CHEMISTRY
Re-withdrawal
Objective 1.
Gaseous substances are given: H2, HCl, CO2, CO, O2, NH3.
1. Determine which of them are lighter than air, and which are heavier (justify the answer).
2. Determine which ones cannot be collected by water displacement.
3. Determine what will happen to these gases if they are passed through a solution of acid, alkali (confirm the answer with the equations of reactions).
Solution.
1. Lighter than air, those whose molar mass is less than 29 g / mol (molar mass of air). This H 2, CO, NH 3. Heavier: HCl, CO 2, O 2.
2. Water displacement can collect gases that are insoluble or poorly soluble in water. This H 2, CO 2, CO, O 2 ... It is impossible to collect gases by displacement of water: HCl, NH 3.
3. Substances with basic properties react with acids:
NH 3 + HCl = NH 4 Cl
Substances with acidic properties react with alkalis:
HCl + KOH = KCl + H2O
Esep 1.
Gas trіzdі zattar berіlgen: H2, HCl, CO2, CO, O2, NH3.
1.Olardyk kaysysy auadan auyr zhane kaysyzy zheңil ekenin anyқtaңyzdar (zhauaptarygyzdy dәleldeңizder).
2. Olardyk kaysyn courts ygystyru adisimen anyktauғa bolmaityyn anyқtaңyzdar.
3. Jaeger olardy siltinin, uyshgyldyk eritindileri argyly atkizgende wasps gazdarmen not bolatynyn anyқtaңyzdar (zhauaptarygyzdy reaction teңdeuleri argyly dәleldeңizder).
Sheshui.
1. Auadan zheңil, yaғni molarlyқ masses 29 g / moldan (auanyң molarlyқ masses) kishі bolatyn gazdar: H2, CO, NH3. Ayyr: HCl, CO2, O2.
2. Courts of ygystyru adisimen of the court of emeytin nemese of the court of az eritin gazdardy aluғa bolda. Olar This is H2, CO2, CO, O2. Courts of ygystyru adisі argyly zhinauғa bolmaityn gazdar: HCl, NH3.
3. Uyshkylmen negizdik Kasiet kursetetin zattar urekettesedi:
NH3 + HCl = NH4Cl
Siltilermen Kyshgyldyk Kasiet kursetetin zattar Urekettesedi:
HCl + KOH = KCl + H2O
CO2 + 2KOH = K2CO3 + H2O or CO2 + KOH = KHCO3
Objective 2.
In early spring, early in the morning, when the ambient temperature was still 0 ° C, and the pressure was 760 mm Hg. Art., three comrades, walking their dogs, saw an empty bottle on the lawn. “It’s empty,” said one of them. "No, it is full to the brim, and I know the formula of the substance with which it is filled," said another. "You're both wrong," said the third.
1. Which of the comrades, in your opinion, was right (justify the answer)?
2. Calculate the amount of substance and the number of particles contained in the bottle if its volume is 0.7 dm3.
3. Calculate the molar mass of the gas contained in the bottle.
Solution.
1. The third is right, since there is air in the bottle (it is not empty - the first is wrong), and air is not an individual substance (the second is also wrong). Air is a mixture of gases:
2. Since the conditions are normal, thenV M = 22.4 l / mol. Let's calculate the amount of substancen = V / V M = 0.7 / 22.4 l / mol = 0.03125 mol. Particle CountN = N A n= 6.02 1023 mol -1 0.03125 mol = 1.88 1022 particles.
3. The molar mass of air can be calculated by knowing the composition of the air. The air contains about 78% N 2, 21% O 2, 0.5% Ar and 0.5% CO 2 ... The average molar mass will beM Wed = x one · M 1 + x 2 M 2 + x 3 M 3 + x 4 · M 4
Esep 2.
Erte kөktemde taңerteң erte қorshaғan ortanyң temperature 0 ° С, height 760 mm son. baғ. bolip tұrғan kede үsh adam өzdernin itterin қydyrtuғa shyқty zhane olar gazondagy boss құtyny (bottle) kөrdі. "Ol bos" - dedі onyn bіreuі. "Zhog, auzyna deyin zattarmen toly" dedy ekinshisi, sebeb ol utynyk ishindegi zattardyң formulaasyn bledі. "Sender ekeulerin de drys tappadyңdar" - dedy ushinshisi.
1. Sizderdin oylarygyzsha, wasps үsh adamnyң kaysysy dұrys oilady (zhauaptaryңdy dәleldeңder)?
2. Jaeger Utynyk (bottlekany) іshіndegі zattyң kөlemi 0.7 dm3 - he teң bolatyny belgіlі bolsa, zat mөlsherin zhune molecularar sanyn tabykyzdar.
3. Utynyk ishindegi gazdyk molarlyk massasyn eseptegizder.
Sheshui.
1. Oshinshi adam dyrys aytty, sebebi onyn ishinde aua bar (ol bose emes, endeshe birinshi adam dyrys tappady), al aua zheke zhemes (sol sebepti ekinshi adam da durys tappady). Aua birneshe gazdardyң ospasynan tarady: N 2, O 2, Ar, CO 2, H 2 O, etc.
2. Yғni zhaғday Kalypty, yendesheV M = 22.4 l / mol. Zat mөlsherin eseptimizn = V / V M = 0.7 / 22.4 l / mol = 0.03125 mol. Sana's moleculeN = N A n = 6,02 · 1023 mol-1 · 0.03125 mol = 1.88 · 1022 blik.
