High School Chemistry Problem Solving !! There should be a detailed process!!

Learn to analyze and write the substances obtained by the reaction step by step. Of course, the premise is that you are familiar with the reaction principle. In this question, sodium aluminum is placed in water, sodium first reacts to produce sodium hydroxide, when sodium hydroxide is generated, aluminum begins to dissolve to produce H2, and sodium metaaluminate is generated, because the question tells you that there is OH-remainder, that is, you are told that the aluminum reaction is finished, at this time the composition of the solution is sodium hydroxide and sodium metaaluminate, and then add hydrochloric acid to the solution, hydrochloric acid first reacts with sodium hydroxide, and then reacts with sodium metaaluminate to form a precipitate, and the maximum amount of precipitation indicates that the sodium metaaluminate reaction is just completed at this time, and there is only sodium chloride and water in the solution.

According to the conservation of elements, n(na+) = n(nacl) = n(hcl) = 1mol

So the mass of sodium is:

If the quality of aluminum is required, the previous "20 ml of C(OH-)= solution" should be used

HCl reacts with NaOH first and then reacts with Naalo2, when the precipitate amount is maximum, there are only NaCl and H2O in the solution, and Al all generate Al(OH)3 precipitate, which is consumed at this time, so N(Na)=, M(Na)=.

The third question can't be done, it is likely that the first two questions are still jointed, and I will start with the first two questions:

1.From the comparison of **, it can be concluded that group A is hydrochloric acid that has not reacted, and Alcoa powder has completely reacted; Groups B and C are the aluminum powder that has not been reacted, and the hydrochloric acid has completely reacted.

Then, the calculation of the amount and concentration of hydrochloric acid starts with group B or group C, because the hydrochloric acid of group B and C must be completed

The amount of hydrogen substance: n(h2)= l (l mol)= mol, conserved by the elements, 2mol of hydrochloric acid needs to be consumed to generate 1mol of hydrochloric acid, the amount of hydrochloric acid: n(hcl)=2*n(h2)= mol, the amount of hydrochloric acid concentration:

c(hcl)= mol/ l=1 mol/l。

2.To find the ratio of magnesium to aluminum, we should find group A with complete reaction of magnesium and aluminum, and let the amount of mg be x mol and the amount of al be y mol.

The amount of substance of hydrogen: n(h2)= l(moll)= mol according to the relationship between the gain and loss of electrons in redox:

1 mol mg can reduce 2 mol h(+), 1 mol h22mol al can reduce 6mol h(+), and 3 mol h2 can be listed

x+3y2=sum of hydrogen produced by al)

24x+27y=sum of al-mass).

Solution: x= y=

The ratio of the amount of magnesium and aluminum substances is 1:1

There is a problem with the number of your questions, according to your group A data, x=0, y=1 120, it is not magnesium aluminum alloy at all, and the mass of group A should be 255mg

3.The solutes in the filtrate were NaCl and Naalo2, and the amount of Mg and Al in group C was X, then 24x 27x

Solve x mol

According to the conservation of the CL element, we get:

The amount of the substance of NaCl The amount of the substance of HCl 1 mol According to the conservation of the Al element, gets:

The amount of Naalo2 The amount of Al of matter mol is conserved by the element Na, get:

The volume of NaOH NaCl Naalo2 molNaOH solution

3 means that at this time the hydrochloric acid has been finished, but the metal is not finished, add Naoh, the aluminum is converted into metaaluminate, magnesium ions precipitate to generate magnesium hydroxide, because there are aluminum ions in the solution, the added sodium hydroxide first reacts with these aluminum ions, which you know the principle, and then magnesium ions, down is metal aluminum, calculate the remaining aluminum in C, the amount of aluminum and magnesium ions that have been generated, and then calculate OH-, the ratio is OH-: metal = 1:4 aluminum ions, 1:

2 magnesium ions, 1:? (I forgot how much, I haven't counted it for a long time) aluminum, got it?

The solubility of Na2CO3 is small, and the distilled water required when the sample is all Na2CO3 is the most.

Sorry, the diagram is not shown in my case and cannot be explained;

However, it is estimated that this figure is first added with hydrochloric acid, no CO2 is produced, and NaCl and NaHCO3 are generated by the reaction of HCl and NaOH and Na2CO3 respectively; Then hydrochloric acid is added to produce CO2, which reacts HCl with NaHCO3 to produce NaCl and CO2; Finally, the addition of hydrochloric acid does not produce CO2 because there is no reaction anymore. Without a graph, no data is not easy to calculate. FYI.

