They fool us. Welder's mask
They brought me an automatic welder's mask etaltech et8f with a complaint - it is unstable. Unfortunately, I didn’t take a picture of it, it’s like this, only the sticker is different:
Let's see the instructions:
It says in black and white that it is powered by solar panels. I open it and...
Two lithium batteries, tightly soldered into the board. Here are some solar panels for you. Unfortunately, there are no mask schemes on the Internet. The board says artotic s777f - This is a Chinese manufacturer of these masks, as usual, a large Chinese factory rivets products, and we only hang a brand - corvette, etalon, craton, caliber ...
Lithium batteries are connected in series, and through the diode they go to the VCC bus. The board has a 27L2C operational amplifier, two BU4551BF quad two-channel analog multiplexers, and one HCF4047 multivibrator. I reverse-engineered the circuit a little, often I had this expression on my face: Oh, but I managed to understand something.
The multiplexers are always powered by VCC. Since they are CMOS, they only draw current during switching. The solar battery is connected to the base of the transistor in such a way that, in the presence of illumination, it opens the transistor and through the transistor with VCC, through the filter, power is supplied to the operational amplifier. The mask has two variable tuning resistors - the degree of dimming and sensitivity. There are two switches inside - the welding-sharpening mode, and the speed of glass growth after the arc stops. Two photodiodes connected in parallel are used as sensors. Moreover, in the “sharpening” mode, they short-circuit, sitting on the ground. It turns out that the solar battery is used only as a sensor. After 2-3-5 years, the batteries will turn sour and the mask will be thrown away, buying a new one. So cunningly the Chinese provide a constant flow of orders. No ionistors and charging circuits are provided.
What else did you find out. Glass is a double sandwich of LCD filters, that is, two glasses are used for guaranteed shading. True, the quality of the glasses is not high, and I clearly saw a difference in shading between the center and the edges. The glass is connected between the outputs Q and! Q of the multivibrator 4047. At the same time, there is a meander on the glass, the amplitude of which is the degree of shading. When changing the degree of shading from minimum to maximum, the amplitude of the meander changes from 4.2V to 6V. To implement this tricky trick, the voltage at the power input of the multivibrator is changed. Why feed the glass with a rectangular voltage - I don’t know, either to reduce the phenomenon of polarization, or for what else. I tried to play around with glass just like that, if voltage is applied to it - it charges like a container and grows for quite a long time when the voltage is removed - it should take 5-7 seconds before it becomes transparent.
UPD. Alternating current to power the LCD light filter is used to eliminate the phenomenon of electrolysis, if you power the glass with direct current, then one of the transparent electrodes will dissolve over time. The supply voltage is different - for fubag optima 11, the glass supply voltage is 24 V alternating with a frequency of 0.5 Hz.
The sensors themselves - photodiodes in a tinted plastic case, are sharpened for IR radiation, so the mask stubbornly did not want to work on an energy-saving lamp. But it reacted sharply to the LCD monitor, and worked well on an incandescent lamp.
That's it. Given the lack of mask control circuits on the Internet at all, it seems like an interesting task to rivet an open source mask control circuit on a microcontroller. With normal solar charging, smart signal processing from sensors and some additional features. For example, by automatic shading tightly if the temperature is below the threshold, it still won’t work quickly in the cold - so we’ll completely shade and become just a welding mask.
Back forward
Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.
Lesson Objectives:
- repeat the algorithm for compiling the OVR by the electronic balance method and reveal the essence of the MPR semi-reaction method.
- to show the advantages in the formation of skills for predicting the direction of OVR in solutions using manganese compounds as an example.
- consolidate skills in compiling OVR equations occurring in various environments.
- to teach how to apply the acquired knowledge to solve specific problems.
Lesson objectives.
- Prepare students for task 36 of the exam in chemistry
- Planned result
Subject:
- know RIA, rules for compiling RIA;
- be able to Determine the nature of the medium, the conditions for the occurrence of OVR, the initial and products of formation, the oxidizing agent and the reducing agent, draw up an electronic balance and use the semi-reaction method, conduct an experiment and draw a conclusion based on the experiment.
Metasubject:
- be able to Organize their activities, determine their goals and objectives, choose the means of achieving the goal and apply them in practice, evaluate the results; establish causal relationships, build logical reasoning, draw conclusions; ability to create models and schemes; the ability to organize educational cooperation and joint activities with the teacher and peers, work individually and in a group.
Personal: Formation of a responsible attitude to learning, readiness and ability of students for self-development and self-education based on motivation for learning and cognition; formation of communicative competence in communication and cooperation with peers in the process of learning activities.
Equipment and reagents:
- personal computer, projector, presentation
- Potassium permanganate solution, crystalline potassium permanganate, sulfuric acid solution, alkali solution, potassium iodide solution, sodium sulfite, 5-10% hydrogen peroxide solution
- Large test tubes placed in a demonstration stand with a white background, a device for obtaining gases, a receiving flask, an iron stand, a spirit lamp, a splinter, matches, test tubes in a universal stand on each table, a glass rod
- Appendix 1 “Compounds of the element manganese: oxidizing and reducing agents, calculation of oxidation states”
- Appendix 2 “Algorithm for compiling OVR equations using the electronic balance method”
- Appendix 3 “Algorithm for compiling OVR equations by the ion-electron method”
- Appendix 4 “Oxidative and reducing properties of hydrogen peroxide depending on the nature of the medium. Instructions for performing a laboratory experiment”.
Lesson type: assimilation of new knowledge using existing knowledge and skills, followed by generalization and systematization.
Forms used in the lesson
- Explanation (explanatory and illustrative)
- reasoning (partial-exploratory)
- general characteristics (problematic)
Methods used in the lesson
- verbal (conversation, explanation)
- visual (experiments, computer presentation, information applications)
- practical (demonstration and independent performance of experiments).
Lesson plan.
- Knowledge update.
- Repetition of the main theoretical concepts of the topic.
- Definition of the environment (acidic, neutral or alkaline) in which the reaction takes place.