3. Auanyk kramyn bile ryp auanyk molarlyk massasyn esepteuge bolady. Aua shamamen tumenegi gazdar kospasynan turady: 78% N 2, 21% O 2, 0.5% Ar more than 0.5% CO 2 ... Ortasha molarlyқ massas teңboladyM Wed = x one · M 1 + x 2 M 2 + x 3 M 3 + x 4 · M 4 = 0.78 28 + 0.21 32 + 0.05 40 + 0.05 44 ≈ 29 g / mol.
Objective 3.
You have calcium carbonate and hydrochloric acid at your disposal. Suggest ways to synthesize at least 6 new substances, including 2 simple ones. In the synthesis, you can use only the starting materials, the products of their interaction, the necessary catalysts and electric current.
Solution.
1. CaCO 3 = CaO + CO 2 (when heated)
2.
3.
4. CaO + H2O = Ca (OH) 2
5. CaCl 2 = Ca + Cl 2 (melt electrolysis)
6. 2 HCl = H 2 + Cl 2 (solution electrolysis)
7. 2H2O = 2H2 + O2 (electrolysis)
8. Ca + H2 = CaH2
9. Ca (OH) 2 + Cl2 = CaOCl2 + H2O (at 0ºC)
10. when heating)
11. Cl2 + H2O = HCl + HClO (at 0ºC)
12. 3 Cl 2 + 3 H 2 O = 5 HCl + HClO 3 (when heated)
Esep3.
Sizderde calcium carbonate y zhәne tұz қyshқyly bar. Wasps zattar argyly 6-dan by whom emes zhaңa zattardy, onyң ішінde 2 zhay zattardy galay aluғa boldy? Sintezde tek kana bastapy zattardy, olardan alynkan onimderdi қoldanuғa bolady, catalyst zhune elektr toғy kazhet.
Sheshui.
1. CaCO 3 = CaO + CO 2 (Kyzdyrganda)
2. CaCO3 + HCl = CaCl2 + CO2 + H2O
3. CaCO3 + CO2 + H2O = Ca (HCO3) 2
4. CaO + H2O = Ca (OH) 2
5. CaCl 2 = Ca + Cl 2 (most electrolysis i)
6. 2 HCl = H 2 + Cl 2 (erіndі electrolysis і)
7. 2 H 2 O = 2 H 2 + O 2 (electrolysis)
8. Ca + H 2 = CaH 2
9. Ca (OH) 2 + Cl2 = CaOCl2 + H2O (0ºC-de)
10. 6Ca (OH) 2 + 6Cl2 = 5CaCl2 + Ca (ClO3) 2 + 6H2O ( Kyzdyrgan kezde)
11. Cl2 + H2O = HCl + HClO (0ºC -de)
12. 3Cl2 + 3H2O = 5HCl + HClO3 (izdyrkan kezde)
Task 4.
The gas mixture containing two hydrogen halides has a hydrogen density equal to 38. The volume of this mixture at n. at. was absorbed by an equal volume of water. For neutralization of 100 ml of the resulting solution, 11.2 ml of 0.4 mol / L sodium hydroxide solution was consumed.
1. Determine which hydrogen halides may have been contained in this mixture.
2. Calculate the composition of the gas mixture in volume percent.
3. Suggest a method for determining the qualitative composition of the gas mixture.
Solution.
1. The mass of 1 mol of the gas mixture at n. at. is 38 2 = 76 g. Thus, the gas mixture cannot be present at the same time HBr and HI ( M(HBr) = 81 g / mol, M(HI ) = 128 g / mol). Also cannot be present at the same time HF and HCl ( M(HF) = 20 g / mol, M(HCl ) = 36.5 g / mol). The mixture must contain hydrogen halide withMless than 76 g / mol and hydrogen halide withMmore than 76 g / mol. Possible mixture compositions: 1) HF and HBr; 2) HF and HI; 3) HCl and HBr; 4) HCl and HI.
The concentration of hydrogen halides in solution is (11.2 · 0.4): 100 = 0.0448 mol / l. This value is in good agreement with the calculated value 1: 22.4 = 0.0446 mol / l for the process of dissolving 1 liter of gas (n.u.) in 1 liter of water (provided that the hydrogen halide molecules are monomeric). Thus, the gas mixture does not contain hydrogen fluoride, which in the gas phase is in the form ( HF) n, where n = 2-6.
Then only two variants of mixtures correspond to the conditions of the problem: HCl + HBr or HCl + HI.
2. For a mixture of HCl + HBr: let x mole - amount HCl in 22.4 liters of the mixture (n.o.). Then the amount HBr is (1- x ) mol. The mass of 22.4 liters of the mixture is:
36.5 x + 81 (1- x) = 76; x = 0.112; 1- x = 0.888.
The composition of the mixture: HCl - 11.2%, HBr - 88.8%.
Likewise for the mixture HCl + HI:
36.5 x + 128 (1- x) = 76; x = 0.562.
Mixture composition: HCl - 56.2%, HI - 43.8%
3. Since both mixtures must contain hydrogen chloride, it remains qualitatively to determine hydrogen bromide or hydrogen iodide. It is more convenient to make this definition in the form of simple substances - bromine or iodine. For the conversion of hydrogen halides into simple substances water solution can be oxidized with chlorine:
2HBr + Cl2 = 2HCl + Br2
2HI + Cl2 = 2HCl + I2
The resulting solutions of halogens can be distinguished by the color of the solution in a non-polar solvent (during extraction) or by the more sensitive reaction of the color of starch.
