A compulsory high school chemistry calculation problem is still thanks

Solution:1The sentence of waiting for the amount of matter tells you to use mol to solve this set of problems.

2.The addition of two chlorides to sulfuric acid is to tell you that these two chlorides are dissolved in sulfuric acid, sulfuric acid is still sulfuric acid, chloride is still chloride, so 50 1000*1 = mol NaOH (oh root) goes to combine with (H root) (neutralization reaction), but sulfuric acid is binary, and the one to be divided by two is sulfuric acid (containing roots).

3.NaOH adds a precipitate, and then adds no precipitate, and there is a thing called AL salt (ALCL3 or AL2SO43) in high school chemistry, just remember it, this is called amphoteric hydroxide. See below for the reasons.

Amphoteric hydroxides are generally hydroxides of amphoteric elements (such as zinc hydroxide, aluminum hydroxide, etc.) and intermediate valence hydroxides of variable metals (such as chromium hydroxide, etc.). (Just memorize aluminum hydroxide in high school).

al(oh)3＋3h+=al3+＋3h2o

al(oh)3＋oh-=alo2-＋2h2o

4.v3- v2- v1 = 100ml, why reduce it, remember the hobby of the questioner, and give the next set if you are not serious. The first way to solve the problem is already very careful, and the AL salt (ALCL3) is definitely there.

5.However, Al(OH)3 deep into 1mol can be 3molNaOH, so the generated AL(OH)3 can be wanted, the second question is V2- V1=600ml, that is, it is used to generate Al(OH)3 precipitation, and there must be a precipitation, pay attention to the "amount of substance", indicating that another substance is also, and the reaction to generate precipitation, reflecting the equation is written, of course, it is the same as the generation of Al(OH)3, in addition to aluminum, there is also trivalent iron is a 3-valent metal. That's him.

The initial addition of 50 ml of NaOH neutralizes the remaining hydrogen ions, otherwise the precipitate will not be produced.

50ml will make the solution neutral, indicating that 100ml of sulfuric acid solution contains hydrogen ions, that is, sulfate. In this way, the amount concentration of sulfuric acid substances is.

Starting from the last phenomenon, this effect shows that there are amphoteric hydroxides, and the only possibility is aluminum hydroxide, that is to say, 100ml of NaOH will dissolve all aluminum hydroxide, so that aluminum hydroxide has, and the precipitation completely consumes hydroxide. At the maximum value of the precipitation, hydroxide is added together, which is additionally consumed by another chloride and is required to be in the form of a precipitation, with the possible result of iron. i.e. chlorides are AlCl3 and FeCl3, each.

1) Neutralize sulfuric acid to neutral (2) (3) Aluminum chloride Ferric chloride each.

The first c is -ch, which can be seen to be sp hybridization, so the A and D options can be excluded; Using the more familiar knowledge of acetylene, we can see that -c c- is sp hybridization, so excluding option c, the answer is determined to be b.

Choose B, methyl carbon is sp3 hybrid, carbonyl carbon is sp2 hybrid, planar triangle, three-dimensional structure, and the last 2 carbons are sp hybrid, straight type.

1.Your question is in a leaky state, but it should be liquid, otherwise it won't count.

co2（g）+2h2o（g）= ch4（g）+2o2（g）△h5=

C (graphite) + O2 (g) = CO2 (g) H2 = —393 5kJ mol

2h2(g)+o2(g)=2h2o(l) △h3= kj/mol

C (graphite) + 2H2 (G) = CH4 (G) ΔH4 = H5 + H2 + H3

Starting 2 1 0

Change the final stateFrom the final state, it can be seen that the concentration of SO3 is.

v=△c/△t

So v=3It is known that the molar mass of the hydrocarbon m= formula m= ·vm, where vm=

The relative molecular mass of this hydrocarbon is 72

2)∵n(h2o)=，n(co2)=11/44=

and n(hydrocarbon)=,n(h)=2·n(h2o)=,n(c)=n(CO2)=

n(h)=，n(c)=

The molecular formula of this hydrocarbon is C5H12

Graphite) + 2H2 (G) = CH4 (G) Heat of reaction ΔH4 can be determined by:

2) +(3)-(1) can be obtained, so the δh4 of this reaction is:

h2+△h3- △h1= —393．5kj/mol+─ =211kj/mol2.

