Ask a chemistry question, it s a high 1, help,

Could it be that your answer is wrong, and everyone seems to do the same, but it's also possible that I'm wrong, so I'll teach you the easiest way, and you should have learned the conservation of matter, which is about the same as the conservation of energy, which is that the masses of the elements before and after the reaction are equal.

The following equations such as 3, 11, 3, 7 and so on are all mass fractions of c or o, which you should learn).

Before the reaction, element c has element o.

After the reaction, only CO2 and CO are left in the container, and if the mass of CO2 is x and the mass of CO is Y, then:

The mass of element c is x*(3 11)+y*(3 7)=oThe mass of the element is x*(8 11)+y*(4 7)=Solve this system of binary linear equations to get y=

C has Mo, O2 has Mo.

c + o2= co2

There is also Mocc + CO2 = 2Co

So there is also Moc that is.

Mention the method and do the math yourself. (Because it's too cumbersome to use the junior high school method.) ）

Calculate how many grams of oxygen are needed to calculate grams of carbon and see how much worse, because a carbon monoxide molecule contains one oxygen atom, which is one less than carbon dioxide, and the mass of carbon monoxide can be calculated with the amount of difference.

carbon dioxide and carbon monoxide.

Make a system of equations to calculate.

To be independent, don't give it answers, just give it ideas, and use your own brains.

What did you ask? Wait for ...... in the issue

Since the volume ratio of the gas is equal to the ratio of the amount of the substance, it is advisable to let the air have 1 mol, then the gas on the right side should have 3 mol, and stay in ** after ignition, indicating that the remaining gas after combustion is also 1 mol, then the total amount of H2 and O2 consumed in the reaction is 2 mol, according to the equation 2H2 + O2 = 2H2O, it can be seen that the consumption of H2 = 4 3; o2 = 2/3

If the remaining 1 mol of gas is H2, the volume ratio of H2 to O2 is (1+4 3): 2 3 = 7:2

If the remaining 1 mol of gas is O2, the volume ratio of H2 and O2 is 4 3 :(1+2 3 )=4:5

Bubbles are formed, and a blue precipitate is formed.

It can be seen that sodium cannot replace the Cu element in the salt solution.

Analysis: When sodium is put into the copper sulfate solution, the sodium element actually reacts with water to produce sodium hydroxide and hydrogen gas, which produces bubbles because of the formation of gas.

The sodium hydroxide solution formed by the reaction of sodium and water will immediately react with the copper sulfate solution to form copper hydroxide precipitate and sodium sulfate solution, so there will be blue precipitate.

Because sodium is put into any solution, the first thing that happens is the reaction of sodium with water in the solution to produce sodium hydroxide and hydrogen gas, so sodium cannot displace the elements in the salt solution.

(colorless bubbles) and (blue flocculent precipitates) are generated, so it can be seen that sodium cannot displace the (copper) element in the salt solution.

Explanation: na threw herself into the water and na was lively (immediately!!) ) reacts with water to form NaOH and hydrogen, and NaOH reacts with CuSO4 to form Cu(OH)2 blue flocculent precipitate, instead of Na displacing copper sulfate with copper, so sodium cannot replace the Cu element in the salt solution!

Under the same conditions as select d, the amount of gaseous substances is the same, and the volume is also the same, so a and c are excluded

After the reaction, the amount of NaCl decreases more than that of water, and the mass fraction of NaCl decreases, so D is chosen

The reactions that can be used to make sodium are (please explain) 2: NaCl+H2O---Na+H2+O2+Cl2 3: NaCl---Na+Cl2

1.K with NaCl2Electrolysis of saturated saline 3Steamed and dried table salt 4Electrolytic molten NaCl

The following substances do not change in the air (please explain).

4: Your salt will go bad.

1. na2o 2. naoh 3. 4. nacl

There are two factories A and B on both sides of a river, and the industrial wastewater discharged by them contains six ions: K+, Ag+, Fe3+, Cl-, OH-, and NO3-.