Also, the original hydrogen halides can be distinguished by the different color of silver halides:
HBr + AgNO 3 = AgBr ↓ + HNO 3 (light yellow precipitate)
HI + AgNO 3 = AgI ↓ + HNO 3 (yellow precipitate)
Esep 4.
Eke halogensutekten teratyn gaz Kospasynyk days of boyinsha tygyzdygy 38. Wasps of Kospanyk қ.zh.-daғy kөlemi sudyk kөlemimen birdey. Alynkan 100 ml eritindin beytaraptanand 11.2 ml 0.4 mol / l sodium hydroxydinin eritindis zhumsaldy.
1. Wasps of қospada қandai galogensutek baryn anyқtaңyzdar.
2. Gas ospasyң ramyn kolemdik percentpen anyқtaңyzdar.
3. Gas қospasynyң sapasyn anyқtaityn zhaғdaydy ұsynyңyzdar.
Sheshui.
1.1 mol gas "rescue" masses қ.zh. Rides: 38 2 = 76 g. M(HBr) = 81 g / mol, M(HI) = 128 g / mol) bola almaida. Sonymen қatar bir mezgilde HF zhәne HCl ( M(HF) = 20 g / mol, M(HCl) = 36.5 g / mol) bola almaida. Қosapada M massasy 76g / moldan az halogensutek boluy kerek. Mumkin bolatyn gas of ospalary: 1) HF for HBr; 2) HF woman HI; 3) HCl for HBr; 4) HCl for HI.
Erіtіndіdegі halogenostekterің concentration (11.2 · 0.4): 100 = 0.0448 mol / l. Bl mәn 1 l of suғa (halogensutek molecules monomerli bolғan zhaғdayda) 1 l of gazda (қ.zh.) eritu process үshin tөmendegi esepteu nәtizhesine zhaқyn: 1: 22.4 = 0.0446 mol / l. Endeshe, Gospasynda gas fluorosutek bolmaydy, sebeb ol gaz fazasynda (HF) n tүrinde bolady, mұndaғy n = 2-6.
Endeshe eseptің sharty tek ekі nұsқаға sәykes keledі: HCl + HBr nemesе HCl + HI.
2. HCl + HBr Spassy Ushin: 22.4 L Spassa (қ.zh.) HCl mulsherі - x. Onda HBr mөlsherі (1-x) mol bolada. 22.4 liters of Span massas:
36.5x + 81 (1-x) = 76; x = 0.112; 1-x = 0.888.
Spa frames: HCl - 11.2%, HBr - 88.8%.
Ospa shin HCl + HI:
36.5x + 128 (1-x) = 76; x = 0.562.
Spa frames: HCl - 56.2%, HI - 43.8%
3. Endeshe bromsutek zhene iodsutek eki ospa da boluy azhet. Bұl anyқtama zhai zat tүrіnde - brom nemese iodine anyқtauғa yңғaily. Haloensutektі zhai zhay aynaldyru shin ony eritindisin chlormen totyқtyru says:
2HBr + Cl2 = 2HCl + Br2
2HI + Cl2 = 2HCl + I2
Haloenderdin alynkan eritindilerin nonpolarly eritkishtegi eritindinin tusi boyinsha (extarction kezindegi) nemese starch acerі argyly anyntauғa bolady.
Sonday-aқ halogensutekterdі kumіs halogenidіndegі әrtүrlі tүsterі argyly anyқtauғa bolady:
HBr + AgNO3 = AgBr ↓ + HNO3 (ashyқ-sary tұnba)
HI + AgNO3 = AgI ↓ + HNO3 (sary tұnba)
Problem 5 (Thermochemical calculations, impurities).
Burning 1.5 g of a zinc sample released 5.9 kJ of heat. Determine if the zinc sample contained non-combustible impurities if it is known that when 1 mol of zinc is burned, 348 kJ of heat is released.
Esep5 ( Қospalar, termohimiyalyқ esepteuler). 1.5 g myrysh ulgisin zhaganda 5.9 kJ zhylu bulindi. 1mol myryshty zhaganda 348 kJ zhylu bulinetyinin bile ryp myrysh ulgisinde zhanbaityn Kospalar barma, zhoupa anyktaakyzdar.
Solution:
Sheshui:
CHEMISTRY
Conclusion
Exercise 1.
Decipher the transformation chain and carry out chemical reactions:
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Additionally it is known:
Substance A- corundum
SubstanceB- the most abundant metal (Me) in the earth's crust
Substance C- a compound containing 15.79% Me, 28.07% S, 56.14% O
Substance E- white gelatinous substance, poorly soluble in water. Product of interaction of substance C with alkali
SubstanceD- the sodium salt of the most common metal, the molecule of which contains 40 electrons.
Solution:
A - Al 2 O 3
B - Al
C - Al2 (SO4) 3
D - NaAlO2
E - Al (OH) 3
For each defined formula of a substance - 1 point
For each correct written equation of a chemical reaction (with the conditions for implementation) - 2 points
TOTAL: 5 1 + 8 2 = 21 points
1 tapsyrma.