- Electronic and ion-electronic method for compiling OVR equations
- Consolidation of acquired knowledge
During the classes
1. Actualization of knowledge.
Preparation for task 36 consists of several elements:
The study of theoretical material, individual consultations with the teacher and the implementation of tasks based on this methodological material.
Before starting work, it is necessary to master the basic terms, definitions, concepts and master the technique of chemical calculations.
In the task, a reaction scheme is proposed, and the formulas of one or two substances are replaced by dots.
All tasks 36 can be divided into three types:
The teacher projects diagrams onto the screen using a video projector Slide 2
2. Repetition of the material covered
In the program of the basic school, you have already touched on the main issues necessary to complete task 36.
You know what chemical reactions are redox reactions and that in OVR one of the participants is oxidized. This is a reducing agent, i.e. it donates electrons and increases its oxidation state. The other one is being restored. This is an oxidizing agent, i.e. it draws a valence pair of electrons to itself, lowers its oxidation state.
slide 3 The teacher projects diagrams onto the screen using a video projector
We carry out the task. Students have an application on their desks Appendix 1
Let's do the exercise:
- determination of the oxidation state of elements by the formula
- the structure of the manganese atom, determine the possible oxidation states of the element, its oxidizing and reducing ability.
- fill in the table according to the types of chemical reactions
- form the conclusion
Students complete the table. They conclude: all substitution reactions and reactions in which simple substances are present belong to OVR. Consider the structure of the manganese atom. They make a conclusion.
3. Determination of the medium (acidic, neutral or alkaline) in which the reaction proceeds.
When you start this task, you, logically, must identify the missing substances. To do this, it is necessary to know the main oxidizing and reducing agents, as well as the products of their reduction or oxidation.
In addition, in order to add missing substances, one should take into account the medium in which the redox reaction occurs.
You can define the environment
A) by products of reduction of the oxidizing agent (for example, manganese)
Permanganates are strong oxidizing agents, and, depending on the pH of the medium:
The teacher projects diagrams onto the screen using a video projector, conducts an experiment
slide 4, 5, 6 Demonstration experiment “Chemical chameleon”
Reduction of potassium permanganate with sodium sulfite in various media.
4. Procedures for selecting coefficients in equations
As for the actual procedure for selecting the coefficients in the equations, the electron balance method can be used, and for reactions in solutions, the so-called semi-reaction method, or electron-ion method, is convenient.
The teacher projects diagrams onto the screen using a video projector Slide 7,8,9
Compilation of OVR equations by the electronic balance method
The electron balance method is based on a comparison of the oxidation states in the initial and final substances, when all the initial substances and reaction products are known. You have already used this method when working in lessons in grades 8-9.
Appendix 2
Whiteboard work: Equalize the reactions using the electron balance method, determine the oxidizing agent and reducing agent. slide 7,8,9
They conclude: Using the electronic balance method, it is convenient to arrange the coefficients if the starting materials and reaction products are known, i.e. complete reaction schemes are given.
Semi-reaction method, or electron-ion.
When using the semi-reaction method (electron-ionic balance), it should be borne in mind that in aqueous solutions, the binding of excess oxygen and the addition of oxygen by the reducing agent occur differently in acidic, neutral, and alkaline media.
The teacher projects diagrams onto the screen using a video projector. Performs experiments.
Students have applications on their desks. Appendix 3 slide 10,11
D demonstration experience. Reduction of potassium permanganate with potassium iodide in various media. "Chemical Chameleon"
The teacher projects diagrams onto the screen using a video projector, students have diagrams on their desks for convenience.
Whiteboard work: Equalize the reactions using MPR, determine the oxidizing agent and reducing agent.
One reaction is performed by the teacher, two are left for independent work of students.
slide 12,13,14
We conclude:
Having considered the method of electron-ion balance or the method of semi-reactions, the following advantages of this method can be distinguished:
5. Consolidation of acquired knowledge
Reactions - disproportionation.
The teacher projects diagrams onto the screen using a video projector. Performs experience.
Demonstration experiment “Chemical chameleon” slide 15, 16
Description:
For the experiment, you need a test tube with a gas outlet tube. Crystalline potassium permanganate (potassium permanganate) was poured into a test tube. When heated, potassium permanganate decomposes, the released oxygen enters through the gas outlet tube into the receiver flask. Oxygen is heavier than air, so it does not leave the flask and gradually fills it. If you lower a smoldering torch into a flask with collected oxygen, then it will flare up brightly, because. oxygen supports combustion.
The reaction equation:
2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2
After the end of the experiment and cooling of the test tube, several milliliters of water are poured into it, the contents are thoroughly shaken and the color of the formed substances is observed (K 2 MnO 4 is green and MnO 2 is dark brown).
K 2 Mn +6 O 4 + H 2 O -> KMn +7 O 4 + Mn +4 O 2 + KOH
When strongly diluted with water, a self-oxidation-self-recovery reaction occurs. The color will change from green to red-violet and a brown precipitate will form.
Independent work in notebooks: Level the reactions using MPR, determine the oxidizing agent and reducing agent. slide 15,16
Form output: These are reactions where the oxidizing agent and reducing agent are the same element that is part of one molecule.
We independently conduct an experiment and write an equation using the half-reaction method
The teacher explains that hydrogen peroxide can exhibit oxidizing and reducing properties in both acidic and alkaline environments.
(peroxides can be both oxidizing and reducing agents, peroxide electrons can move from one molecule to another:
H 2 O 2 + H 2 O 2 \u003d O 2 + 2H 2 O.)
Students perform a laboratory experiment and draw a conclusion about the manifestation of hydrogen peroxide oxidizing and reducing properties depending on the environment.
Note. For the experiment, a 3% solution of hydrogen peroxide is used, which can be purchased at a pharmacy, as well as a solution of potassium permanganate.
The technique of the experiment is simple and does not require much time. .