The wastewater of plant A is obviously alkaline, so the three ions contained in the wastewater of plant A are K+, Cl-, and OH-

Plant B's wastewater contains three other ions. If + into a certain amount of iron powder can ** where the metal ag

1.Why are the three ions contained in the wastewater of plant A K+, Cl-, and Oh-??

Precipitate: AGCL, AGoH, Fe(OH)3,

There may be NO3- ions, and KOH is the base.

2.Why + into a certain amount of iron powder can ** metal ag?

Why is NO3- not a reducing ion?

ag++fe---ag+fe2+

no3- will no longer be restored.

Why can Koh+HCl=KCl+H2O be used as H+ +OH- =H20 to represent the chemical reaction equation?

k++oh-+h++cl-==k++cl-+h2o

Discard the same ions on both sides.

Finally, the first step in writing the ionic equation is to write the chemical equation of the reaction, so how do you write the product?

For example: Ca(HCO3)2 vs. NaOH Ca(HCO3)2+2NaOH===CaCO3 +H2O+Na2CO3

Mg(HCO3)2 with excess NaOH

Mg(HCO3)2 Kei excess clarified lime water.

Spawns to write out:

The two sides subtract ions, precipitates, water, and gases that cannot be discarded.

(1) It is absolutely not possible to replace Na in NaCl solution with metal K.

Because K or Ca reacts directly with water (even if Na Na is replaced, it reacts with water.)

2) The equation is: 2NaCl + 2H2O = 2NaOH + Cl2 + H2 Condition: Energized (3).

3) Steamed dried salt or table salt, (4) It can be electrolyzed into sodium and chlorine. Active metals can be electrolyzed in this way.

2 will deteriorate in the air, 3 will lose water and become powder, sodium chloride will only deliquescent but itself does not change.

3. This is to investigate the problem of ion coexistence, the most important thing is that the solution is not electrical, but also depends on what the solution is, acidic or alkaline. You have to look carefully.

The n of nitrate is already the most **, not the highest, so there is no reduction!

The latter is the ion equation, and the essence of the acid-base neutralization reaction is that the hydrogen ions react with the hydroxide ions to form water.

Writing products can have a relationship with the amount between the two, for example, the first one is the reaction of acid salt and alkali, then there are HCO3-+OH-=CO32-+ and H2O, you have to read the book, remember the steps and rules of writing ion equations! Also, I wish you progress in your studies!

One: 1: On the contrary, K is made with KCL and sodium because the boiling point of potassium is lower than that of sodium, and potassium is volatilized in the melting state.

2: Electrolysis of saturated brine to obtain chlorine and sodium hydroxide 3: Evaporation of the promise to get crystallized powder 4: Considering the strong reducing property of sodium, electrolysis to melt NaCl is the best choice. 、

2: It is divided into chemical changes and physical changes1, which is oxidized to sodium peroxide, and then reacts with water and carbon dioxide in the air2, NaOH absorbs carbon dioxide and water to form sodium carbonate crystal water3, and the crystals will be weathered after long-term exposure to air.

Three: the alkaline must contain hydroxide, and OH- and silver ions and iron ions form a precipitate, so the positive ions can only be potassium ions. In this case, B must contain silver ions, then it cannot contain chloride ions, and the elimination method knows that chloride ions are in A.

And because B is in positive and negative balance, B contains NO3-negative ions.

2. After adding iron powder, it is first redox with iron ions to form ferrous ions, and the excess iron continues to replace silver, because iron is more reducible than silver.

3；The n element in nitrate is +5 valence, which is the most important and can no longer be oxidized, that is, it is not reducible. Nitrite can.

Fourth: whether it can be written as an ionic form depends on whether the reactants and products are strong electrolytes and have low solubility, or whether there are precipitates, gases and other substances that are out of equilibrium system. If so, the substance cannot be written as ionic.

Therefore, it can be seen that in example 1, magnesium hydroxide is mainly generated (because magnesium hydroxide is less soluble than magnesium carbonate, the reaction proceeds in the direction of generating less solubility). 3. There are both calcium carbonate and magnesium hydroxide. are all written as full chemical formulas.

Other sodium salts are related to strong electrolytes and high solubility, which can be written as ionic formulas.