Help solve a few chemistry problems, urgently

Chemical energy, internal energy, internal energy, electrical energy.

16g h2o d2o 3g

No. ch4+o2=co2+h2o

No O2 prior purity before experimentation.

Upstairs is right, you can also use the pH measurement method to see if there is acid HBR

1.Chemical energy, internal energy, internal energy, electrical energy.

5.No O27Add Cl2 to see if there is BR2

Chemical energy, internal energy, internal energy, electrical energy.

1) d Sodium carbonate crystals react with carbon dioxide in the air to form sodium dicarbonate. (2)

If a produces more gas than B with the same mass, it means that a has more valence electrons than b in the same mass. then a has less relative atomic mass than b. And because their valence electrons are all +2 valence, it means that they are in the same main group. The metallicity of the same family is strengthened from top to bottom. ac3)b

1b caustic soda is sodium hydroxide exposed in the air that reacts with carbon dioxide to form sodium carbonate, which can be obtained according to the chemical equation.

The mass of hydrogen produced by the reaction of a metal of equal mass in a sufficient amount of acid is directly proportional to the valence of the metal and inversely proportional to the relative atomic mass of the metal.

The phenomenon of sodium carbonate and hydrochloric acid in 3b is different when they are added dropwise to each other, and when acid is added dropwise into sodium carbonate, the acid is not enough.

First. Sodium bicarbonate is generated, so there is no gas first, on the contrary, when sodium carbonate is added to hydrochloric acid drop by drop, the acid is sufficient, so sodium bicarbonate will not be generated, and gas will be generated directly, so these two can be identified, and then it is simple.

Add the remaining two to sodium carbonate respectively, and the white precipitate is calcium chloride, and the non-phenomenon is sodium nitrate.

Identification is complete. Hope.

b. Caustic soda reacts with water and carbon dioxide in the air.

A releases more hydrogen with the same mass, the more amount of matter, so the smaller the relative atomic mass (C is not sure).

b Hydrochloric acid and sodium carbonate are different and can be identified. Add sodium carbonate to the remaining two, and the precipitate is calcium chloride.

d Because only Ca(OH)2 of these four substances is slightly soluble, the solubility is inherently small.

So the mass fraction of solutes in the solution is minimal.

The solubility of calcium hydroxide decreases with increasing temperature.

So the less solute there is.

Answer: D Because among these four substances, only Ca(OH)2 has the smallest solubility, which is only in grams at 0 and nearly grams at 100. Therefore, this question cannot be calculated in a general way by using 10 (10+90)*100%.

The remaining substances are much larger than it. In the case of A and B, B is smaller than A, because B contains crystal water.

The first type: a C atom with an H atom, a benzene ring, and two ethyl groups.

The second type: a c atom with an H atom, a benzene ring, a methyl group, and a propyl group.

The third type: a C atom with an H atom, two methyl groups, and a benzene ring (there are two -CH2- between the benzene ring and the C atom).

Fourth: a c atom with an h atom, a benzene ring (there is a -ch2- between the benzene ring and the c atom), a methyl group, and an ethyl group.

The reduced mass is the mass of O2.

1.That is, oxygen is generated.

2.After the full reaction, there must be KCl and mNO22KCLO3=mNO2=2KCl+3O2 X

The solution is x= i.e., the remaining kcl is and mno2 is.

The mass ratio is kcl:mno2=:15

cuo+h2=cu+h2o

The reduced mass of the solid is the mass of the O element in the cuo.

Reduced is.

The mass fraction of O in cuo is 16 (16+64) = so the mass of cuo is.

It turns out that the quality of the CU is.

Maybe the equation numbers are not aligned, just understand it yourself.

KCLO3 = 2KCl + 3O2 (MnO2 catalysis) Obviously, the amount of reduction is the O2 generated, so the generation of oxygen, and, and KCL, the mass is calculated by yourself!

2、cuo+h2=cu+h2o

The reduction is o, and, and the rest is cu.

1.The less mass is the mass of oxygen, i.e., potassium chloride and manganese dioxide.

1:16 reaction Since Cl changes from +1 to -1 in Naclo, giving 32 E, B is correct.

Na2Sx in C is similar to Na2O2 in structure, and it can be seen that there should be non-polar covalent bonds (same as [O-O]2-) and ionic bonds.

In D, NaClo is reduced and is an oxidizing agent.

In a, x should be 5, and the total sulfur valency is -2, and the loss of 32e becomes +30, because it is +6 valence, so there are 5 sulfur.

In summary, choose B

Trim Equation:

The valency of Sx2- -xs(+6) increases by 6x+2 valence and decreases the valency of clo- -cl- by 2 valence.

The least common multiple is 6x+2 electrons for transfer of 6x+2 electrons.

