High School Chemistry Problems, Senior One Chemistry Problems

I choose ぐCUSO4 is a blue solution, and although the other three groups cannot be identified, it does not meet the "colorless solution" in the question.

B BaCl2+H2SO4 generates BaSO4 precipitate, and the other two solutions are insoluble. MgCl2+NaOH generates Mg(OH)2 precipitate, and two other solutions are added, the one that dissolves the precipitate is H2SO4 and the other is BaCl2. However, MgCl2 and NaOH cannot be distinguished.

c.Agno3 produces white, light yellow and yellow precipitates with other three solutions, HCl produces white precipitates with AgnO3, NABR produces pale yellow precipitates, and Ki produces yellow precipitates.

D and the other three groups have no obvious phenomenon is nano3, in the remaining three groups, and the other two groups have precipitated BACL2, let the remaining Na2CO3 and H2SO4 are A and B, respectively, and add B to the filtrate of BaCl2 and A to generate precipitate bubbles, then A is H2SO4 and B is Na2CO3.

Answer: Use the mutual addition and observe the precipitation.

c.Silver halide is different in silver.

Silver chloride white, silver bromide, pale yellow, silver iodide, yellow.

d.Sulfuric acid reacts with sodium carbonate to form gas, which passes into the solution of the two to form barium chloride.

E element element can be used as a semiconductor material, so E is thought to be Si;

b element gaseous hydride and the most ** oxide hydrate can react, either oxygen reaction, or neutralization reaction; And B is in front of E, and it will not be a metal (metal has no gaseous hydride), and O and F have no most ** oxide hydrate, and CH4 and carbonic acid do not react, then B can only be N, and the generated M is NH4NO3 (formed by the combination of NH3 and HNO3), which happens to meet the "42 electrons" after testing.

The sum of the outermost electrons of b, d, and e is 10, so the outermost electron number of d is 1, and in the middle of n and si, d can only be na;

A and d are the same as the main group, and a may be h and li in a short period; But the atomic radius of a is less than b (i.e., n), so a can only be h;

The atomic number of c is between n and na, and the atomic radius is less than n, which can only be o or f; However, if C is F, then we are not familiar with the substances formed by Na, H, and F. And when C is O, it happens to match the reaction of SiO2 and NaOH, so C is determined to be O.

Words like this: a-h; b-n；c-o；d-na；e-si

In the third phase IVA group, NH4NO3 contains coordination bonds (NH4+ internal), polar covalent bonds (NH4+ internal and NO3- internal), and ionic bonds (between NH4+ and NO3-).

Of course, the coordination bond is also counted as a covalent bond, but it is safer to write it separately because the coordination bond is an important point of investigation

The gaseous hydrides formed are NH3, H2O, and SIH4, and the stability is H2O NH3 SIH4 (the stronger the non-metallic nature of the element, the more stable the gaseous hydrides); The 18E- compound formed by H and O is H2O2 [18 electrons are very important, and it is best to summarize them yourself, which can be compared with 10 electrons; The electronic formula is inconvenient to write, I believe you will, the key is that H2O2 is a covalent compound, there is a single bond between two O atoms, and there are 2 electrons above and below each O

It can be compared to the reaction of CO2 and NaOH].

The solution is used as an electrolyte solution, and H2 and O2 undergo a galvanic cell reaction, that is, a fuel cell.

Total reaction: 2H2 + O2 = 2H2O

Positive reaction: O2+4(E-)2H2O=4(Oh-).

Negative electrode reaction: 2H2-4(E -)4(OH -)=4H2O

Remember positive electrode reduction, negative electrode oxidation; Note that the product of the negative electrode reaction here must not be written as H+, because the electrolyte solution is alkaline, and H+ cannot exist stably, so directly add alkali to water].

There are many variations of this problem, and it can be transformed into various other metals.

It is most convenient to analyze the relationship between valence and valence.

Mg-Al, just react, the generation of H2 is, that is, the valency of H element decreases, that is, Mg2+, Al3+ increase together.