Ainalular tizbegin ashyp, chemistry, reaction teudeulerin zhazykyzdar:
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Kossymsha belgіli bolғany:
Aback- corundum
Bback– zher sharynda eң kөp taralғan metal (Me)
WITH costs - 15.79% Me, 28.07% S, 56.14%
E zaty - aқ koimalzhyk zat, sud nashar eridі. Zattyk siltimen urekettesuinin onimі S
D zaty- eң kөp taralғan metals sodium tұzy, molecules 40 electronnan thrady.
Sheshui:
A - Al2O3
B - Al
C - Al2 (SO4) 3
D - NaAlO2
E - Al (OH) 3
Urbir zattyk formulaasyn anyқtaғanғa - 1 ұpaidan
Darys zhazylkan rbir khimyalyқ reaction teңdeuine (sharty kөrsetilgen) - 2 ұpaidan
BARLES: 5 1 + 8 2 = 21 ұpay
Task 2.Six numbered beakers (beakers) contain solids (in the form of powders): sodium bicarbonate, sodium chloride, zinc sulfate, potassium phosphate, calcium carbonate, ferrous sulfate ( II ). Using the reagents and equipment available on the table, determine the contents of each vial (beaker). Bring chemical formula of each substance and write the equations of the chemical reactions.
Reagents: 2 M HCl, 2 M NaOH, H 2 O distilled, 2M solution AgNO 3
Equipment:a rack with test tubes (7-10 pieces), a spatula, pipettes.
Solution:
Stages of work | Observations | Reaction equations, conclusions |
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Dissolve samples of substances in water | One substance did not dissolve | This is CaCO3 |
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Add dissolved and undissolved substances to samples HCl | Gas evolves in two test tubes. | NaHCO3 + HCl = CaCO3 + HCl = |
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Add sodium hydroxide solution to the samples of substances (not an excess) | In two test tubes, precipitates of green (marsh) color and white amorphous fall out. | These are FeSO4 and Zn (NO3) 2 FeSO4 + NaOH = Zn (NO3) 2 + NaOH = |
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Add silver nitrate to the samples of substances drop by drop | White cheesy and yellow precipitates fall out in two test tubes. | These are NaCl and K3PO4 NaCl + AgNO3 = K3PO4 + AgNO3 = |
For the determination of each substance, 1 point.
For the reaction equation - 2 points
Total: 6 1 + 6 2 = 18 points
Note: If all the coefficients are not placed in the reaction equation, but the essence of the chemical reaction is reflected - 1 point
2 tapsyrma.Alty nomelengen bukste (chemical glass) atty zat bar (ntaқ tүrіnde): sodium bicarbonates, sodium chloride, myryshg sulfates, potassium phosphates, calcium carbonates, temir (II) sulfates. Stoldaғy reaktivterdi zhane құraldardy paidalana otyryp, urbir bukstegi zatty anyқtaңyzdar. Urbir zattyk khimyalyқ formulaasyn zhne khimyalyқ reaction teңdeulerin zhazyңyzdar.
Reagent:2M HCl, 2M NaOH, distilledene H2O, 2M AgNO3 erіtіndісі
Құral-zhabdyқtar: test tubes bar tripod (7-10 dan), spatula (ұstaғysh), pipette alar.
Sheshui:
Zhumys stagtar | Құbylys | Teudeuler reaction |
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Zattyk son of the court eritu | Bir zat ta erigen zhoқ | Bұl CaCO3 |
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Yerigen zhane Yerimegen zattyk sons of the son of NSI Kosu | EKI test tube gas blіnedі | NaHCO3 + HCl = CaCO3 + HCl = |
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Zattyk synamasyna sodium hydroxydin kosu (az mulsherde) | Ekі prrobirkada zhasyl tusti (saz balshyk turizdі) zhune aқ tusti amorphty tұnba paida bolady | Bұl FeSO4 wne Zn (NO3) 2 FeSO4 + NaOH = Zn (NO3) 2 + NaOH = |
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Synamaғa tamshylatyp kүmis nitratyn kosamyz | Ekі test tube ақ іrіmshіk tәrіzdі zhune sary tұnba tүsedі. | Bұl NaCl zhәne K3PO4 NaCl + AgNO3 = K3PO4 + AgNO3 = |
Urbir zatty anyyқtaғanғa 1 ұpaidan.
Urbir reaction teudeuine - 2 ұpaidan.
Barlyғy: 6 1 + 6 2 = 18ұpai
Eskertu: Jaeger reaction teңdeuinde barlyқ coefficient koyylmaғan bolsa, biraқ khimyalyқ reactionaryң mәni anyқtalғan bolsa - 1 ұpay beruge bolsa
If a dry gas outlet tube is needed for the experiment, then proceed as follows. A rubber tube with a glass tip is put on the free end of the gas outlet tube. When testing the instrument for leaks, the removable tip will get wet and the flue tube will remain dry.
You can collect gas in a vessel different methods... The two most common are the air displacement method and the water displacement method. Each of them has its own advantages and disadvantages, and the choice of the method is largely determined by the properties of the gas to be collected.
Air displacement method
Any gas can be collected using this method, but this raises the problem of accurately determining the moment when all the air from the receiving vessel will be displaced by the collected gas.
