Chemistry lab questions! Hurry, can you tell me about it? There are pluses!!

Removing the oxide film on the surface of magnesium with sandpaper is a physical change, and it is found that magnesium is silvery-white metal, which is a physical property, clamping the magnesium belt with a crucible pliers, igniting it on the flame of an alcohol lamp, and finding that it can burn This is the chemical property of magnesium, and emitting a dazzling white light when burning, producing a white powdery solid, the above description belongs to the properties of magnesium.

Physics, physics ,,, chemistry, it is better to be clear.

Physical, metallic, flammable, reaction process and result.

Use sandpaper to remove the oxide film on the surface, this is physical, and it does not change the substance, that is, the oxide film substance on the surface is separated from the magnesium element!

I put it in an alcohol lamp and burned, and this was a chemical change.

Compounds of oxygen and magnesium are produced!

Its solid powder is magnesium oxide, and the white light emitted belongs to the physical imagination! This is still an exothermic reaction, but in fact, it is a chemical energy converted into heat energy! It can only be said to be a kind of energy transformation, and it can be said to be a kind of physical energy transformation imagination!

In addition to the oxide film, there is no change in the essence of the substance, and no new substance is generated, so it is a physical change;

Magnesium does not need to participate in chemical reactions to be a silvery-white metal, which is a physical property;

It can be burned to form magnesium oxide and generate new substances, which is a chemical change;

The above sentences are objective facts and should be a description of the phenomenon.

When doing the magnesium strip combustion test, the oxide film on the magnesium surface is removed with sandpaper, which is a physical change; Magnesium is found when the silvery-white metal, which is a physical property; The magnesium belt was clamped with a crucible pliers and ignited on the flame of an alcohol lamp, and it was found that it could burn, which is the chemical property of magnesium's combustion chamber emits a dazzling white light, producing a white powdery solid, which is a reaction phenomenon described above.

Sandpaper removes"Physical changes Chemical changes in the oxide film.

Physical. Chemical properties. Phenomenon.

Physical change; Physical. Chemical properties, description of the combustion test of magnesium.

1.Eugenol is volatile and can be distilled with steam without being destroyed, and eugenol is almost insoluble in water, can be miscible with organic solvents, and is converted into sodium salt and dissolved in water when encountering sodium hydroxide solution, and can be free after acidification. Therefore, pure eugenol can be extracted by steam distillation, extracted with sodium hydroxide, acidified and then extracted with ethyl acetate.

2.Volatile oil can also be extracted by organic solvent extraction, supercritical method, and microwave extraction.

1. Advantages: easy operation, simple equipment, easy color development, etc., using thin layer chromatography, the measured compound (or mixed compound) and the standard are spotted on the same thin layer chromatographic plate, chromatographic analysis with the same agent, the same color rendering method (fluorescence, sulfuric acid chromogenic agent or other), the same RF value and the same spot color, established as the same compound.

2. Spotting is an important part of thin layer chromatography analysis, and it is also the most time-consuming link, especially in quantitative path layer chromatography analysis, the time required for spotting accounts for 30-60% of the entire analysis time. Therefore, it is necessary to optimize the spotting technique.

The general requirement is that the spotting points are as concentrated as possible, the area is constant, and the spotting volume is accurate.

The choice of spotting technique is usually related to the sample volume, the type of layer, the purpose of the analysis (qualitative or quantitative), the form, etc.

An appropriate spotter should be selected. A volumetric micropipette is convenient, as can a microsyringe (with a flat needle) or a volumetric capillary dispenser. The position can be determined on a special dot pattern fitted with a measuring tape, or it can be measured yourself.

In conventional blist-layer chromatography, the dispensing volume is generally a picoliter, and for high-performance thin layers, the sample volume is mostly 50 to 500 nanoliters. If the multi-spot method is used, the amount of spotting should be larger, but the total volume should not exceed three times the amount that can be transferred at one time. When the sample liquid is not much, it can be ordered at one time, and when it is more, it can be ordered in stages, and it is required to dry while spotting and drying, so as to avoid excessive diffusion of the diameter of the origin.