Add NaOH to generate precipitated Mg(OH)2 and Al(OH)3, which is equivalent to the positive charge being OH-bound, so OH- is required, that is, NAOH is, 200ml

The mass of the precipitate, Mg(OH)2 and AL(OH)3, is the mass of Mg-Al plus the mass of Oh- = 6+ g

According to the conservation of mass, it just means that the reaction is just complete, and there is no need to calculate how much each contains, convert the liters of hydrogen to moles, multiply by 2 and divide by 3mol l to get the milliliter of sodium hydroxide. The maximum mass indicates complete precipitation, multiply the number of moles of sodium hydroxide by 17g mol plus 6 to the maximum mass.

Let mg x mol, al y mol.

Al, Mg dissolved in hydrochloric acid becomes AlCl3, MgCl2 mol h2

is conserved by electrons 2x+3y=

Mass: 24x+27y=6. The solution is x= y=4 30.

Add NaOH, AlCl3 , MgCl2 will become Mg(OH)2 4 30 mol Al(OH)3 is required, and the maximum precipitation amount is (

Sodium Oxide and Water: Na2O + H2O = 2NaOH Sodium Peroxide and Water: 2Na2O2 + 2H2O = 4NaOH + O2 We assume that the sodium peroxide in the title contains xg, then the sodium oxide contains (70-x)g

We found that of the two reactions, only one of the reactions resulted in the loss of the mass of the solution, and that was sodium peroxide and water.

According to the equation, x g sodium peroxide reacts with water to form sodium hydroxide (40 39) x g and loses mass (8 39) x g.

70-x)g of sodium oxide reacts with water to form sodium hydroxide (40 31)*(70-x)g, and the solution has no mass loss.

Therefore the equations can be listed.

40 39) x + (40 31) * (70-x) = solution: x = 39g, so the mixture contains 39g of sodium peroxide and 31g of sodium oxide

, solution: (8 points) (1) Let the amounts of Na0 and Na0 in the original solid mixture be n( ) and n(y) respectively

Because nao 2naoh nao 2naoh

n(x) 2n(x) n(y) 2n(y)

So so there is m(nao) = min in the original solid mixture).

m(nao)=min)).

2) From the above calculations, it can be seen that the amount of substances with NaOH in the solution is 2mol, and the amount of HCI required for complete neutralization is also 2mol (2 points).

Therefore, the concentration of the substance of the hydrochloric acid solution is: 2mol m0l l (2 minutes).

Answer: The mass of sodium peroxide and sodium oxide in the original mixture is 39g and 31g; The concentration of hydrochloric acid used for complete neutralization is 5mol l

(2) Assuming that the volume is a cm3 and the mass is pv= grams, then m(hcl) = grams.

n(hcl)=m(hcl )/m= (mol)

C(HCL)=N(HCl) V= mol l so is mol l

There is c=n v to calculate the concentrated hydrochloric acid required; Glass rods, volumetric flasks, droppers, beakers.

b(3)n(na2co3)= mol

na2co3+2hcl =2 nacl +h2o +co2↑

x, so x= *2= so 250ml HCl solution.

n(hcl)=(250/

c=n/v= mol/l

4）na2co3+2hcl =2 nacl +h2o +co2↑

y so y =

m(na2co3)= g

na2co3%=(

21. CO2 will first be combined with Ca(OH)2, alkaline Ca(OH)2>NaOH

1) The first stage: Ca(OH)2+CO2=CaO3+H2O; 2naoh + co2 =na2co3+h2o

After the first stage of reaction, there is still a residual CO2 of mol

The second stage is Na2CO3+H2O+CO2=2NaHCO3; cao3+h2o +co2=ca(hco3)2

2) Look at the reaction equation and know that the first stage ions are reduced, and the reduction is Ca2+, (too, oh-also.) When CO2 is the case, it continues to increase and returns to the starting point before the reaction: +Ca2+)+mol

CO2 is ,, all fill in the blanks; CO2 is , fill in the blank, and CO2 is , fill in.