Before collecting gas by displacement air, you need to find out whether it is heavier or lighter than air. The position of the receiving vessel will depend on this (Fig.). To do this, calculate the relative density of the gas in air by the formula: D air. (X) = Mr (X) / 29, where Mr is the relative molecular weight of the collected gas, 29 is the relative molecular weight of air. If the calculated value turns out to be less than one, then the gas is lighter than air, and the receiving vessel must be positioned with the opening downward (Fig. 57, a). If the relative density of the gas in the air is more than unity, then the gas is heavier than air, and the receiving vessel should be positioned with the opening upward (Fig. 57, b).
Rice. 57. The position of the receptacle (1): a - for gas, which is lighter than air; b - for gas that is heavier than air.
The filling of the vessel can be controlled in different ways, depending on the type of gas being collected. For example, colored nitric oxide (IV) is easily detected by its reddish-brown color. To detect oxygen, a smoldering splinter is used, which is brought to the edge of the vessel, but not brought inside.
Water displacement method.
By using this method, it is much easier to control the filling of the receiving vessel with gas. However, this method has a serious limitation - it can not be used if the gas dissolves in water or reacts with it .
To collect gas by displacement of water, it is necessary to have a wide vessel, for example, a crystallizer, 2/3 filled with water. A receiving vessel, for example a test tube, is filled to the top with water, closed with a finger, quickly turned upside down and lowered into a crystallizer. When the opening of the test tube is under water, the opening of the test tube is opened and a gas outlet tube is introduced into the test tube (Fig. 58).
Rice. 58. A device for collecting gas by displacing water: 1 - receiving tube filled with water; 2 - crystallizer.
After all the water has been displaced from the test tube by the gas, the opening of the test tube close under water stopper and removed from the crystallizer.
If the gas that is collected by the water displacement method is obtained by heating, the following rule must be strictly observed:
Do not stop heating the tube with initial substances if the gas outlet tube is under water!
Registration of experiment results
The form of recording the results obtained during the performance of a chemical experiment is not regulated by anyone. But the protocol of the experiment must necessarily include the following points: the name of the experiment and the date of its conduct, the purpose of the experiment, a list of equipment and reagents that were used, a drawing or diagram of the device, a description of the actions that were performed during the work, observations, equations of the ongoing reactions, calculations , if they were performed while performing the work, conclusions.
The form of the report on the carried out practical work.
Record the date of the experiment and the name of the experiment.
Formulate the purpose of the experiment yourself.
Write down briefly everything you did.
Draw the experiment or draw the device that you used. Try to keep the drawing clear. Be sure to add explanatory labels to the figure. Use colored pencils or felt-tip pens to depict colored substances.
Write down your observations, i.e. describe the conditions and signs of chemical reactions.
Write the equations for all the chemical reactions that occurred during the experiment. Don't forget to place your odds.
Learn from experience (or work).
You can draw up a report on the work as a sequential description of actions and observations, or in the form of a table:
Experience no ...
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Experience description |
Experiment drawing |
Signs of reactions |
Conclusions. Reaction equations |
When solving experimental problems related to the recognition and identification of substances, it is convenient to draw up the report in the form of another table:
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Procedure |
Reagent |
Tube number |
Conclusion |
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Topic 1. Basic concepts and laws of chemistry.
Laboratory experiments.
Examples of physical phenomena.
Experiment No. 1. Heating glass (glass tube)
in the flame of an alcohol lamp.
Equipment and reagents: glass tube, spirit lamp, matches, asbestos mesh.
1. Grasp the ends of the glass tube with both hands.
2. Bring the middle of the tube into the flame of an alcohol lamp. Remember that the top of the flame is the hottest.
3. Rotate the tube, keeping the spirit lamp in the flame (fig. 59).
4. When the glass becomes very hot (after 3-4 minutes), try to bend the tube without using excessive force.
Rice. 59. Bending the glass tube.
Place the glass tube over the asbestos mesh. Be careful: hot glass on outward appearance no different from cold!
1) Has the glass changed?
2) Did you get a new substance by heating the glass tube?
Experience No. 2. Melting of paraffin.
Equipment and reagents: crucible or glass plate, spirit lamp, matches, crucible tongs or test tube holder, asbestos mesh, paraffin.
Instructions for performing the experiment.
1. Place a small piece of paraffin wax in a crucible (or on a glass plate).
2. Take the crucible (or glass plate) with crucible tongs (or fix it in the tube holder).
3. Place the wax crucible (or glass plate) on top of the spirit lamp flame. Watch the changes carefully.
4. After the paraffin has melted, place the crucible (or glass plate) on the asbestos mesh and extinguish the alcohol lamp.
5. When the crucible (or glass plate) has cooled, consider the substance that is in the crucible (or on the glass plate).
1) Has the paraffin wax changed?
2) Did you get a new substance when heating paraffin?
3) What is this phenomenon: physical or chemical?
Examples of chemical phenomena.
Experience No. 3. Annealing a copper plate or wire
in the flame of an alcohol lamp.
Equipment and reagents: spirit lamp, matches, crucible tongs or test tube holder, asbestos mesh, copper wire or plate.
Instructions for performing the experiment.
1. Take the copper plate (or copper wire) with crucible tongs.
2. Place a copper plate on top of the spirit lamp flame and heat it up.
3. After 1-2 minutes, remove the plate from the flame and peel off from it with a knife or a splinter the formed black plaque on a blank sheet of paper.
4. Repeat heating and clean off the resulting plaque again.
5. Compare the resulting black coating with the copper strip.
1) Has the copper plate changed during incandescence?