Large volumes of dilute sample solutions, preferably sprayed onto the plate in a narrow band. Strip spotting improves separation and reduces systematic errors when quantifying with densitometry.

In the qualitative thin layer, the sample can be pinched continuously on the starting line, which can still show the advantages of band-like separation.

In point quota, the dispensing volume must be the same. If different sample sizes are required, different concentrations of calibration standards must be applied. However, this is not necessary with the strip dispensing technique, which allows different volumes of calibration standard solutions to be used without loss of quantitative accuracy.

If the same dispensing device (e.g., a capillary dropper) is used for all samples, and the volume determination error in the sample preparation is less than 1%, a high reducibility can be obtained. If the volumetric device is different, it should be noted that the counting should be as accurate as possible.

It volatilizes when it is opened.

3. Because if the spotting point is immersed under the liquid level of the agent, the sample on the spot will be dissolved in the agent, so that the spotting will be invalid and the chromatography will not be obtained. Therefore, the spotting point should not be leached below the liquid level.

1.Two people.

2..Excessive dosage not only wastes the drug, but also has an impact on the pH of the solution, which may affect the accuracy of the results.

If there is no indication of how much liquid to drop, 1 2ml is generally dropped

3.This way the readings will be more accurate.

4.Accurate; If the color of the solution does not change within half a minute, the titration endpoint has been reached.

1 digit can only be accurate to one decimal place.

2. A large number of indicators may change the pH of the solution

3 There is no scale below, who knows how much is used, so it is necessary to re-add 4 accurately, strictly speaking, slightly excessive, the indicator just changes color, and the color does not change after standing for 1 minute.

1..Acid-base standard solutions can also be prepared with pipettes and analytical balances for greater accuracy. However, general experiments do not require such high accuracy, so it is enough to measure hydrochloric acid with a graduated cylinder and weigh solid NAOH with a tray balance. Leave two decimal places.

2.Too much dosage is a waste of medicine, and the second is that it has an impact on the pH of the solution, which may affect the accuracy of the results; The dosage is generally 2-3 drops;

3.This is for the sake of easy reading, in fact, it is possible to use the remaining part of the solution for titration, as long as it is sufficient, but in this case the reading should be the difference between the two readings.

4.Within the allowable error range, the color change of the indicator in the solution can be observed to confirm the titration endpoint; Add slowly dropwise, and the solution changes color when the last drop is dropped, and the color of the solution does not change within half a minute, the titration endpoint has been reached.

A is wrong, the F ion and the Fe ion form a complex, and there are no Fe ions in the solution.

Assuming that sulfurous acid is more acidic than carbonic acid, carbon dioxide is produced when sulfur dioxide is introduced into a sodium carbonate or sodium bicarbonate solution (strong acid to weak acid). Then as long as it is proved that carbon dioxide is indeed produced, the conclusion that sulfurous acid is stronger than carbonic acid can be obtained. Let's talk about the role of other medicines:

Potassium permanganate: absorbs excess sulfur dioxide; Magenta: proves that sulfur dioxide has been completely absorbed; Clarified lime water:

Proof of the presence of carbon dioxide. The reaction between magenta and sulfur dioxide is itself reversible, and under heating conditions, sulfur dioxide will decompose again, which will make it impossible to be sure that the gas that becomes turbid in the clarified lime water is carbon dioxide. Hope it helps!

Magenta is used to verify the presence of SO2 and clarified lime water is used to verify SO2 or CO2

Because SO2 dissolves in water to form sulfurous acid. Then the introduction of sodium bicarbonate or sodium carbonate solution will also produce sulfurous acid.

The purpose of the experiment was to verify that sulfurous acid is stronger than carbonic acid, so when sodium bicarbonate or sodium carbonate solution is introduced, strong acid and weak acid will produce CO2 gas. CO2 does not react with acidic potassium permanganate, but SO2 reacts, so SO2 is removed after passing acid potassium permanganate, leaving only CO2. Then the magenta does not change color, indicating that the acidic potassium permanganate in the front has removed the SO2, so the discoloration of the clarified lime water in the next step is only caused by CO2.