The teacher projects diagrams onto the screen using a video projector. For convenience, students have an application on their desks. Appendix 4 Slide 17
Laboratory work: “Reduction of permanganate with hydrogen peroxide” “Chemical chameleon” - Turning a raspberry solution into a colorless one”
They conclude: In this case, hydrogen peroxide exhibits reducing properties, and potassium permanganate - oxidizing.
6. Homework: Slide 18
Panteleev Pavel Alexandrovich
The paper gives explanations for the appearance of color in various compounds, and also investigates the properties of chameleon substances.
Download:
Preview:
Color chemistry. Substances-chameleons
Section: natural science
Completed by: Panteleev Pavel Nikolaevich,
Student 11 "A" class
Secondary school №1148
them. F. M. Dostoevsky
Lecturer: Karmatskaya Lyubov Aleksandrovna
1. Introduction. Page 2
2. Nature of color:
2.1. organic substances; Page 3
2.2. inorganic substances. Page 4
3. The influence of the environment on the color. Page 5
4. Substances-chameleons. Page 7
5. Experimental part:
5.1. The transition of chromate to dichromate and vice versa; Page 8
5.2. Oxidizing properties of chromium (VI) salts; Page 9
5.3. Oxidation of ethanol with a chromium mixture. Page 10
6. Photochromism. Page 10
7. Conclusions. Page 13
8. List of used sources. Page 14
1. Introduction.
At first glance, it may seem difficult to explain the nature of color. Why do substances have different colors? How does color even come about?
It is interesting that creatures live in the depths of the ocean, in the body of which blue blood flows. One of these representatives is holothurians. At the same time, the blood of fish caught in the sea is red, like the blood of many other large creatures.
What determines the color of various substances?
First of all, the color depends not only on how the substance is colored, but also on how it is illuminated. After all, in the dark everything seems black. The color is also determined by the chemical structures that prevail in the substance: for example, the color of the leaves of plants is not only green, but also blue, purple, etc. This is due to the fact that in such plants, in addition to chlorophyll, which gives the green color, other compounds predominate.
Blue blood in holothurians is explained by the fact that they have vanadium instead of iron in the pigment that provides the color of the blood. It is its compounds that give the blue color to the liquid contained in holothurians. In the depths where they live, the oxygen content in the water is very low and they have to adapt to these conditions, so compounds have arisen in organisms that are completely different from those of the inhabitants of the air environment.
But we have not yet answered the above questions. In this work, we will try to give complete, detailed answers to them. To do this, a number of studies should be carried out.
The purpose of this work will be to give an explanation of the appearance of color in various compounds, as well as to investigate the properties of chameleon substances.
In accordance with the goal, tasks were set
In general, color is the result of the interaction of light with the molecules of matter. This result is explained by several processes:
* the interaction of magnetic vibrations of the light beam with the molecules of matter;
* selective absorption of certain light waves by molecules with different structures;
* exposure to rays reflected or passed through a substance on the retina or on an optical device.
The basis for explaining color is the state of electrons in a molecule: their mobility, the ability to move from one energy level to another, to move from one atom to another.
Color is associated with the mobility of electrons in a molecule of a substance and with the possibility of electrons moving to still free levels when absorbing the energy of a light quantum (elementary particle of light radiation).
Color arises as a result of the interaction of light quanta with electrons in the molecules of matter. However, due to the fact that the state of electrons in the atoms of metals and non-metals, organic and inorganic compounds is different, the mechanism for the appearance of color in substances is also different.
2.1 Color of organic compounds.
For organic matter, which have color (and not all of them have this property), molecules are similar in structure: they are usually large, consisting of tens of atoms. For the appearance of color in this case, it is not the electrons of individual atoms that matter, but the state of the system of electrons of the entire molecule.
Ordinary sunlight is a stream of electromagnetic waves. A light wave is characterized by its length - the distance between adjacent maxima or two adjacent troughs. It is measured in nanometers (nm). The shorter the wave, the greater its energy, and vice versa.
The color of a substance depends on which waves (rays) of visible light it absorbs. If sunlight is not absorbed by the substance at all, but reflected and scattered, then the substance will appear white (colorless). If the substance absorbs all the rays, then it appears black.
The process of absorption or reflection of certain rays of light is associated with the structural features of the substance molecule. The absorption of a light flux is always associated with the transfer of energy to the electrons of a substance molecule. If the molecule contains s-electrons (forming a spherical cloud), then a lot of energy is required to excite them and transfer them to another energy level. Therefore, compounds with s-electrons always appear colorless. At the same time, p-electrons (forming a cloud in the shape of a figure eight) are easily excited, since the connection they make is less strong. Such electrons are found in molecules that have conjugated double bonds. The longer the conjugation chain, the more p-electrons and the less energy is required to excite them. If the energy of the visible light waves (wavelengths from 400 to 760 nm) is sufficient to excite electrons, then the color that we see appears. The rays expended on the excitation of the molecule will be absorbed by it, and the unabsorbed rays will be perceived by us as the color of the substance.
2.2 Color of inorganic substances.
For inorganic substancescolor is due to electronic transitions and charge transfer from an atom of one element to an atom of another. The decisive role here is played by the outer electron shell of the element.
As in organic substances, the appearance of color here is associated with the absorption and reflection of light.
In general, the color of a substance is the sum of the reflected waves (or those that have passed through the substance without delay). At the same time, the color of a substance means that certain quanta are absorbed by it from the entire range of wavelengths of visible light. In molecules of colored substances, the energy levels of electrons are located close to each other. For example, substances: hydrogen, fluorine, nitrogen - seem to us colorless. This is due to the fact that visible light quanta are not absorbed by them, since they cannot transfer electrons to a higher level. That is, ultraviolet rays pass through these substances, which are not perceived by the human eye, and therefore the substances themselves have no color for us. In colored substances, for example, chlorine, bromine, iodine, the electronic levels are closer to each other, so the light quanta in them are able to transfer electrons from one state to another.
An experience. Influence of a metal ion on the color of compounds.
