Two high school chemistry questions, urgent, urgent, urgent, two high school chemistry questions.

The formation of No2 is to decrease the price of 1*2 to generate O2 and increase the price of O2 by 4*1, so the valency of the metal is reduced by 2 valence.

When 3mol H2SO4 is generated according to the measurement number, 3molCu2S should be reflected, and a total of 6 valence is increased, so the nitrogen element is reduced by 6 valence to form 2molNO

11 solution: from the valence of nitrate 2 mol + 4N to 2 valence 1mol oxygen to 4 valence, so the metal to 2 valence (how many mol do not know).

2. Denitric acid becomes nitric oxide, and 1mol decreases by 3 There is 6 mol +1cu to +2cu valence covariance +6, so 2mol is reduced

1，b2，c

1,NO3-,generate NO2,N to get 1*2 electrons, generate O2,O to lose 4*1 electrons, then 2 of the 4 electrons lost are given to N, and 2 are given to metal salts, and the electron valency is reduced.

2. The generation of 3mol sulfuric acid reacts to 3mol Cu2S, Cu becomes 2-valent, loses 1 electron, a total of 6mol electrons are transferred from Cu, and NO3 is reduced by Cu and S from 5+ to 2+, so the NO3- reduced by Cu is 2mol

Lower. The number of electrons lost = the number of electrons gained, by the increased valency = the decreasing valency.

The valency of the n element decreases from +5 to +4, giving 1 electron, and having 2 nitrogen dioxide, so getting 2 electrons.

The valency of the element o increases, -2 rises to 0, 2 electrons are gained, there is 1 oxygen molecule, so 4 electrons are obtained.

The number of electrons lost = the number of electrons gained, so 2 more electrons should be gained, and the valency of the metal element decreases.

To be reduced, it should be reduced in valency, only n is reduced in valency, nitric oxide is a reduction product, there is 10mol of nitric oxide.

1.First of all, the valency rise and fall of redox chemical reactions must be conserved. For O, the valence increases by a total of 2n*2

2 valence to 0 valence, so that the amount of matter of O2 is n, then the oxygen atom 2N). For n, the valency decreases by 2n*1

From 5 valence to +4 valence, the amount of substance n is 2n), so the valency of the metal must be reduced by n before it will be conserved, and b is chosen.

2.For this reaction, it was first found that N in Hno3 not only exerts oxidation, but also produces Cu(No3)2 with Cu in a part of NO3-. The actual reaction is related to 3Cu2S-(22-6*2)NO3.

Of course, according to the valence rise and fall, it must be conserved. Analyzing the total increase of the valency of Cu (6mol), the valency of S increased by 3*8=24mol, and the corresponding valency of N decreased (24+6) 3==10mol

It would be nice to know the nature of chemical reactions.

1. KCl+H2SO4+KMNO4=mnSO4+K2SO4+CL2+H20 (thanks upstairs, there is indeed one missing).

2. A: Na, B: S, C: O2

a:na2o ;

b: na2o2 ;

c:naoh ;

d: na2so3 ;

e:na2so4;

f:so2;

g:so3;

x:na2s

The molecular weight of CH4 is about 16, and it is estimated to be methane from the perspective of the amount of hollowing picos. . . Hydrocarbons are, anhydrous calcium chloride absorbs water, that is, it takes away the H,,H elements in the original combustion of Douqi, and the same is C, so it is methane.

C2H4 concentrated sulfuric acid absorbs water, that is, it takes away the H in the original combustion, 20 of the 50ml is itself, and 30 is water, the ratio of hydrogen to oxygen is 2:1, so H4, sodium hydroxide absorbs CO2, 40ml, that is to say, there are two Cs

1. Water = hydrogen.

Dioxygen deficit digests carbon = carbon.

The carbon-to-hydrogen ratio is 1:4

It could only be methane.

2. CXHY+(x+y 4)o2==xco2+y 2h2o1 x+y 4 x y 2

20 20x 10y

After passing through sulfuric acid, the water vapor is gone, then 10y=50, y=5 after passing through sodium hydroxide, the carbon dioxide is gone, then 20x=40, x=2.

Therefore, it is the simplest C2H5

Then it can be confirmed that the gaseous hydrocarbon is butane C4H10

1) CAC2+2H2OCA(OH)2+C2H2 B water storage, so that when the C2H2 gas generated in the flask enters B, the same volume of water as it enters the graduated cylinder C to measure the amount of acetylene generated Prevent the action between calcium carbide and residual water, prevent the generated C2H2 from escaping into the atmosphere, and facilitate the smooth dropping of liquid.

2) Too much C2H2 is generated, exceeding the volume of B, and its volume cannot be determined Too little C2H2 is generated, and the measurement error is large.

3) The reaction rate is fast, a large amount of heat is released, and the resulting Ca(OH)2 suspension clogs the duct.

4) Saturated salt water.

5) In question (5), the state is, 25, and to calculate the amount of matter needs to be transformed into a standard state, 0

The right Feifei Fei heating tube other people gothic overdischarge.