2) Was a new substance formed when the copper plate was heated?
3) What is this phenomenon: physical or chemical?
Experience No. 4. The action of hydrochloric acid on chalk or marble.
Equipment and reagents: a 50 ml beaker, marble (small pieces or chips), hydrochloric acid solution (1: 3), matches.
Instructions for performing the experiment.
1. Place 2-3 small pea-sized pieces of marble in a beaker. Be careful not to break the bottom of the glass.
2. Pour enough hydrochloric acid into the glass so that the pieces of marble are completely covered with it. What are you watching?
3. Light a match and place it in a glass. What are you watching?
4. Perform a drawing of the experiment, write down your observations.
1) Was a new substance formed when hydrochloric acid was added to the marble? What kind of substance is it?
2) Why did the match go out?
3) What is this phenomenon: physical or chemical?
Types of chemical reactions.
Kipp apparatus used to produce hydrogen, carbon dioxide and hydrogen sulfide. The solid reagent is placed in the middle spherical reservoir of the apparatus on a plastic annular liner that prevents the solid reagent from entering the lower reservoir. As a solid reagent for producing hydrogen, zinc granules are used, carbon dioxide - pieces of marble, hydrogen sulfide - pieces of iron sulfide. The poured solid pieces should be about 1 cm 3. It is not recommended to use powder, as the gas current will be very strong. After loading the solid reagent into the apparatus, a liquid reagent is poured into the apparatus through the upper throat (for example, a dilute solution of hydrochloric acid in the production of hydrogen, carbon dioxide and hydrogen sulfide). The liquid is poured in such an amount that its level (with the gas valve open) reaches half of the upper spherical expansion of the lower part. Gas is passed for 5-10 minutes in order to displace air from the apparatus, after which the gas valve is closed, a safety funnel is inserted into the upper throat. The gas outlet tube is connected to the device where the gas needs to be passed.
When the valve is closed, the released gas displaces the liquid from the spherical expansion of the apparatus, and it stops working. When the tap is opened, the acid again enters the tank with the solid reagent, and the apparatus starts to work. It is one of the most convenient and safest methods for producing gases in the laboratory.
Collect gas into a vessel can different methods... The two most common methods are the water displacement method and the air displacement method. The choice of method is dictated by the properties of the gas to be collected.
Air displacement method... Almost any gas can be collected using this method. Before sampling gas, it is necessary to determine whether it is lighter than air or heavier. If the relative density of the gas in the air is greater than unity, then the receiving vessel should be held with the opening upward, since the gas is heavier than air and will sink to the bottom of the vessel (for example, carbon dioxide, hydrogen sulfide, oxygen, chlorine, etc.). If the relative density of the gas in the air is less than unity, then the receiving vessel should be held with the opening downward, since the gas is lighter than air and will rise up the vessel (for example, hydrogen, etc.). The filling of the vessel can be controlled in different ways, depending on the properties of the gas. For example, to determine oxygen, a smoldering torch is used, which, when brought to the edge of the vessel (but not inward!), Flares up; when determining carbon dioxide, the hot torch goes out.
Water displacement method... This method can only collect gases that do not dissolve in water (or slightly dissolve) and do not react with it. To collect the gas, a crystallizer is required, which is 1/3 filled with water. The receiving vessel (most often a test tube) is filled up to the top with water, closed with a finger and lowered into the crystallizer. When the opening of the vessel is under water, it is opened and a gas outlet tube is introduced into the vessel. After all the water has been forced out of the vessel by the gas, the hole is closed under water with a stopper and the vessel is removed from the crystallizer.
Checking gas for purity... Many gases burn in air. If you ignite a mixture of combustible gas with air, an explosion will occur, so the gas must be checked for purity. The test consists in burning a small portion of gas (about 15 ml) in a test tube. To do this, the gas is collected in a test tube and ignited from the flame of an alcohol lamp. If the gas does not contain air impurities, then combustion is accompanied by a light pop. If a sharp barking sound is heard, then the gas is polluted with air and needs to be cleaned.
Test "Nitrogen and its compounds"
Option 1 1. Strongest Molecule: a) H 2; b) F 2; c) O 2; d) N 2. 2. Coloring of phenolphthalein in ammonia solution: a) crimson; b) green; c) yellow; d) blue. 3. The oxidation state is +3 at the nitrogen atom in the compound: a) NH 4 NO 3; b) NaNO 3; c) NO 2; d) KNO 2. 4. Thermal decomposition of copper (II) nitrate produces:a) copper (II) nitrite and O 2 ; b) nitric oxide (IV) and О 2 ; c) copper (II) oxide, brown gas NO 2 and O 2; d) copper (II) hydroxide, N 2 and O 2. 5. What ion is formed by the donor-acceptor mechanism? a) NH 4 +; b) NO 3 -; c) Cl -; d) SO 4 2–. 6. Indicate strong electrolytes: a) nitric acid; b) nitrous acid; c) an aqueous solution of ammonia; d) ammonium nitrate. 7. Hydrogen is released during the interaction: a) Zn + HNO 3 (dil.); b) Cu + HCl (solution); c) Al + NaOH + H 2 O; d) Zn + H 2 SO 4 (dil.); e) Fe + HNO 3 (conc.). 8. Write the equation for the reaction of zinc with very dilute nitric acid if one of the reaction products is ammonium nitrate. Indicate the coefficient in front of the oxidizing agent. 9.Give names to substances A, B, C. Option 2 1. By displacing water it is impossible to collect: a) nitrogen; b) hydrogen; c) oxygen; d) ammonia. 2. The reagent for the ammonium ion is a solution of: a) potassium sulfate; b) silver nitrate; c) sodium hydroxide; d) barium chloride. 3. With the interaction of НNО 3 (conc.) gas is formed with copper shavings: a) N 2 O; b) NH 3; c) NO 2; d) H 2. 4. Thermal decomposition of sodium nitrate forms: a) sodium oxide, brown gas NO 2, O 2; b) sodium nitrite and O 2; c) sodium, brown gas NO 2, O 2; d) sodium hydroxide, N 2, O 2. 5. Oxidation degree of nitrogen in ammonium sulfate: a) –3; b) –1; c) +1; d) +3. 6. Which of these substances does concentrated HNO react with? 3 under normal conditions? a) NaOH; b) AgCl; c) Al; d) Fe; e) Cu. 7. Specify the number of ions in the abbreviated ionic equation for the interaction of sodium sulfate and silver nitrate: a) 1; b) 2; at 3; d) 4. 8. Write the equation for the interaction of magnesium with dilute nitric acid if one of the reaction products is a simple substance. Indicate the coefficient in front of the oxidizer in the equation. 9. Write the reaction equations for the following transformations:
Give names to substances A, B, C, D.
