of thousands of times greater than the mass of individual atoms. But still, their masses on macroscopic scales (grams and kilograms) are extremely small.
Relative molecular weight.
Since the masses of molecules are very small, it is convenient to use in calculations not the absolute values of the masses, but relative ones.
By international agreement, the masses of all atoms and molecules are compared with the masses of a carbon atom (the so-called carbon scale of atomic masses).
The relative molecular (or atomic) mass M r of a substance is the ratio of the mass m 0 of a molecule (or atom) of a given substance to the mass of a carbon atom:
The relative atomic masses of all chemical elements are accurately measured. By adding the relative atomic masses of the elements that make up the molecule of a substance, we can calculate the relative molecular weight of the substance. For example, the relative molecular weight of carbon dioxide CO 2 is approximately 44, since the relative atomic mass carbon is almost equal to 12, and oxygen is approximately 16: 12 + 2 16 = 44.
The comparison of atoms and molecules with the mass of a carbon atom was adopted in 1961. main reason This choice is that carbon is included in a huge number of different chemical compounds. The factor is introduced so that the relative masses of atoms are close to integers.
Molar mass of water:
If the molecules in a liquid are tightly packed and each of them fits into a cube of volume V 1 with a rib d, then .
The volume of one molecule: , where: Vm one mole N A is Avogadro's number.
The volume of one mole of liquid: , where: M- its molar mass is its density.
Molecule Diameter:
Calculating, we have: 
Relative molecular weight of aluminum Mr=27. Determine its main molecular characteristics.
1.Molar mass of aluminum: M=Mr. 10 -3 M = 27. 10-3
Find the concentration of molecules, helium (M = 4. 10 -3 kg / mol) under normal conditions (p = 10 5 Pa, T = 273K), their root-mean-square velocity and gas density. From what depth does an air bubble float up in a pond if its volume doubles?We do not know whether the temperature of the air in the bubble remains the same. If it is the same, then the ascent process is described by the equation pV=const. If it changes, then the equation pV/T=const.
Let us estimate whether we make a big error if we neglect the change in temperature.
Suppose that we have the most unfavorable result. Let the weather be very hot and the water temperature on the surface of the reservoir reaches +25 0 C (298 K). At the bottom, the temperature cannot be lower than +4 0 C (277 K), since this temperature corresponds to the maximum density of water. Thus, the temperature difference is 21K. In relation to the initial temperature, this value is %%. It is unlikely that we will meet such a reservoir, the temperature difference between the surface and the bottom of which is equal to the named value. In addition, the bubble rises quickly enough and it is unlikely that during the ascent it will have time to fully warm up. Thus, the real error will be much smaller and we can completely neglect the change in air temperature in the bubble and use the Boyle-Mariotte law to describe the process: p 1 V 1 \u003d p 2 V 2, where: p1- air pressure in the bubble at depth h (p 1 = p atm. + rgh), p 2 is the air pressure in the bubble near the surface. p 2 = p atm.
 |
(p atm + rgh)V =p atm 2V; ;
|
The glass turned upside down is filled with air. The problem states that the glass begins to sink only at a certain depth. Apparently, if it is released at a depth less than some critical depth, it will float (it is assumed that the glass is located strictly vertically and does not tip over).
The level, above which the glass floats, and below which it sinks, is characterized by the equality of forces applied to the glass from different sides.
The forces acting on the glass in the vertical direction are the downward force of gravity and the upward force of buoyancy.
The buoyant force is related to the density of the liquid in which the glass is placed and the volume of liquid displaced by it.
The force of gravity acting on a glass is directly proportional to its mass.
It follows from the context of the problem that as the glass sinks, the upward force decreases. A decrease in the buoyancy force can occur only due to a decrease in the volume of the displaced liquid, since liquids are practically incompressible and the density of water at the surface and at some depth is the same.
A decrease in the volume of the displaced liquid can occur due to compression of the air in the glass, which, in turn, can occur due to an increase in pressure. The change in temperature as the glass sinks can be ignored if we are not interested in too high an accuracy of the result. The corresponding justification is given in the previous example.
The relationship between the pressure of a gas and its volume at a constant temperature is expressed by the Boyle-Mariotte law.
The fluid pressure really increases with depth and is transmitted in all directions, including upwards, equally.
Hydrostatic pressure is directly proportional to the density of the liquid and its height (depth of immersion).