The amount of OH- and H+ added is equal: 1molOH- binds to 1molH+, 1molmg begins to bind to 2molCl-, binds to 2moloh- when the precipitate reaches the maximum, 1molAL begins to bind to 3molCl-, and when the precipitate reaches the maximum, it binds to 3moloh-, 1molCl- to 1molH+, so 1molH+ to 1moloh+.

cuo is a black solid powder, and the dark red substance is cu. According to the metal oxidation order table, Al first replaces Cu2+ (which is the replacement between metal elements), and then replaces H+, so the gas is H2.

The equation is correct.

From the title, we can know that the alloy of Al and Mg is 3G, at this time, we use extreme value analysis, assuming that all 3g are Al or are all Mg, we can calculate that HCl is excessive, which is a common analysis method in chemistry.

Because the alloy will be dissolved by HCl, the number of H+mol of the dissolved alloy is equal to the number of moles of the alloy, but because the alloy is dissolved in all the precipitates, oh- must first react with the remaining H+, plus the number of moles with the alloy, that is, the H+ of the previous reaction, that is, when oh-=H+.

cuo is a black solid powder, and the dark red substance is cu. According to the metal oxidation order table, Al first replaces Cu2+ (which is the replacement between metal elements), and then replaces H+, so the gas is H2.

The equation is correct agree with the upstairs.

The copper generated by the equation is preceded by 3 written 3, which is 3cuAluminum and copper chloride only produce aluminum chloride and copper, there is no copper oxide, don't ask why. The dark red substance is copper. Chloride ions can accelerate the destruction of alumina films.

Many materials such as stainless steel, aluminum, etc., can resist corrosion by passivation films, as long as chloride ions are in direct contact with these materials, corrosion will occur.

Because the metal in the passivated state still has a certain ability to react, that is, the dissolution and repair (repassivation) of the passivation film is in a state of dynamic equilibrium. When the medium contains reactive anions (such as chloride ions), the equilibrium is disrupted and dissolution prevails. The reason is that chloride ions can preferentially and selectively adsorb on the passivation film, expelling oxygen atoms, and then combine with the cations in the passivation film to form soluble chlorides, resulting in the formation of small pits (pore size mostly 20 30 m) at specific points of the newly exposed base metal, these small pits are called pore cores, which can also be understood as the active centers of pore formation.

The presence of chloride ions has a direct destructive effect on the blunt state of the metal.

Let's start by writing down the equation.

In this way, the amount of substances containing solute sodium hydroxide in the 5% sodium hydroxide solution is calculated as the amount of Na2SO4 and H2O left in the solution after the reaction

According to the ratio, after the reaction is calculated, Na2SO4 and Na2SO4 are calculated to be and 9GH2O

and then according to the mass fraction formula.

2, manganese dioxide =

mNO2+4HCl=mnCl2+Cl2+2H2OAccording to the equation, the excess of HCl is judged, so the amount of Cl2 generated is determined by Mno2.

So the generated cl2

There is still a solution left after the reaction.

There is also MnCl2 in the solution

So there is a total of cl- left here.

1 10 is cl-

cl-+ag+＝agcl ↓

Therefore, the agCl precipitate can be generated.

1. Solution: n= of 25g copper sulfate crystals

As it happens to react with Cuso4 5h2o + 2naoh =cu（oh）2 ↓+na2so4+5h2o

Therefore, (1) the required sodium hydroxide (2) can be found from the chemical formula to find the amount of sodium sulfate substance generated, and then replace it with mass, and then use sodium hydroxide to find the mass of the previous solution, +25g, that is, the current mass, and then it can be calculated.

4hcl =mncl2 + cl2↑+h2o

Manganese dioxide is the concentrated HCl required for a complete reaction, so manganese dioxide can be completely reacted to form Cl2

The chloride ions remaining in the solution after the reaction are 9...2mol So take out one-tenth of the AgCl precipitate formed after reacting with a sufficient amount of AgNO3 with a mass of X

then the quality is.

1.Solution: (1) If n(na)>=n(al).

23n(na)+27n(al)=m

1/2n(na)+3/2n(al)=n/

According to the above binary equations, n(na) and n(al) can be obtained, and if the assumption is satisfied, then the masses of the two can be obtained according to the molar masses of the two.

2) If n(na)23n(na)+27n(al)=m

1/2n(na)+3/2n(na)=n/

According to the above binary equations, n(na) and n(al) can be obtained, and if the assumption is satisfied, then the masses of the two can be obtained according to the molar masses of the two.

2.Solution: From the meaning of the question, it can be seen that the amount of Fe before and after the reaction does not change, and the Al element becomes the O element, so the mass of the Al element is equal to the mass of the O element in Fe2O3, and the mass fraction of Fe in the original alloy is the mass fraction of Fe in Fe2O3, that is, 70%.

In the 1t, na, al only na reacts with water, and the reaction equation is 2na+2h2o=2naoh+h2 (arrow, you can't type it on your phone, forgive me), n(h2)=n, so n(na)=n, and then calculate m(na)=23n, and then launch (al), I won't count the specifics.

The 2T, Fe, Al and dilute hydrochloric acid are reacted to generate FeCl2 and AlCl3, and then add excess NaOH to obtain Fe(OH)2 precipitate and Naalo2 solution respectively, the precipitate is burned and becomes Fe2O3, according to the title, the mass of O3 in AgFe2O3 is the mass of Al, then Fe% = 56 * 2 160, the landlord himself calculates it, I don't write the equation either, there should be it in the book or notes