3x+1)naclo+na2sx+(2x-2)naoh= (3x+1)nacl+xna2so4+(x-1)h2o

The ratio of the amount of Na2Sx to Naclo is 1:16, then 3x+1=16x=5

X should be 5 in Ba

Na2Sx in C is structurally similar to Na2O2, and it can be seen that there should be non-polar covalent bonds ([s-s-s-s-s]2-) and ionic bonds.

Naclo oxidant, Na2SX is the reducing agent.

a, x=5b, correct, c, should be a non-polar covalent bond d, the former is a reducing agent and the latter is an oxidizing agent. Equation: 16NaClo + Na2SX + (2X-2) NaOH = 16NaCl + XNa2SO4 + (X-1)H2O is conserved here with oxygen atoms, and 16+2x-2=4x+x-1 is solved to obtain x=5

The above is very well done, and I will just talk about my quick way of doing this kind of multiple-choice question.

Let the average valency of s be y, according to the electron conservation column: 16x2=(6-y)x, the first equation.

The second equation: xy=-2, because in Na2Sx, two Nas are positive bivalent.

Two equations can be used to find x=5 and y=

Question 1, the reason is that the reaction is a chemical reaction with the participation of elemental substances, that is, the reactants are compounds, and the products are generated by elemental substances, so these four reactions must have valency changes (that is, they must be redox reactions).

There can be different situations for the first question.

Scenario 1: Q2: M compound is H2SO4, X is H2O2, Y is SO2, Z is H2S (elements: A is hydrogen, B is oxygen, C is sulfur).

Question 3: H2O2+SO2 H2SO4 (This reaction also belongs to redox reaction) Question 4: H2O2+H2S S+2H2O

Scenario 2: Q2: M compound is H2SO4, X is SO3, Y is H2O, Z is H2S (element: A is sulfur, B is oxygen, C is hydrogen).

Question 3: H2O+SO3 H2SO4 (This reaction is not a redox reaction)Q4: SO3+3H2S 4S+3H2O

1. 1,2,4,5

2.By 'm is a colorless viscous oily liquid compound'

It is easy to think of sulfuric acid, then a, b, and c are one of O2, S, and H2 respectively, and X is So2, Y is H2O2, and Z is H2S

Addendum: A is H2, B is O2, and C is S

First, add water to dissolve, and then add BaCl2

Because the resulting barium sulfate is insoluble in water and insoluble in acid, it is temporarily placed, and then an excess potassium carbonate solution is added to generate calcium carbonate, and then filtered, because there is still an excess of potassium carbonate in the solution, hydrochloric acid is added to acidify it into carbon dioxide and potassium chloride.

Finally, the evaporation crystallization is the potassium chloride we need.

A operation 3 must be before 5, otherwise the barium ions cannot be removed.

After filtration, hydrochloric acid is added, otherwise the precipitate is melted and dissolved, and the reaction is excessive carbonate.

First, according to the constant number of atoms before and after the chemical reaction, the equation for x is calculated as n2h4.

The valency is -2

Because the valency of the whole N2H4 is 0, H4 is +4, and N is -2

The brownish-yellow substance produced is iodine, because iodine ions are oxidized into iodine element, it proves that there must be a substance that is oxidizing, and potassium permanganate dissolved in water will have permanganate ions, and the high manganate ions have strong oxidation and can oxidize iodine ions into iodine element. So a certain solid is potassium permanganate.

This is inorganic chemistry in college After a rough look, the question is relatively simple, but I haven't touched it for a year, and I forgot it.

The inner capacity is almost ==

Feel free to help you see something

Analysis: The d orbital can hold a total of 10 electrons to be semi-filled, and there must be 5 electrons. Take b

For example, the outermost arrangement of Fe is 3D64S2, so Fe3+ is 3D5, and according to Hunt's rule, it is half full.

10, a analysis: the reaction quotient of 1 is to reach the equilibrium, an equilibrium is δg = 0 (under constant temperature and pressure), don't ask me why, wait until you learn physical chemistry to know.

11 Analysis: What is DG? I guess it's the Gibbs function. If the galvanic cell works spontaneously, it should be that the Gibbs function change should be less than zero, not increased.

13 Correct.

Analysis: After a cursory look, I think it's very correct.

14, wrong.

Analysis: Take 4s and 3d as examples, they are a group of energy levels, but not equivalent orbitals, 4s < 3d

Inorganic learning is the basis of other chemistry

Oxidation is the reduction of the valency of S The occurrence of the nature of acidity The combination of H and OH- so 1 choose A

Electron transfer is a valence change, so you just need to mark the elements of the valence change.

There are no requirements for electronic transfer, you have to pay attention to the book, and you will do more questions.

h2so4：500*

v=m density=

500*61=439ml

So 61ml of 98% concentrated sulfuric acid and 439ml of water are needed