High School Chemistry Topics I m in a hurry, come help me

Reacting with non-oxidizing acids, iron produces only divalent ferrous ions).

b (should be fe(2+)+2h(+)no3(-)=fe(3+)+2h2o +no)).

c (The trim is mismatched.) It should be 2Fe(3+) Cu=2Fe(2+) Cu(2+)).

So all three are wrong. Did you make a mistake by copying one?

2.3FeCl2+4Hno3(dilute)==2FeCl3+Fe(NO3)3+NO+2H2O

3 (Dilute nitric acid is not enough to oxidize Cl-, at least it will not oxidize Cl- salt or HCl solution to Cl2 and release it.)

In addition, in the explanation of the mechanism of strong oxidation of aqua regia, a popular view is that a certain concentration of nitric acid can partially oxidize Cl- to neutral Cl atoms, and reduce itself to nitrosyl chloride (NOCL), which is strongly oxidizing. However, it is said that in recent years, it has been found that there are loopholes in the chemical kinetics of this statement (I googled it, but I didn't find any information). ）

4。Potassium manganate is unstable and only exists in strong alkaline conditions, if potassium manganate must be used to react with hydrogen peroxide in strong alkaline conditions, hydrogen peroxide can also be reacted under strong alkaline conditions, potassium manganate is reduced to manganese dioxide, and hydrogen peroxide is oxidized to oxygen, but the reaction is very difficult.

In the case of potassium permanganate, it is 2kmNO4 + H2O2 = 2mNO2 + 2KOH + O2

2kmno4+h2o2+3h2so4=k2so4+2mnso4+4h2o+3o2↑

2mno4- +5h2o2 + 6h+ = 2mn2+ +5o2 + 8h2o

A is false, and the reaction produces 2-valent iron ions.

b correct. C false, copper is less reactive than iron and cannot produce such a reaction.

Fe(2+)+4H(+)NO3(-)=Fe(3+)+2H2O +NOn, Cl is more oxidizing than N.

No oxygen will be generated, and the product should only be water and MN ions.

The general non-oxidizing acid in A cannot oxidize Fe to Fe3+C, and Fe3+ can oxidize Cu, but it is not trimmed.

3.No, because the resulting Cl2 is more oxidizing than Hno3.

4.The reduction product should be mno2, and h(-) should be h(+) There is also a problem with the balance, and the balance of gain and loss electrons must be satisfied.

2mNO4(2-) 6H2O2 +8H(+)=2mNO2+ 3O2 [Gas Generation Bond] +10H2O

1 Option B A can only generate 2-valent iron ions, and 3-valent iron ions are more oxidizing than hydrogen ions.

c Ions are not conserved 2Fe(3+)+Cu=2Fe(2+)+Cu(2+)2 The first question has.

3 The charge is marked incorrectly mno4(-)h(+).

Catalyst MNO2

2kmNO4+H2O2+6HCl=2mnCl2+3O2 [gas bond] +4H2O+2KCl

The equal sign says mno2

1. All three are wrong.

Item A, ferrous ions Fe + 2H+ = Fe2+ +H2 should be generated

Item b, the number of charges on both sides of the equation is not balanced The correct one is: 3fe2+ +4h+ +no3- = 3fe3+ +2h2o + no

C, the charge number on both sides of the equation is not balanced, the product is written correctly, the trivalent iron has strong oxidation, and copper can be oxidized into copper ions.

The correct one is: 2Fe3+ +Cu = 2Fe2+ +Cu2+

fe2+ +4h+ +no3- = 3fe3+ +2h2o + no

3. Yes, dilute nitric acid cannot oxidize Cl-

4. The fault is that permanganate ions (mNO4-) and hydrogen ions (H+) are written incorrectly, and the trim is incorrect. The product is written correctly, hydrogen peroxide is oxidized by potassium permanganate to produce oxygen.

The correct ionic equation is: 2mNO4- +5H2O2 + 6H+ = 2mN2+ +5O2 + 8H2O

The chemical equation is 2kmNO4 + 5H2O2 + 3H2SO4 = 2MnSO4 + K2SO4 + 5O2 + 8H2O

False, the hydroxide ion iron oxide element only generates positive bivalent; b is false, the number of electrons obtained by nitrogen does not match the number of electrons lost by iron; c Yes, this is the method of using ferric chloride ** scrap copper in industry;

gas generation) +2H2O

3.No; In fact, even concentrated nitric acid cannot, remember the composition of aqua regia?