Reading well is much more useful than begging on your knees.

waist, got up and cleaned up briefly, leaned against the bed, and lit a cigarette. Thinking about that strange Taoist priest.

The question is, do you want to be superstitious or not, go for it.

Stoichiometry is easier to do than mathematics and physics, and the key is that you should grasp it well.

1.The chemical equation of the reaction.

2.If you can understand these two points, it will be much easier to do the calculation problem, and if you know the chemical equation, the key is that those quantities can be substituted into the calculation.

It's up to you to understand, and sometimes the data has to be processed.

The mixture is to be turned into a pure substance, and the difference between the mass before and after the reaction is the difference method.

There is also the law of conservation.

There are only a few conservations in chemistry.

1. Conservation of mass.

2. The valence rise and fall are equal.

3.The gains and losses of electrons are equal.

4. Conservation of mass of elements, etc., if you want to subdivide a lot of them, usually listen to the teacher talk about practice problems, especially when calculating problems, you should listen more.

Take a look at the teacher's analysis method, if you can solve it step by step!

That's a bit much, I hope it helps!

Don't study it, you can't do math.

Don't be afraid, just memorize the chemical equations, and practice more, more practice is definitely the key.

College Entrance Examination Resource Network There are a lot of fine questions! I'm also high all my life! 100% accurate.

According to the clever use of chemical equations, the core idea of the solution is that mass is conserved in chemical reactions, and there is the most basic proportional (quantity) relationship between each reactant and the product.

Example 1] A mixture of H and Co has the same density.

Re-introduce the excess o

2. Finally, the solid mass in the container increases.

A GB GC GD G Analysis: This problem should be solved according to the following relation:

The added mass of the solid is the mass of H2.

The added mass of the solid is the mass of CO.

Therefore, in the end, the weight of the container has increased, and A should be chosen.

Just analyze the estimate of this question. The key to solving the problem is to find out the relationship between the amount in the reaction.

Example 2] Fes2 reacts with nitric acid and has Fe

3+ and H2SO

4. If the ratio of the amount of Fes2 and Hno3 in the reaction is 1 8, then Hno

The only reduction product of 3 is.

a．no2b．no

c．n2o d．n2o

3 Analysis: This question is calculated using the redox relation. FES in the reaction

The ratio of the amount of 2 to Hno3 is 1 8, due to the formation of Fe(NO3).

3, then FES

The ratio of 2 to the amount of reduced Hno3 is 1 5.

Let the valence of n element be x, and the following redox relationship can be listed and analyzed:

The key to this question is to find the implicit relationships.

2. Equations or systems of equations.

According to the relationship between the conservation of mass and proportionality, the establishment of unknowns according to the problem conditions, and the solution of column equations or systems of equations are the most commonly used methods in chemical calculations, and their problem-solving skills are also the most important calculation skills.

Example 3] MCE

1999—24) electrolysis of m(NO

3) X aqueous solution when weight is increased on the cathode

g, BL oxygen is produced at the anode at the same time (standard condition), so that the atomic weight of m is known.

Analysis: Equations, or systems of equations, are the most commonly used and least error-prone chemical calculations.

Electrode reaction of the cathode and anode:

Cathode: 4mx++4xe=4m

Anode: 4xOH - 4xe = 2xH2O + XO2

Let the atomic weight of m be y

The correct answer is c.

Example 4] There is a mixture of a certain alkali metal M and its corresponding oxide for a total of 10

g, after fully reacting with a sufficient amount of water, carefully evaporate the solution to give 14

g anhydrous crystals. The alkali metal m may be.

A Lithium B Sodium C Potassium D Rubidium.

The atomic weights of lithium, sodium, potassium, and rubidium are:

Analysis: The reaction formula relationship between alkali metal and its oxide reaction with water is: 2m+2h

2o=2moh+h

2↑ m2o+h2

o=2moh

There are several ways to solve this problem.

There is a special solution: Let the atomic weight of m be x

The atomic weights of lithium, sodium, potassium, and rubidium are analyzed, and the correct answers to the question are b and c.