Instruments and reagents: four test tubes, water, iron(II), cobalt(II), nickel(II), copper(II) salts.
Execution of experience. Pour 20-30 ml of water into test tubes, add 0.2 g of iron, cobalt, nickel and copper salts each and mix until dissolved. The color of the iron solution became yellow, cobalt - pink, nickel - green, and copper - blue.
Conclusion: As is known from chemistry, the structure of these compounds is the same, but they have a different number of d-electrons: for iron - 6, for cobalt - 7, for nickel - 8, for copper - 9. This number affects the color of the compounds. Therefore, you can see the difference in color.
3. The influence of the environment on the color.
Ions in solution are surrounded by a solvent shell. The layer of such molecules directly adjacent to the ion is calledsolvation shell.
In solutions, ions can act not only on each other, but also on the solvent molecules surrounding them, and those, in turn, on the ions. Upon dissolution and as a result of solvation, a color appears in a previously colorless ion. Replacing water with ammonia deepens the color. Ammonia molecules are more easily deformed and the color intensity is enhanced.
Now Let us compare the color intensity of copper compounds.
Experience No. 3.1. Comparison of color intensity of copper compounds.
Instruments and reagents: four tubes, 1% CuSO solution 4, water, HCl, ammonia solution NH 3, 10% solution of potassium hexacyanoferrate(II).
Execution of experience. Place 4 ml of CuSO into one test tube 4 and 30 ml H 2 O, in the other two - 3 ml CuSO 4 and 40 ml H 2 O. Add 15 ml of concentrated HCl to the first tube - a yellow-green color appears, to the second - 5 ml of a 25% ammonia solution - a blue color appears, in the third - 2 ml of a 10% solution of potassium hexacyanoferrate(II) - we observe a red- brown sediment. Add CuSO solution to the last test tube 4 and leave to control.
2+ + 4Cl - ⇌ 2- + 6H 2 O
2+ + 4NH 3 ⇌ 2+ + 6H 2 O
2 2 + 4- ⇌ Cu 2 + 12 H 2 O
Conclusion: With a decrease in the amount of reagent (substance involved in a chemical reaction) required for the formation of the compound, the color intensity increases. When new copper compounds are formed, charge transfer and color change occur.
4. Substances-chameleons.
The concept of "chameleon" is known primarily as a biological, zoological term denotinga reptile that has the ability to change the color of its skin when irritated, change the color of the environment, etc.
However, "chameleons" can also be found in chemistry. So what's the connection?
Let's go back to chemistry:
Chameleon substances are substances that change their color in chemical reactions and indicate changes in the environment under study. We highlight the general - a change in color (coloration). This is what connects these concepts. Chameleon substances have been known since ancient times. The old manuals for chemical analysis recommend using a "chameleon solution" to determine the content of sodium sulfite Na in samples of an unknown composition. 2 SO 3 , hydrogen peroxide H 2O2 or oxalic acid H 2 C 2 O 4 . "Chameleon solution" is a solution of potassium permanganate KMnO 4
, which during chemical reactions, depending on the medium, changes its color in different ways. For example, in an acidic environment, a bright purple solution of potassium permanganate becomes colorless due to the fact that from the MnO permanganate ion 4
-
a cation is formed, i.e.positively charged ion Mn 2+ ; in a strongly alkaline medium from bright violet MnO 4
- it turns out the green manganate ion MnO 4
2-
. And in a neutral, slightly acidic or slightly alkaline environment, the final reaction product will be an insoluble black-brown precipitate of manganese dioxide MnO 2
.
We add that due to its oxidizing properties,those. the ability to donate or take electrons from atoms of other elements,and visual color change in chemical reactions, potassium permanganate has found wide application in chemical analysis.
So, in this case, the "chameleon solution" (potassium permanganate) is used as an indicator, i.e.a substance that indicates the presence of a chemical reaction or changes that have occurred in the medium under study.
There are other substances called "chameleons". We will consider substances containing the element chromium Cr.
Potassium chromate - inorganic compound, metal saltpotassium and chromic acid with the formula K 2 CrO 4 , yellow crystals, soluble in water.
Potassium bichromate (potassium bichromate, potassium chromium peak) - K 2Cr2O7 . Inorganic compound, orange crystals, soluble in water. Highly toxic.
5. Experimental part.
Experience No. 5.1. The transition of chromate to dichromate and vice versa.
Instruments and reagents: potassium chromate solution K 2 CrO 4 , potassium bichromate solution K 2Cr2O7 , sulfuric acid, sodium hydroxide.
Execution of experience. Sulfuric acid is added to a solution of potassium chromate, as a result, the color of the solution changes from yellow to orange.
2K 2 CrO 4 + H 2 SO 4 \u003d K 2 Cr 2 O 7 + K 2 SO 4 + H 2 O
I add alkali to a solution of potassium bichromate, as a result, the color of the solution changes from orange to yellow.
K 2 Cr 2 O 7 + 4NaOH \u003d 2Na 2 CrO 4 + 2KOH + H 2 O
Conclusion: In an acidic environment, chromates are unstable, the yellow ion turns into a Cr ion 2 O 7 2- orange, and in an alkaline medium, the reaction proceeds in the opposite direction:
2Cr 2 O 4 2- + 2H + acidic medium - alkaline medium Cr 2 O 7 2- + H 2 O.
Oxidizing properties of chromium (VI) salts.
Instruments and reagents: potassium bichromate solution K 2Cr2O7 , sodium sulfite solution Na 2 SO 3 , sulfuric acid H 2 SO 4 .
Execution of experience. To solution K 2Cr2O7 , acidified with sulfuric acid, add a solution of Na 2 SO 3. We observe a color change: the orange solution turned green-blue.
Conclusion: In an acidic environment, chromium is reduced by sodium sulfite from chromium (VI) to chromium (III): K 2 Cr 2 O 7 + 3Na 2 SO 3 + 4H 2 SO 4 \u003d K 2 SO 4 + Cr 2 (SO 4) 3 + 3Na 2 SO 4 + 4H 2 O.