Answers
Option 1 1 - G; 2 - a; 3 - G; 4 - v; 5 - a; 6 - a, d; 7 - c, d; 8 – 10,
2Ag + + SO 4 2– = Ag 2 SO 4;
8 – 12, 9.A - NO, B - NO 2, C - HNO 3, D - NH 4 NO 3,
Gaseous substances from the course of inorganic and organic chemistry
In preparation for the upcoming exams, graduates of the 9th and 11th grades need to study the issue of gaseous substances ( physical properties, methods and methods of obtaining, their recognition and application). Having studied the topics of the specification of the OGE and USE exams (on the websitewww. fipi. ru ), we can say that there is practically no separate question on gaseous substances (see table):
| Unified State Exam | №14 (Characteristic Chemical properties hydrocarbons: alkanes, cycloalkanes, alkenes, dienes, alkynes, aromatic hydrocarbons (benzene and toluene). The main methods of obtaining hydrocarbons (in the laboratory);№26 (Rules for working in the laboratory. Laboratory glassware and equipment. Safety rules for working with caustic, flammable and toxic substances, means household chemicals... Scientific research methods chemical substances and transformations. Methods for separation of mixtures and purification of substances. Metallurgy concept: general ways obtaining metals. General scientific principles of chemical production (on the example of industrial production of ammonia, sulfuric acid, methanol). Chemical pollution environment and its consequences. Natural sources hydrocarbons, their processing. High molecular weight compounds. Polymerization and polycondensation reactions. Polymers. Plastics, fibers, rubbers) |
So, in option number 3 (Chemistry. Preparation for the OGE-2017. 30 training materials for the demo version of 2017. 9th grade: teaching aid / edited by V.N. Doronkin. - Rostov n / a: Legion, 2016. - 288 p.) Students are asked to answer the following question (No. 13):
Are the following judgments about the methods of obtaining substances correct?
A. Ammonia cannot be collected by displacing water.
B. Oxygen cannot be collected by displacing water.
1) only A is true
2) only B is true
3) both statements are true
4) both judgments are wrong
To answer the question, the guys should know the physical and chemical properties of ammonia and oxygen. Ammonia interacts very well with water; therefore, it cannot be obtained by displacing water. Oxygen dissolves in water, but does not interact with it. Therefore, it can be obtained by displacing water.
In option number 4 (Chemistry. Preparation for the exam-2017. 30 training options for the demo version for 2017: study guide / edited by V.N. Doronkin. - Rostov n / a: Legion, 2016. - 544 p. ) students were asked to answer the following question (No. 14):
From the proposed list, select two substances that are formed when a mixture of solid potassium acetate and potassium hydroxide is heated:
1) hydrogen;
2) methane;
3) ethane;
4) carbon dioxide;
5) potassium carbonate
Answer: 2 (decarboxylation reaction)
Moreover, in order to pass the exam, children need to know what is the raw material for obtaining a particular gaseous substance. For example, in the same book edited by Doronkin, question # 26 (option 8) reads like this:
Establish a correspondence between the substance obtained in the industry and the raw materials used to obtain it: for each position marked with a letter, select the corresponding position marked with a number:
Write down the selected numbers in the table under the corresponding letters:
Answer:
In option number 12, students are asked to recall the field of application of some gaseous substances:
Establish a correspondence between the substance and its area of application: for each position indicated by a letter, select the corresponding position indicated by a number:
Answer:With the children taking the chemistry exam in the 9th grade, in the preparation for the exam, we fill in the following table (in the 11th grade we repeat and expand it):
Hydrogen
The lightest gas, 14.5 times lighter than air, with air in the ratio of two volumes of hydrogen to one volume of oxygen forms "detonating gas"
1. By interaction of alkali and alkaline earth metals with water:
2 Na + 2 H 2 O = 2 NaOH + H 2
2. Interaction of metals (up to hydrogen) with hydrochloric acid (any concentration) and dilute sulfuric acid:
Zn + 2 HCl = ZnCl 2 + H 2
3. Interaction of transition (amphoteric) metals with a concentrated alkali solution when heated:
2Al + 2NaOH ( end ) + 6H 2 O = 2Na + 3H 2
4. Decomposition of water by electric current:
2H 2 O = 2H 2 + O 2
By the characteristic sound of an explosion: a vessel with hydrogen is brought to the flame (a dull clap - pure hydrogen, a "barking" sound - hydrogen mixed with air):