Having written down as the initial equation the equation characterizing the equilibrium state of the glass, successively substituting into it the expressions found during the analysis of the problem and solving the resulting equation with respect to the desired depth, we come to the conclusion that in order to obtain a numerical answer, we need to know the values of water density, atmospheric pressure, mass glass, its volume and free fall acceleration.
All of the above reasoning can be displayed as follows:
Since there is no data in the text of the task, we will set it ourselves.
Given:
Water density r=10 3 kg/m 3 .
Atmospheric pressure 10 5 Pa.
The volume of the glass is 200 ml = 200. 10 -3 l \u003d 2. 10 -4 m 3.
The mass of the glass is 50 g = 5. 10 -2 kg.
Free fall acceleration g = 10 m/s 2 .
Numerical solution:
|
The problem of lifting a balloon, like the problem of a sinking glass, can be classified as a static problem.
The ball will begin to rise in the same way as the glass sinks, as soon as the equality of the forces applied to these bodies and directed up and down is violated. The ball, like the glass, is subject to the force of gravity directed downwards and the buoyant force directed upwards.
The buoyant force is related to the density of the cold air surrounding the ball. This density can be found from the Mendeleev-Clapeyron equation.
The force of gravity is directly proportional to the mass of the ball. The mass of the ball, in turn, consists of the mass of the shell and the mass of hot air inside it. The mass of hot air can also be found from the Mendeleev-Clapeyron equation.
Schematically, the reasoning can be displayed as follows:
From the equation, one can express the desired value, estimate the possible values of the quantities necessary to obtain a numerical solution to the problem, substitute these quantities into the resulting equation and find the answer in numerical form.
A closed vessel contains 200 g of helium. The gas goes through a complex process. The change in its parameters is reflected in the graph of the dependence of the volume on the absolute temperature.1. Express the mass of gas in SI.
2. What is the relative molecular weight of this gas?
3. What is the molar mass of this gas (in SI)?
4. What is the amount of the substance contained in the vessel?
5. How many gas molecules are in the vessel?
6. What is the mass of one molecule of a given gas?
7. Name the processes in sections 1-2, 2-3, 3-1.
8. Determine the volume of gas at points 1,2, 3, 4 in ml, l, m 3.
9. Determine the gas temperature at points 1,2, 3, 4 at 0 C, K.
10. Determine the gas pressure at points 1, 2, 3, 4 in mm. rt. Art. , atm, Pa.
11. Plot this process on a graph of pressure versus absolute temperature.
12. Plot this process on a pressure versus volume graph.
Solution instructions:
1. See condition.
2. The relative molecular weight of an element is determined using the periodic table.
3. M=M r 10 -3 kg/mol.
7. p=const - isobaric; V=const-isochoric; T=const - isothermal.
8. 1 m 3 \u003d 10 3 l; 1 l \u003d 10 3 ml. 9. T = t+ 273.10.1 atm. \u003d 10 5 Pa \u003d 760 mm Hg. Art.
8-10. You can use the Mendeleev-Clapeyron equation, or gas laws Boyle-Mariotte, Gay-Lussac, Charles.
Answers to the problem
| m = 0.2 kg | |||||||
| M r = 4 | |||||||
| M = 4 10 -3 kg/mol | |||||||
| n = 50 mol | |||||||
| N = 3 10 25 | |||||||
| m = 6.7 10 -27 kg | |||||||
| 1 - 2 - isobaric | |||||||
| 2 - 3 - isochoric | |||||||
| 3 - 1 - isothermal | |||||||
| № | ml | l | m 3 | ||||
| 2 10 5 | 0,2 | ||||||
| 7 10 5 | 0,7 | ||||||
| 7 10 5 | 0,7 | ||||||
| 4 10 5 | 0,4 | ||||||
| № | 0 С | To | |||||
| № | mmHg. | atm | Pa | ||||
| 7.6 10 3 | 10 6 | ||||||
| 7.6 10 3 | 10 6 | ||||||
| 2.28 10 3 | 0.3 10 6 | ||||||
| 3.8 10 3 | 0.5 10 6 | ||||||
 |
 |
||||||
We advise you to read
Psychological characteristics of children in adolescence
Transferring a child to another school - the procedure and necessary documents Whether to transfer a child to another school
Chlamydia urogenital - description, causes, symptoms (signs), diagnosis, treatment Treatment of urogenital chlamydia
The benefits and significance of hydroamino acid threonine for the human body L threonine what