Experience No. 5.4. Oxidation of ethanol with a chromium mixture.
Instruments and reagents: 5% potassium dichromate solution K 2Cr2O7 , 20% sulfuric acid H 2 SO 4 , ethyl alcohol (ethanol).
Performing the experiment: To 2 ml of a 5% solution of potassium bichromate, add 1 ml of a 20% solution of sulfuric acid and 0.5 ml of ethanol. We observe a strong darkening of the solution. We dilute the solution with water to better see its shade. We get a yellow-green solution.
To 2 Cr 2 O 7 + 3C 2 H 5 OH + H 2 SO 4 → 3CH 3 -COH + Cr 2 O 3 + K 2 SO 4 + 4H 2 O
Conclusion: In an acidic environment, ethyl alcohol is oxidized with potassium bichromate. This produces an aldehyde. This experience shows the interaction of chemical chameleons with organic substances.
Experience 5.4. clearly illustrates the principle by which indicators operate to detect alcohol in the body. The principle is based on the specific enzymatic oxidation of ethanol, accompanied by the formation of hydrogen peroxide (H 2 O 2 ), causing the formation of a colored chromogen,those. organic matter containing a chromophore group (chemical group consisting of carbon, oxygen, nitrogen atoms).
Thus, these indicators visually (on a color scale) show the alcohol content in human saliva. They are used in medical institutions, when establishing the facts of alcohol consumption and intoxication. The scope of indicators is any situation when it is necessary to establish the fact of alcohol consumption: conducting pre-trip inspections of vehicle drivers, identifying drunk drivers on roads by traffic police, using them in emergency diagnostics as a means of self-control, etc.
6. Photochromism.
Let's get acquainted with an interesting phenomenon, where a change in the color of substances also occurs, photochromism.
Today, glasses with chameleon glasses are unlikely to surprise anyone. But the history of the discovery of unusual substances that change their color depending on the light is very interesting. In 1881, the English chemist Phipson received a letter from his friend Thomas Griffith describing his unusual observations. Griffith wrote that the front door of the post office, located opposite his windows, changes color during the day - darkens when the sun is at its zenith, and brightens at dusk. Intrigued by the message, Phipson examined lithopon, the paint that had been used to paint the post office door. His friend's observation was confirmed. Phipson was unable to explain the cause of the phenomenon. However, many researchers are seriously interested in the reversible color reaction. And at the beginning of the 20th century, they managed to synthesize several organic substances called "photochromes", that is, "light-sensitive paints". Since the time of Phipson, scientists have learned a lot about photochromes -Substances that change color when exposed to light.
Photochromism, or tenebescence, is the phenomenon of a reversible change in the color of a substance under the action of visible light, ultraviolet.
Exposure to light causes in a photochromic substance, atomic rearrangements, change in the population of electronic levels. In parallel with a change in color, a substance can change its refractive index, solubility, reactivity, electrical conductivity, and other chemical and physical characteristics. Photochromism is inherent in a limited number of organic and inorganic, natural and synthetic compounds.
There are chemical and physical photochromism:
- chemical photochromism: intramolecular and intermolecular reversible photochemical reactions (tautomerization (reversible isomerism), dissociation (cleavage), cis-trans-isomerization, etc.);
- physical photochromism: the result of the transition of atoms or molecules into different states. The change in color in this case is due to a change in the population of the electronic levels. Such photochromism is observed when only powerful light fluxes are exposed to the substance.
Photochromes in nature:
- Mineral tugtupit able to change color from white or pale pink to bright pink.
Photochromic materials
There are the following types of photochromic materials: liquid solutions and polymer films (macromolecular compounds) containing photochromic organic compounds, glasses with silver halide microcrystals uniformly distributed in their volume (silver compounds with halogens), photolysis ( decay by light) which causes photochromism; Alkaline and alkaline earth metal halide crystals activated with various additives (e.g. CaF 2 /La,Ce; SrTiO 3 /Ni,Mo).
These materials are used as light filters with variable optical density (that is, they regulate the flow of light) in eye protection and devices from light radiation, in laser technology, etc.
Photochromic lenses
Photochromic lens exposed to light, partially covered with paper. A second level of color is visible between the light and dark parts, since photochromic molecules are located on both surfaces of the lens polycarbonate and others plastics . Photochromic lenses typically darken in the presence of UV and brighten in its absence in less than a minute, but the full transition from one state to another occurs from 5 to 15 minutes.
Conclusions.
So, the color of various compounds depends on:
* from the interaction of light with the molecules of matter;
* in organic substances, color occurs as a result of the excitation of the electrons of the element and their transition to other levels. The state of the system of electrons of the entire large molecule is important;
* in inorganic substances, color is due to electronic transitions and charge transfer from an atom of one element to an atom of another. An important role is played by the outer electron shell of the element;
* the color of the compound is affected by the external environment;
*An important role is played by the number of electrons in the compound.
List of sources used
1. Artemenko A. I. "Organic chemistry and man" (theoretical foundations, advanced course). Moscow, "Enlightenment", 2000.
2. Fadeev G. N. "Chemistry and color" (a book for extracurricular reading). Moscow, "Enlightenment", 1977.
A welder who is exposed to the harmful ultraviolet rays of the welding arc must take care of his health, and even more so about the safety of his eyesight. Standard shields are not able to provide the level of protection that a chameleon helmet has.
A mistake when choosing a mask for welding a chameleon can lead not only to facial burns, but also to loss of vision.
The fact that the filter is darkened does not mean the end of exposure to harmful rays. Therefore, the question of how to choose the right chameleon welding mask will be answered by the reviews of welders who have been using this type of protection for a long time. How to choose a chameleon welding mask for comfortable work?
Unlike a standard shield, the welding chameleon has taken welder protection to a new level. The principle of operation of such a mask is liquid crystal polarization. During provocation, they change direction and interfere with UV exposure. The masks of the expensive price segment use multi-layer protection, which ensures the most uniform darkening. And an additional filter provides blocking of infrared radiation.