4.The product should be mno2, because hydrogen peroxide cannot reduce permanganate to bivalent, only strong reducing agents can, it should be:

2mno4(-)3h2o2+2h(+)==2mno2+4h2o+3o2 (gas mark).

Summary. There are 1 electron in each of the 2 sp hybrid orbitals of the C atom, 2 orbitals have 2 electrons in the 4 sp3 hybrid orbitals of the O atom, and 2 orbitals each have 1 electron, so that the 2 SP hybrid orbitals of the C atom overlap with the 1 sp3 hybrid orbital of the 2 O atoms, respectively, and the other 2 P orbitals of the C atom and the two sp3 hybrid orbitals of the other 2 O atoms overlap "shoulder to shoulder", and the structure is O=C=O

Dear <> happy to answer for you, take a look at this high school chemistry: yes, what is the chemistry question.

This question is equivalent to a small amount of ammonium aluminum sulfate, set it as 1, then there is a letter bureau 2mol sulfate, barium from the slippery clear let the son is excessive, so the sulfate should be all precipitated, that is, 2mol barium sulfate, 2mol barium ions are needed, and the <> is pro

What's the question? Mistaken person? Question 1.

The first question is: Carbon dioxide can't be PP hybridization?

The <>-pro-carbon dioxide is a hybrid of 2 sps.

Why. There is 1 electron in each of the 2 sp hybrid orbitals of the C atom, 2 orbitals have 2 electrons in the 4 sp3 hybrid orbitals of the O atom, and 2 bond-space trembling orbitals each have 1 electron, so that the 2 SP hybrid orbitals of the C atom overlap with the 1 sp3 hybrid orbital of the 2 O atoms, and the other 2 P orbitals of the C atom and the two sp3 hybrid orbitals of the other 2 O atoms overlap "shoulder to shoulder", and the structure is O=C=O

Isn't that a difficult question!

First question: The liquid in step 1 adds up to almost 50 ml, so choose C, half of the volume, and the heating back to the mold is more smooth.

The second question: choose C, the organic matter and water can be steamed out into the condenser tube, the most important thing is that the water can be steamed out, so that the "water separator" can play a role.

The third question: the role of sodium carbonate, memorize the relevant content of "preparation of ethyl acetate" in the book, there are three main points; Fill in the last blank with "dry".

The fourth question: the wrong choice is because if the upper mouth enters and the lower mouth goes out, the ruler book water in the mezzanine is not filled; c Wrong because the separating funnel is deflated by an inverted funnel operated by a, not a glass plug on the top; D is wrong because there should still be an aqueous layer underneath, and it can't be an organic layer that is released first.

Question 5: Using grams as a basis, calculate the theoretical yield of that ester, the data here is prepared (cinnamic acid is just mol), and then divide the actual denier yield by the theoretical yield to get the yield.

Do the rest yourself.

(1) CaCO3 = High Temperature = CAO+CO2

Cao + SO2 = High temperature = CaSO3

2caso3 + o2 = 2caso4

2) Choose D as the answer

A false. n is the most ** is +5 valence, and the hydrate corresponding to the oxide is nitric acid, which is a strong acid.

B false. Oxides have salt-forming oxides, not salt-forming oxides, and salt-forming oxides can react with strong alkalis to form oxygenates, which is also a characteristic reaction to judge whether an oxide belongs to salt-forming oxides.

For example, if CO2 reacts with NaOH to form the oxynate Na2CO3, then CO2 is a salt-forming oxide.

For example, CO and NaOH cannot form oxygenates under general conditions, and sodium formate can only be generated under very harsh conditions such as high temperature and high pressure, which is also an organic salt. Therefore, CO is considered to be a non-salt oxide.

The oxides formed by N, C and Si are: NO2, NO, N2O4, NO2, etc.; co2，co ； sio2

Among them, CO and NO are non-salt oxides, which are considered not to react with NaOH; SiO2 reacts with NaOH at room temperature to form NaSiO3, but the reaction rate is very slow.

C false. The stronger the non-metallic nature of the element, the greater the stability of the gaseous hydride formed.

Such as: halogen stability hf hcl hbr hi

n, c, si, non-metallic: c n si

From this, it can be concluded that the stability of the gaseous hydride formed is: NH3 CH4 SIH4

However, the spatial structure of CH4 is very special, and the spatial structure of CH4 is a regular tetrahedral type with 4 H atoms as the 4 vertices and C atoms as the center, which has a high degree of spatial symmetry, large C-H bond energy, and good CH4 stability.