134 are all gases that can be dissolved directly into water, mostly 30ml ***ml (water) and 2 no to dissolve in water and react with oxygen, this reaction with oxygen is less than 30, no matter how much it is, anyway, it is less than 3, and 3 is equal to 14, so the answer comes out, is your answer wrong, or do you want to verify that the answer is wrong?

In the same state, the number of moles of the same volume of gas is the same, and 1, 3, 4 are all soluble in water, so 1 3 4, and 2 will react to generate 8ml of NO2 + 18ml of NO, no is insoluble in water, so the number of moles dissolved in water in 2 is less than 1, 3, 4, so the concentration is 2<1=3=4

2 in 26mlNO is generated by reacting with 4MLO2 with water.

Xno+Yo2+zh2O=MHno3+NNO According to the volume ratio of NO to O2, that is, the molar ratio can be obtained when the volume ratio is 1:1 to generate Hno3 of X, and when it is greater than 1:1, Hno3 of Y and NO of (X-Y) are generated, less than 1:

At 1, Hno3 of x and O2 of (y-x) are generated

Assuming that the molar volume of the gas under this condition is VM, then (1), (3), and (4) are inverted in water, and the volume of the resulting solution is equal to the volume of the dissolved gas or the volume of the gas decreased, so the concentration is 1 VM, and for (2), after the mixture of NO and O2, it becomes a mixture of 8mlNO2 and 16mLNOn, which is dissolved in water, and the reaction occurs 3NO2+H2O==2HNO3+NO, so the remaining gas volume is 62 3ml, The volume of the resulting solution is (30-62 3) ml, and the amount of nitric acid produced is 16 3 vm (mmol), so the concentration is 4 7 vm, so the answer is c1 = c3 = c4 is greater than c2

26ml in your 2 should be の, right?

Actually, I guess you have a problem with 26no and 4o2, and finally the volume becomes.

4NO+O2 +2H2O = 4HNO3 The gas changed from 5 to 4 and the solute nitric acid was still 4, so the concentration was several times the others.

The answer should be C1 = C3 = C4 > C2, right, just write the chemical formula, NO and O2 will react, and then the gas flies, so the concentration is smaller;

NH3 is completely soluble in water, nitric oxide reacts with oxygen to form nitric acid, nitrogen dioxide dissolves in water and also forms nitric acid, the concentration of these three is equal, and HCl is partially soluble in water.

The NO in 2 reacts with O2 first, that is, 4NO+2O2=4NO2, and the volume becomes smaller after the reaction.

The concentration in 1 2 4 is 1, the concentration in 2 is the concentration of nitric acid, the nitric oxide is excessive, the amount of nitric acid in oxygen is 16 (3*, the volume of the solution is 28 3, compared to 4 7*

The answer is c1=c3=c4>c2, right?

1 and 4 needless to say, dissolve completely. 3 is NO2 and oxygen in water and air to form HNO3. The amount of substance is the same as 1,4 (which can be known from the reaction formula), and the NO in 2 is less than 30ml.

Therefore, the final Hno3 obtained is less than that of 3, and the concentration is low. Similarly, changing no to no2 has no effect on this question, so the conclusion is the same. The examination is how NO, NO2 and water react in the presence of aerobic and anaerobic, which is a must for the exam.

According to the conservation of nitrogen atoms, the reddest nitrogen atoms are all turned into solutions, and the ones in (3) become no and run away some of them and are smaller than the others, 2) the conclusion is different, one is equal to four is greater than two is greater than three.

To compare the magnitude of their concentrations, first calculate the amount of solute in the four solutions.

First of all, 1 and 4 are judged, because their NH3 and HC do not change when dissolved in water, so their concentrations are equal. l

Reaction in solution No. 2: 4NO+3O2+2H2O====4HNO3

In solution No. 3, NO2 and H2O react: 3NO2 + H20===NO (gas) + 2HNO3.

Convert the volume of all gases into moles, then n(nh3) = n(hcl) = n(no2) = n(hno3) in solution 4 > n(hno3) in solution no. 3.

The main investigation should be the conversion of nitrogen oxides and the calculation of the number and concentration of gases, which is a relatively low difficulty topic.