Sensors are built into the body of the helmet, which detect the arc and provide permanent eye protection. The whole structure is enclosed in a block, which is protected from both sides with the help of plastic light filters. You can carry out related work (grinder, hammer) without removing the protective helmet from your head. Plastic filters require replacement over time, as they are consumables. The key point of the protective process is the speed of the light filter. The response time of professional models is 1 millisecond.
The protective properties of a chameleon directly depend on the ambient temperature. If the temperature is below minus 10 degrees, the filter operation slows down. Conscientious manufacturers indicate the maximum operating temperature in the product passport. Adjustments can be made during the workflow. The buttons have a convenient location and are easily controlled by tactile contact.
It's important to know! The mask must be stored in a heated room, otherwise its resource is reduced.
Filter classification
The light filter is the main element of the chameleon helmet. The European standard EN 379 dictates the parameters of light filters according to the regulation, which denotes qualities through a slash: 1/1/½. So, let's analyze in detail the meaning of each marking point.
Secrets of choosing a protective mask
A chameleon helmet can be equipped with filters, or it can be sold without them.
According to the regulatory and technical documentation, the material for manufacturing should not be a current conductor, be resistant to metal splashes, and also prevent radiation from penetrating inside, thereby ensuring the welder's face is safe. Most modern masks meet these requirements.
The body of domestically produced masks is mainly made of fiber or plastic. European and American samples are distinguished by their original design, and can be made in the form of an animal's head. There is an option made of leather, used mainly in cramped conditions.
In addition to the appearance, professionals advise how to choose a chameleon mask for welding according to certain parameters.
Adjusting the fastening of the mask on the head determines the comfort of using the product in the future. A comfortable viewing angle depends on the proximity of the filter to the eyes of the welder. If you decide to purchase diopter lenses, you need to get a filter with a wide viewing window, this will eliminate the need to raise the mask. Simply put, the view of the weld can be carried out over the lens.
Professional advice: Buy only those chameleon shields that are certified and have a warranty period, do not buy fakes!
Professional advice: The light filter is aimed at working with argon-arc welding, it can protect both from electric arc welding and from working with semi-automatic devices.
Popular models that the market offers
The leading countries in the production of masks and filters are Taiwan and China. But sometimes their quality of products leaves much to be desired: the filters do not work correctly, which negatively affects the vision of the welder. The domestic manufacturer provides products of sufficient quality, but sometimes the filter does not work correctly when working with argon-arc welding.
The Korean brand OTOS, sometimes sold under the French brand GYSMATIC, has a weak point - the filter. There were cases of delamination, as well as the appearance of spots and microcracks.
The masks offered by Europe are higher in price, but their quality is consistently high. The filter of one sample may not be suitable for another product. Next, several brands that produce quality masks that have corresponding quality certificate:
Professional advice. If during welding there is discomfort in the form of burning, fatigue and tearing of the eyes, you should stop using such a mask. Most likely a low quality product.
Now you know all the secrets of the chameleon shield. Not only the health of the welder's eyes, but also the quality of the current work depends on high-quality protection.
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1 ALL-RUSSIAN SCHOOLCHILDREN OLYMPIAD IN CHEMISTRY SCHOOL STAGE. 11 GRADE Tasks, answers and assessment criteria Task 1. Chameleon element The diagram below shows the transformations of compounds of one chemical element: Substances B, D and E are insoluble in water, and a solution of substance D changes color under the action of sulfuric acid. Determine the substances A E and write the equations of the reactions presented in the diagram. Task 2. Properties of homologues Below are the schemes of thermal decomposition of three organic substances A, D and F, which are the closest homologues: A B + C D E + C E G + H 2 O Determine unknown substances if it is known that aqueous solutions of compounds A, B, D, E and F turn litmus red. Give trivial and systematic names of substances A-E. Write the equation for the reaction of compound G with benzene in the presence of aluminum chloride. Problem 3. Synthesis of vanadate In a muffle furnace at a temperature of 820 C and a pressure of 101.3 kPa, 8.260 g of a stoichiometric mixture of vanadium(v) oxide and sodium carbonate was calcined. Salt was formed, and a gas with a volume of 3.14 liters was released (under the conditions of the experiment). 1) Calculate the composition of the mixture in mass fractions. 2) Determine the formula of the resulting salt. Write the reaction equation. 3) The obtained salt belongs to the homologous series of salts, in which the homologous difference is NaVO 3. Establish the formula of the ancestor of this series. 4) Give examples of formulas of two salts of this homologous series. one