The spatial structure of NH3 is a tetrahedral structure with N atom as the center, 3 H atoms and a lone pair of electrons of N as the 4 vertices, in which 4 atoms in the NH3 molecule form a trigonal pyramid. Due to the large repulsion of lone pairs, the Nh3 molecule and a lone pair of N's lone pairs do not form a tetrahedron, and the repulsion of n-H bonds by lone pairs of electrons decreases the N-H bond energy and NH3 stability.

The stability of CH4 is stronger than that of NH3.

Therefore, the stability of the gaseous hydrides formed is from large to small: CH4 NH3 SIH4

d correct. C in CH4, NH3, SIH4 is -4 valence; n manifest -3 valence; Si explicit -4 valence, all of which are reducible, and O2 can undergo redox reaction, that is, CH4, NH3, and SiH4 can all be burned in O2 and release a large amount of heat, so the heat of reaction is greater than 0

1.Because of the high temperature generated when coal is burned. Causes limestone to decompose into calcium oxide.

The reaction equation of this step is: CaCO3 = high temperature = Cao + CO2 and under high temperature conditions, calcium oxide will react with sulfur dioxide to form calcium sulfite. The reaction equation for this step is:

Cao + SO2 = High temperature = CaSO3Therefore, the amount of sulfur dioxide in the exhaust gas can be reduced.

2.The answer is C A counter-example of the answer is: The hydrate of the most ** oxide of n is HNO3 and nitric acid is a strong acid.

The counter-example of the answer b is: carbon monoxide, the oxide of c, is not a salt oxide and cannot react with a strong base d The answer is obviously wrong and does not need to be prompted again.

Hope the answer will help you.

Chemistry enthusiasts sincerely answer for you.

1.Limestone caCO3=high temperature=cao+CO2 cao+so2=caso3

2.Option A, false, Hno3 is a strong acid.

Option B, wrong, acidic oxides can react with bases, and such as CO, NO are not acidic oxides C options, right.

d option, false, the reaction is exothermic, the heat of reaction is less than 0

Got e-

o2 336ml

In the first question, you can use nitric acid as a catalyst.

Couple the equation and remove no2 & no to get the equation cu+o2 =cuo, which you like, and directly follow the equation ratio to get the answer.

If the second question is simple.

w=40%The second question is an elementary school arithmetic problem.

Why do you need equations? Just for the tip ear!

When a certain amount of concentrated Hno3 is put in, the copper is completely dissolved, and the color of the gas produced changes from dark to light, and a total of 672ml of gas (standard condition) is collected, and a certain volume of oxygen is introduced to make the gas completely soluble in water. Calculation:

1) What is the total number of electrons obtained by nitric acid in the reaction of copper with nitric acid?

2) What is the volume of oxygen introduced under standard conditions?

1) The amount of electrons of the substance is.

2) The amount of O2 is .

2.9 grams of magnesium and aluminum alloy were put into 40ml of NaOH solution, and the gas produced by the reaction was measured as standard). When a sufficient amount of hydrochloric acid is added, all the remaining metals are dissolved, and gas is measured to be produced (standard condition). Seeking:

1) the amount and concentration of the substance of the NaOH solution;

2) The mass fraction of magnesium in the alloy.

24x 27y 9 (mass relationship).

2x 3y electrons conserved).

Solution: x; y＝

According to the title, if you want to produce hydrogen under standard conditions (i.e. one mole), you need to mix 20g.

The relationship between the reaction of common metals and acids to produce hydrogen should be remembered very clearly.

To produce hydrogen under standard conditions, the mass of each metal required is as follows:

46 grams of sodium, 24 grams of magnesium, 18 grams of aluminum, 65 grams of zinc, 56 grams of iron.

Since 20 grams of mixed metals are required, one of the two components must have a mass greater than 20 grams and the other less than 20 grams for the reaction of the two components alone. less than 20 grams, only aluminum. So it must contain aluminum.

In addition, it can also be considered from the perspective of conservation of electrons of gain and loss. To generate liters of hydrogen under standard conditions, 2 molar electrons are obtained.

According to the valency of these metals when they form cations, see which metal has a mass of less than 20 when it loses 2mol of electrons.