This shows that the previous H2SO3 does replace CO2, indicating that H2SO3 is more acidic than H2CO3.

There is 1L of NaOH solution Preparation of Na2CO3 solution (1) Take 500ml of NaOH solution to absorb excess CO2 gas until CO2 gas is no longer dissolved;

2) Carefully boil the solution for 1-2min to remove the dissolved CO2 gas in the solution;

3) Add the other half (500 ml) of NaOH solution to the resulting solution and mix the solution thoroughly.

After the above, a very pure sodium carbonate solution is obtained.

The reason is that (1) NaHCO3 is generated, in which the dissolved CO2 gas is removed when boiling, and the NaHCO3 in the experimental procedure (3) happens to be neutralized by NAOH and completely converted into Na2CO3

This is a question of the degree of reaction, too little CO2 may not be reacted to NaOH too much, too much NaOH will produce Na(HCO3)2, the chemical equation calculates, 1L of NaOH is exactly the amount of CO2 required to convert all into Na2CO3, and a very simple equation can be calculated.

Excess CO2 is introduced into the solution

naoh+co2=nahco3

The resulting solution is evaporated, and the remaining crystals are heated and burned.

2NaHCO3=Na2CO3+CO2 +H2O At this time, the pure Na2CO3 solid is obtained, and if you want a solution, you can add water to prepare it according to the required proportion.

NaOH is divided into 2 equal parts, one part is filled with a sufficient amount of CO2, and then the other part NaOH is mixed with the reaction solution.

From the point of view of chemical methods, b:

Excess sodium carbonate is added to the mixed solution first.

When BaCl2+Na2CO3====2NaCl+BaCO3 (precipitate) to barium carbonate precipitate is no longer generated, filter out the supernatant (containing sodium chloride and sodium carbonate) and add excess hydrochloric acid to the supernatant.

2HCl+Na2CO3====2NaCl+H2O+CO2 (gas) until no more bubbles are produced, evaporate the filtrate, excess hydrochloric acid and water will evaporate, leaving sodium chloride crystals and then adding excess hydrochloric acid to the barium carbonate precipitate.

2HCl+BaCO3====BaCl2+H2O+CO2 (gas) until the precipitate disappears completely, and barium chloride crystals can be obtained after evaporation.

As for d, there is no reaction due to the coexistence of sodium ions and hydroxide ions with barium ions and chloride ions, respectively. Moreover, the addition of sodium hydroxide to the mixture does not significantly affect the solubility of sodium chloride or barium chloride, so the answer may be wrong.

You should believe in yourself and don't always assume that the answer is correct.

There's a problem with this question, and the crystals can't come out even if you add reagents.

I also choose B.

d does not react. A will introduce k at the end

c will introduce SO4

So choose B!

The purpose of the experiment is to measure the content of iron, the iron ions in the original compound are complexed with oxalate and are almost impossible to ionize, so potassium permanganate is added to oxidize the oxalate ions, and the iron ions are dissociated into Fe3+, and then zinc powder is added, because the amount of zinc powder is not easy to control, so the excess zinc powder is added to reduce all Fe3+ to Fe2+, and then the zinc powder is filtered to remove. Finally, potassium permanganate was used to accurately titrate the amount of Fe2+, which is the amount of iron in the sample.

I think so: K3[Fe(C2O4)]3·XH2O undergoes a redox reaction with dilute H2SO4, (PS: the chemical formula you will write) +2 valent Fe is oxidized to +3 valence, and the dropwise addition of KMno4 (reaction test indicator) on the way will affect the +2 valence Fe of Fe(C2O4)3-acid group in potassium trioxalate ferrite, that is, the second reaction with him, that is, K3[Fe(C2O4)]3·XH2O and KMno4 have a second reaction, Then Zn powder is added to react with Fe(mNO4)3, and Fe is reduced to continue to react with dilute H2SO4. Understand??