2 Task 4. Hydration of hydrocarbons During the hydration of two non-cyclic hydrocarbons with an unbranched carbon chain containing the same number of carbon atoms, saturated monohydric secondary alcohol and a ketone are formed in a molar ratio of 1 2. When the initial mixture of hydrocarbons with a mass of 15.45 g is burned, reaction products with a total mass are formed 67.05 g. It is known that when the initial mixture of hydrocarbons is passed through an ammonia solution of silver oxide, no precipitate is formed. 1) Determine the molecular formulas of hydrocarbons. Give the necessary calculations and reasoning. 2) Set the possible structure of hydrocarbons. 3) Give the equations for the reactions of hydration of the desired hydrocarbons, indicating the conditions for their implementation. Problem 5. Identification of an oxygen-containing compound An organic molecule contains a benzene ring, carbonyl and hydroxyl groups. All other carbon-carbon bonds are single, there are no other cycles and functional groups. 0.25 mole of this substance contains 1 hydrogen atoms. 1) Determine the molecular formula of organic matter. Please provide relevant calculations. 2) Set the structure and give the name of the organic compound, if it is known that it does not precipitate with bromine water, reacts with a silver mirror, and when oxidized with potassium permanganate in an acidic medium, it forms terephthalic (1,4-benzenedicarboxylic) acid. 3) Give the reaction equations for the interaction of the desired compound with an ammonia solution of silver oxide and potassium permanganate in an acidic medium. Task 6. Preparation and properties of an unknown liquid Substance X is a colorless transparent liquid with a characteristic pungent odor, miscible with water in any ratio. In an aqueous solution of X, litmus turns red. In the second half of the 17th century, this substance was isolated from red wood ants. Several experiments were carried out with substance X. Experience 1. A little substance X was poured into a test tube and concentrated sulfuric acid was added. The test tube was closed with a stopper with a gas outlet tube (see figure). On slight heating, the evolution of Y gas was observed without color and odor. Gas Y was set on fire, a beautiful blue flame was observed. When Y burns, gas Z is formed. 2
3 Experiment 2. A small amount of substance X was poured into a test tube with a solution of potassium dichromate, acidified with sulfuric acid, and heated. The color of the solution changed, gas Z evolved from the reaction mixture. Experiment 3. A catalytic amount of powdered iridium was added to substance X and heated. As a result of the reaction, X decomposed into two gaseous substances, one of which is Z. Experiment 4. We measured the relative vapor density of substance X in air. The value obtained turned out to be noticeably greater than the ratio of the molar mass of X to the average molar mass of air. 1) What substances X, Y and Z are discussed in the condition of the problem? Write the reaction equations for the transformation of X into Y and Y into Z. 2) What safety rules and why should be observed during experiment 1? 3) How and why does the color of the solution change in experiment 2? Illustrate your answer with a chemical reaction equation. 4) Write the reaction equation for the catalytic decomposition of X in the presence of iridium (experiment 3). 5) Explain the results of experiment 4. 3
4 Solutions and grading system In the final assessment of 6 tasks, 5 solutions for which the participant scored the highest scores are counted, that is, one of the tasks with the lowest score is not taken into account. Task 1. Chameleon element A K 3 (or K) B Cr (OH) 3 (or Cr 2 O 3 xh 2 O) C Cr 2 (SO 4) 3 G K 2 CrO 4 D Cr 2 O 3 E Cr Equations reactions: 2K 3 + 3H 2 SO 4 \u003d 2Cr (OH) 3 + 3K 2 SO 4 + 6H 2 O 2Cr (OH) 3 + 3H 2 SO 4 \u003d Cr 2 (SO 4) 3 + 6H 2 O 2K 3 + 3KClO \u003d 2K 2 CrO 4 + 3KCl + 2KOH + 5H 2 O 2Cr (OH) 3 \u003d Cr 2 O 3 + 3H 2 O Cr 2 O 3 + 4KOH + 3KNO 3 \u003d 2K 2 CrO 4 + 3KNO 2 + 2H 2 O Cr 2 O 3 + 2Al \u003d 2Cr + Al 2 O 3 2Cr + 6H 2 SO 4 \u003d Cr 2 (SO 4) 3 + 3SO 2 + 6H 2 O Evaluation criteria: Formulas of substances A E Equations of reactions of 0.5 points (total 3 points ) by y (total 7 points) (put 0.5 points for unbalanced reactions) Task 2. Properties of homologues A oxalic (ethanedioic) acid HOOC COOH B formic (methane) acid HCOOH C carbon dioxide (carbon monoxide (IV)) CO 2 G malonic (propanedioic) acid HOOC CH 2 COOH D acetic (ethanoic) acid CH 3 COOH E succinic (butanedioic) acid HOOC CH 2 CH 2 -COOH F succinic anhydride 4
5 Reaction equation: Evaluation criteria: Formulas of substances A Zh Trivial names of substances A E Systematic names of substances A E Equation of the reaction of substance G with benzene 0.5 points each (3.5 points in total) 0.25 points each (1.5 points in total) ) by 0.25 points (total 1.5 points) 3.5 points Task 3. Synthesis of vanadate 1) The amount of substance and mass of sodium carbonate can be found through the volume of released carbon dioxide: ) = PV / RT = 101.3 3.14 / (8,) = 0.035 mol. m (na 2 CO 3) \u003d νm \u003d 0, \u003d 3.71 g. Composition of the mixture: ω (na 2 CO 3) \u003d 3.71 / 8.26 \u003d 0.449 \u003d 44.9%; ω (v 2 O 5) \u003d 0.551 \u003d 55.1% 2) We determine the vanadate formula from the molar ratio of the reagents: ν (v 2 O 5) \u003d m / M \u003d (8.260 3.71) / 182 \u003d 0.025 mol. ν (na 2 CO 3) : ν (v 2 O 5) = 0.035: 0.025 = 3.5: 2.5 = 7: 5. Reaction equation: 7Na 2 CO 3 + 5V 2 O 5 = 7CO 2 + 2Na 7 V 5 O 16 Vanadate formula Na 7 V 5 O 16. (Any formula of the form (Na 7 V 5 O 16) n is accepted) 3) There must be one vanadium atom in the first member of the homologous series. To find the corresponding formula, it is necessary to subtract 4 homological differences from the formula Na 7 V 5 O 16: Na 7 V 5 O 16 4NaVO 3 \u003d Na 3 VO 4. 4) The closest homologues of the first member of the series Na 4 V 2 O 7 and Na 5 V 3 O 10. Evaluation criteria: Amount of substance CO 2 Mass of sodium carbonate Mixture composition Salt formula Reaction equation Formula of the first term of the series Formulas of two homologues 3 points 2 points (0.5 points for each formula) 5