Alas, you can't do it without that diagram... It's true!!

(2) When 20 is known, the solubility of Na2CO3 is and the solubility of NaOH is .

then the mass of the undeteriorated NaOH in the sample is (8g); The degree of deterioration of NaOH is ( ) is expressed by mass fraction]; The volume of hydrochloric acid consumed in the reaction with NaOH is (50) ml.

Solution: The second question is that I don't know what to do right. I think so, he needs to find the maximum, I assume it's all sodium hydroxide, or it's all sodium carbonate, and see how much water is needed at most. Calculate yes.

The third question is that since the gas is so, sodium carbonate is and sodium hydroxide is 8g. Therefore, it is 8g that has not deteriorated. Metamorphic sodium hydroxide can be calculated according to the conservation of atoms, and sodium carbonate is sodium, so there is sodium hydroxide metamorphosis. That's 4G. So the proportion of metamorphosis is 4 12=

The last question should be noted that when the gas is produced, the reactant is not sodium carbonate but sodium bicarbonate, and since carbon dioxide consumes hydrochloric acid, it is also consumed when sodium bicarbonate is generated, for a total of 25 ml. So the reaction with sodium hydroxide is 50ml.

Conservation of electron gains and losses, this seems to be one of the chemistry questions in the national paper of the 07 college entrance examination, and the answer should be b

The oxidant is Cu(io3)2, Cu changes from 2 to 1, drops one valence, that is, one electron, and i in io3 is 5 valence, becomes 0 valence, and 5 electrons are obtained, because of two io3, a total of 10 electrons are obtained, plus one electron obtained by cu, a total of 11 electrons, 1mol can get 11mol electrons.

The answer to answer A is the principle of normalization in chemistry (the principle of normalization refers to the reaction in which the valence state of different compounds of the same element is different, and the valence state is the same after the reaction.

The reactants reflected in the centering should be one of the more ** states and the other of the lower valence states. The product is an intermediate valence state.

The principle is that there can be no crossover between valences. )

In the centering reaction, if there are several valences of an element, and any valence cannot exceed (greater than or less than) the intermediate valence after conversion, such as -2, 0, +1, +2, +5, then the elements of -2 valence can only be converted to 0 or +1, the elements of +5 valence can only be converted to +2 or +1, the elements of 0 valence can only be converted to +1, and the elements of +2 valence can only be converted to +1, that is, the +1 valence is an intermediate valence state in this reaction, and the most valence greater than +1 is converted into a valence between +1 and the original valence, which is expressed as [+1, original price) less than +1 price is converted to the price between +1 price and the original price at most, expressed as (original price, +1) in the interval, that is, any price conversion can not exceed (greater than or less than) the middle price can be judged according to this principle!!

When the precipitation of an ion is complete, the concentration of an ion is less than 10 (-5).

The third question is not precipitation-dissolution equilibrium, it belongs to ionic equilibrium, or chemical equilibrium. Your second question has been written out of the equation of dichromate and chromate conversion between each other, chromate hydrogenation ions can be reversibly generated dichromate and water, according to the principle of equilibrium movement, the temperature decreases, the equilibrium is moved in the direction of exothermy, the orange in the question is deepened, that is, the concentration of dichromate ions increases, indicating that the equilibrium moves in the direction of generating dichromate, that is, the direction of the positive reaction moves, then you can know that the direction of the positive reaction is also the exothermic reaction, hope to adopt!

Check the homework help! There are parsed ones in it.

Big brother, don't you have a mobile phone? I don't know about the homework class?

The first question is the criss-cross method.

It can be seen that the second question of answer A has an x of 29

29x+30（

x=Answer B: There's a problem with the third question, right? . . I won't answer it yet...

Question 4: You're wrong, right? . . The answer is 50%! It should be 65cu...

Question 5 aThere may be only 2 kinds, 35, 37i.e. 35 + 35 35 + 37 37 + 37

b.(9+3)/(9+6+1)=

c.There may be only 2 kinds, 35, 37i.e. 35 + 35 35 + 37 37 + 37

The average amount of formula =

Only if the ratio of the quantities of the three molecular substances is 1:1:1, C is correct, so the answer B

First question: 193>191

So 193 iridium is more, so choose A