2H 2 + O 2 2H 2 O
Hydrogen burner, production of margarine, rocket fuel, production of various substances (ammonia, metals such as tungsten, hydrochloric acid, organic substances)
Oxygen
Colorless gas, odorless; in the liquid state it has a light blue color, in the solid state it is blue; soluble in water better than nitrogen and hydrogen
1. By decomposition of potassium permanganate:
2 KMnO 4 = K 2 MnO 4 + MnO 2 + O 2
2. By decomposition of hydrogen peroxide:
2 H 2 O 2 2 H 2 + O 2
3. Decomposition of berthollet's salt (potassium chlorate):
2KClO 3 = 2KCl + 3O 2
4. Decomposition of nitrates
5. Decomposition of water under the influence of electric current:
2 H 2 O = 2 H 2 + O 2
6. The process of photosynthesis:
6 CO 2 + 6 H 2 O = C 6 H 12 O 6 + 6O 2
Flashing of a smoldering splinter in a container with oxygen
In metallurgy, as an oxidizer for rocket fuel, in aviation for breathing, in medicine for breathing, in blasting operations, for gas cutting and welding of metals
Carbon dioxide
Colorless gas, odorless, 1.5 times heavier than air. Under normal conditions, one volume of carbon dioxide dissolves in one volume of water. At a pressure of 60 atm, it turns into a colorless liquid. When liquid carbon dioxide evaporates, part of it turns into a solid snow-like mass, which is pressed in industry - "dry ice" is obtained.
1. In the industry by calcining limestone:
CaCO 3 CaO + CO 2
2. The action of hydrochloric acid on chalk or marble:
CaCO 3
+ 2HCl = CaCl 2
+ H 2
O + CO 2
With the help of a burning splinter that goes out in an atmosphere of carbon dioxide, or by the cloudiness of lime water:
CO 2 + Ca(OH) 2 = CaCO 3 ↓ + H 2 O
For creating "smoke" on stage, storing ice cream, in fizzy drinks, in foam fire extinguishers
Ammonia
A colorless gas with a pungent odor, almost 2 times lighter than air. Do not inhale for a long time, because it is poisonous. Easily liquefies at normal pressure and temperature -33.4 O C. When liquid ammonia evaporates from the environment, a lot of heat is absorbed; therefore, ammonia is used in refrigeration units. Let's well dissolve in water: at 20 o C about 710 volumes of ammonia are dissolved in 1 volume of water.
1. In industry: at high temperatures, pressure and in the presence of a catalyst, nitrogen reacts with hydrogen, forming ammonia:
N 2 +3 H 2 2 NH 3 + Q
2. In the laboratory, ammonia is obtained by the action of slaked lime on ammonium salts (most often ammonium chloride):
Ca (OH) 2 + 2NH 4 Cl CaCl 2 + 2NH 3 + 2H 2 O
1) by smell;
2) on a change in the color of wet phenolphthalein paper (became crimson);
3) by the appearance of smoke when bringing a glass rod moistened with hydrochloric acid
1) in refrigeration units; 2) production mineral fertilizers;
3) production of nitric acid;
4) for soldering; 5) obtaining explosives; 6) in medicine and in everyday life (ammonia)
Ethylene 
At normal conditions- a colorless gas with a faint odor, partially soluble in water and ethanol. Let's well dissolve in diethyl ether and hydrocarbons. It is a phytohormone. Possesses narcotic properties. The most produced organic matter in the world.
1) In industry by dehydrogenation of ethane:
CH 3 -CH 3 CH 2 = CH 2 + H 2
2) In the laboratory, ethylene is obtained in two ways:
a) depolymerization of polyethylene:
(-CH 2 -CH 2 -) n nCH 2 = CH 2
b) catalytic dehydration of ethyl alcohol (white clay or pure aluminum oxide and concentrated sulfuric acid are used as a catalyst):
C 2 H 5 OHCH 2 = CH 2 + H 2 O
Oxygen
+
Downside down
+
Upside down
Carbon dioxide
+
Downside down
-
Ammonia
+
Upside down
-
Ethylene
+
Upside down and oblique
-
Thus, in order to successfully pass the OGE and the Unified State Exam, students need to know the ways and methods of obtaining gaseous substances. The most common of these are oxygen, hydrogen, carbon dioxide, and ammonia. In the 11th grade textbook, children are invited practical work No. 1, which is called "Receiving, collecting and recognizing gases." It offers five options - obtaining five different gaseous substances: hydrogen, oxygen, carbon dioxide, ammonia and ethylene. Of course, in a lesson lasting 45 minutes, it is simply unrealistic to complete all 5 options. Therefore, before starting this work, students at home fill out the above table. Thus, the children at home, when filling out the table, repeat the methods and methods of obtaining gaseous substances (chemistry course of grades 8, 9 and 10) and come to the lesson already theoretically knowledgeable. Graduates receive two marks for one topic. The work is big in terms of volume, but the guys are happy to do it. And the incentive is - a good mark in the certificate.