6 Problem 4. Hydration of hydrocarbons 1. If a monohydric saturated alcohol is formed during the hydration of a hydrocarbon, then the initial compound in this reaction is the alkene C n H 2n. The ketone is formed during the hydration of the alkyne C n H 2n 2. H + C n H 2n + H 2 O C n H 2n + 2 O 0.5 points Hg 2+, H + C n H 2n 2 + H 2 O C n H 2n + 2 O 0.5 points Equations for the combustion reactions of alkene and alkyne: C n H 2n + 1.5nO 2 nco 2 + nh 2 O 0.5 points C n H 2n 2 + (1.5n 0.5) O 2 nco 2 + (n 1)H 2 O 0.5 points According to the condition, the molar ratio of alcohol and ketone is 1 2, therefore, alkene and alkyne are taken in the same ratio. Let the amount of alkene substance be x mol, then the amount of alkyne substance is 2x mol. Using these notations, we can express the amount of substance of the combustion reaction products: ν (co 2) III \u003d nx + 2nx \u003d 3nx mol, ν (h 2 O) \u003d nx + 2x (n 1) \u003d (3n 2) x mol. Molar masses: M (C n H 2n) \u003d 14n g / mol, M (C n H 2n 2) \u003d (14n 2) g / mol. Let us write the expressions for the mass of the initial mixture and the mass of the combustion products: 14n x + (14n 2) 2x = 15, nx + 18 (3n 2)x = 67.05 Solution of this system of equations: x = 0.075, n = 5. Therefore, the initial hydrocarbons have molecular formulas: alkene C 5 H 10, alkyne C 5 H 8. 4 points 2) Hydration of two alkenes of composition C 5 H 10 with an unbranched carbon chain leads to the formation of secondary alcohols. These alkenes are pentene-1 and pentene-2. There is only one alkyne of the composition C 5 H 8, which does not have a terminal arrangement of the triple bond and for this reason does not react with an ammonia solution of silver oxide, this is pentyn-2. 3) Equations for the reactions of hydration of pentene-1 and pentene-2: CH 2 \u003d CHCH 2 CH 2 CH 3 + H 2 O CH 3 CH (OH) CH 2 CH 2 CH 3 CH 3 CH \u003d CHCH 2 CH 3 + H 2 O CH 3 CH 2 CH(OH)CH 2 CH 3 and CH 3 CH \u003d CHCH 2 CH 3 + H 2 O CH 3 CH (OH)CH 2 CH 2 CH 3 6
7 Reactions of addition of water to alkenes occur in the presence of acid catalysts, such as sulfuric or phosphoric acids. Alkyne hydration reaction equation: CH 3 C CCH 2 CH 3 + H 2 O CH 3 CH 2 C (O) CH 2 CH 3 and CH 3 C CCH 2 CH 3 + H 2 O CH 3 C (O) CH 2 CH 2 CH 3 The addition of water to alkynes occurs in the presence of mercury(II) salts and strong acids. Task 5. Identification of an oxygen-containing compound 1) The general formula of compounds having a benzene ring, carbonyl and hydroxyl groups C n H 2n 8 O 2. The amount of hydrogen substance in 0.25 mol of this organic substance is: ν (n) \u003d 1, / 6 , = 2 mol. 1 mol of this compound contains 8 mol of hydrogen: ν(n) = 2 / 0.25 = 8 mol. Using these data, you can determine the number of carbon atoms in the desired compound and, accordingly, its molecular formula: 2n 8 = 8; n = 8; molecular formula of the compound C 8 H 8 O 2. 4 points 2) The compound reacts with an ammonia solution of silver oxide with the release of metallic silver (silver mirror reaction), therefore, the carbonyl group in it is aldehyde. With an aqueous solution of bromine, this compound does not precipitate, therefore, the hydroxyl group is not phenolic, that is, it is not directly connected to the benzene ring. As a result of oxidation, 1,4-benzenedicarboxylic acid is formed, therefore, the aldehyde and hydroxymethyl groups are located in the para position with respect to each other: 4-hydroxymethylbenzaldehyde 3) The reaction equation with an ammonia solution of silver oxide: p-hoch 2 C 6 H 4 CHO + 2OH p-hoch 2 C 6 H 4 COONH 4 + 2Ag + 3NH 3 + H 2 O 4 points 7
8 The equation for the oxidation reaction with potassium permanganate in an acidic medium: 5p-HOCH 2 C 6 H 4 CHO + 6KMnO 4 + 9H 2 SO 4 5p-HOOC C 6 H 4 COOH + 3K 2 SO 4 + 6MnSO H 2 O Task 6. Obtaining and properties of an unknown liquid 1) X formic acid, Y carbon monoxide, Z carbon dioxide. HSO 2 4, t HCOOH H 2 O + CO 2CO + O 2 \u003d 2CO 2 3 points (y for each correct substance) (0.5 points for each correct equation) 2) Carbon monoxide is a poisonous substance. When working with it, care should be taken, work under draft, preventing gas from entering the working area. Care should also be taken when working with concentrated sulfuric and formic acids. These are caustic substances that can cause severe burns. Do not allow these substances to come into contact with the skin, especially the eyes should be protected. 3) Dichromate ions Cr 2 O 2 7, which have a bright orange color, are reduced by formic acid to Cr 3+ chromium cations, the color of which is green: 3HCOOH + K 2 Cr 2 O 7 + 4H 2 SO 4 = 3CO 2 + Cr 2 ( SO 4) 3 + K 2 SO 4 + 7H 2 O 2 points Ir H 2 + CO 2 4) HCOOH 5) Hydrogen bonds form between formic acid molecules, due to which quite stable dimers exist even in the gaseous state: For this reason, the vapor density formic acid is greater than the value that can be calculated from the condition that all molecules in the gas phase are single. 2 points 8
Option 4 1. What type of salts can be attributed to: a) 2 CO 3, b) FeNH 4 (SO 4) 2 12H 2 O, crystalline hydrate, c) NH 4 HSO 4? Answer: a) 2 CO 3 basic salt, b) FeNH 4 (SO 4) 2 12H 2 O double
Option 2 1. What type of salts can be attributed to: a) (NO 3) 2, b) KFe (SO 4) 2 12H 2 O; c) CHS? Answer: a) (NO 3) 2 basic salt, b) KFe (SO 4) 2 12H 2 O double salt, crystalline hydrate,
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TRIAL USE IN CHEMISTRY (Krasnogvardeisky district, February 15, 2019) Option 2 Assessment system for examination work in chemistry Part 1 of the task Answer max for the correct answer 1 14 1 2 235 1 3 14 1 4 25 1 5 214 1
