Middle School Chemistry Olympiad 20, Middle School Chemistry Olympiad Questions

Ultrafine aluminum nitride powder is widely used in large-scale integrated circuit production and other fields. The principle of its preparation is as follows:

Due to the incomplete reaction, aluminum nitride products often contain carbon and alumina impurities. In order to determine the content of the relevant ingredients in this product, the following two experiments were carried out:

1) Weigh the sample, add it to the excess NaOH concentrated solution to heat it and evaporate it dry, and ALN reacts with the NaOH solution to generate it.

naalo2, and release ammonia gas.

The chemical equation for the above reaction is ;

Find the mass fraction of a1n in this sample.

2) Another sample is taken and placed in the reactor, and the mass of the gas is measured after full reaction at high temperature, and Aln does not react with O2). Find the mass of the impurity carbon in this sample.

nacl、ba(no3)2、cuso4、na2so4、na2co3

Choose D for this topic

Analysis: Add enough water to produce precipitation, add HCl, filter into a colorless solution, so there must be no copper sulfate, and there is an insoluble acid precipitate, so there must be barium nitrate and sodium sulfate; Because HCl is added in the front, it interferes with the CL- test in the back, so it may contain Cl-.

First, we need to extract important information.

Sodium chloride, barium nitrate, copper sulfate, sodium sulfate, sodium carbonate.

Add dilute hydrochloric acid, the precipitate is insoluble, and the colorless filtrate is obtained after filtration).

It can be concluded that copper sulfate definitely does not exist, so a is not right, and sodium chloride and dilute hydrochloric acid do not react, and whether they exist or not should be seen below.

Option B has solubility and knows not A and D

Looking at the valency again, since it is a solution, the total number of charges adds up to 0

Therefore, it should be a positive divalent ion, so B is chosen

1) According to the periodic law, the stronger the non-metal, the greater the stability of the hydride. From the information in the question, we know the flammability of hydride in the third cycle: SiH4 ph3 H2S HCl, so it is easy to deduce that the order of hydride flammability in the second cycle is CH4 CH3 H2O, HF.

2) According to the toxicity of known compounds: PH3 NH3, CCL4 CF4, create their own laws, and draw the conclusion that "the same type of compound, the greater the relative molecular weight, the greater the toxicity of the compounds", and then apply this law to obtain H2S H2O; cs2＞co2.

Answer. 1）ch4、nh3；h2s、hcl

3) CFCL3 (or CF3CL).

4) creating a hole in the ozone layer of the atmosphere; a

Let's look at condition 1 first: due to the addition of NaOH heating to generate gas in the whole.

High School Chemistry. The only thing that can be learned to react with alkali to form gas is NH4+ ammonium to form ammonia, so NH4+ must exist.

Look again at condition 2: due to joining.

There is a precipitate after BaCl2, and it is found that the precipitate decreases but does not disappear completely after adding HCl, indicating that it must be contained.

Carbonate. And.

Sulfate. It must not contain Ba++ ions, otherwise it cannot exist in large quantities as a solution.

The rest is the calculation: it's going to be used here.

The amount of matter. The concept will be easier to solve.

By precipitation all come from.

baso4, you can calculate so4--.

The amount of matter. for, and.

baco3 mass is, so.

The amount of CO3- is the amount of the substance.

And NH3 is the one that gets the ammonium root.

The amount of matter. For.

is now known.

Positive charge. for, and known.

Negatively charged. For, apparently there is more negative charge than positive charge, because.

The charge is conserved, so there must be an additional cation present, so Na+ must be present, as for Cl- it is not certain whether it is present or not.

So the answer is BC.

That's what the junior high school chemistry competition should pay attention to.

Redox reactions. Target. Balancing.

Question, this is a difficult junior high school competition, it takes time to understand, and the exam is almost inevitable; The rest is to love such inference questions, like to scare you with new knowledge and computational mash-ups, which can actually be solved very well with general class knowledge. It is recommended to turn down the high school textbook and read it.

The amount of substance, ionic reactions, galvanic cells.

And. Electrolytic cell.

I don't remember if I had this), it will help you a lot. Junior high school Olympia is also the content of the high school examination.

Select B, according to 1, it means that there must be NH4+, ammonia can be detected with moist blue litmus test paper, if the test paper turns blue, this means that the gas is ammonia.

According to 2, it can be shown that there must be CO32-, SO42-, because the addition of BAC2 to the solution has a precipitation, indicating that BACO3 or BASO4 is produced, because these two substances are insoluble in water, adding hydrochloric acid to the precipitate, and the precipitate is found to be partially dissolved, it means that both substances exist, because BAC3 is soluble in hydrochloric acid, and Baso4 is insoluble in hydrochloric acid. Therefore, B is chosen

The gas in BC200ml solution must be NH3, and if it is, then NH4 is.

The precipitate is two soluble as BaCO3, , , and is not tolerated as BaSO4, that is, CO3 is and SO4 is, and the Cl added at this time is, and the original solution must not contain BA

Since the sum of the charge numbers of + and - is 0, Na*1+NH4*1< SO4*2+CO3*2+CL*1 must contain Na

1. The gas generated by adding NaOH is ammonia grams, and there must be ammonium ions.

2. The two precipitates obtained by adding BaCl2 are BaSO4, and if BaCO3 must have CO32- and SO42-, there must be no Ba2+

If there are not enough ammonium ions, then the cations must still have Na+. I choose BC

I think option B

Because according to 2, it can be known that the precipitated ions reacted with the be ions may be carbonate or sulfate, but if the hydrochloric acid precipitate decreases later, it means that a part of the precipitate can be dissolved by hydrochloric acid, then only barium carbonate has this property!

Answer A, ammonium must exist!

BNH4+ must be pure, because only NH4+ can produce gas with NaOH. Sulfate, carbonate, must be present, because after adding barium ions, a precipitate is produced, and the precipitate is partially dissolved, indicating that there must be these two ions. Other ions cannot be determined.

Because io3 - iodate is contained in iota, iodine is +5 valence, and iodine is +3, +5 two valences, so the cation is i3+, so iodine iodate as a salt, its chemical formula is i(io3)3. Then rewrite it as I4O9 in the form of oxide.

The chemical formula of iodine iodate should be i(io3)3, i.e. i4O9 option D

d, Because the iodine element in io3- is +5 valence iodine, while the other iodine element must be +3 valence. So there is i(io3)3-, which is i4o9

The chemical formula of potassium iodate is kio3, so the chemical formula of iodate is io3- (where iodine is +5 valence), so the chemical formula of iodine iodate is i(io3)3 (where the first iodine element is +3 valence, and the second is +5 valence), so D is chosen

Oxygen is often -2 valence, and b and c can be excluded by zero valency. Therefore, A, D are chosen