Senior 1 Chemistry Chapter 1 Ionic Reactions Contents

Step 1: Write the chemical equation for the reaction.

For example: cuso4+bacl2=cucl2+baso4 Step 2: Write the substances that are easy to ionize with water and easily in ionic form (only strong electrolytes can be written), and insoluble substances or substances that are difficult to ionize and gases are still expressed by chemical formulas (including weak electrolytes and non-electrolytes, which cannot be written as ionic equations).

For example: Cu2+ +SO42+ +Ba2+ +2Cl-=Cu2+ +2Cl-+BaSO4

Step 3: Remove the ions on both sides of the equation that do not participate in the reaction.

For example: ba2+ +so42- =baso4 Step 4: Check whether the number of atoms of each element on both sides of the ion equation and the total number of charges are equal.

The easiest way to do this is to write down the chemical equation for it. Then disassemble into an ionic type, noting that the insoluble matter cannot be disassembled.

The sediment gas cannot be disassembled.

1.Ba(OH)2 +K2CO3 ===BaCO3 precipitate + 2koh CaCl2 +K2CO3===CaCO3 precipitate +2kCl

BA2+ +CO32- ===BAC3 precipitate Ca2+ +CO32- ===CaCO3 precipitate.

2.BaCO3 + 2Hno3 ===Ba(NO3)2 +CO2 gas sign + H2O

BACO3 +2H+ ====BA2+ +CO2 gas symbol + H2O

CaCO3 + 2Hno3 ===Ca(NO3)2 +CO2 gas sign + H2O

CaCO3 +2H+ ====Ca2+ +CO2 gas sign + H2O

3.CaCl2 +2 AgNO3 === AgCl precipitation + Ca(NO3)2

Cl-+AG+ ***** agCl precipitation.

It is judged that the original white powder must contain the substance: CaCl2 K2CO3, and may contain the substance: Ba(OH)2

Definitely contains K2CO3 and CaCl2 and may contain Ba(OH)2.

ca2++co32-=caco3 caco3+2h+=ca2++co2+h2o cl-+ag+=agcl

There is a packet of white powder, in which there may be 3 colorless solutions of Han Youcai Ba(OH)2CaCl2K2CO3, 1Use a reagent to identify them and write the chemical and ionic equations.

With litmus test solution, drop added to Ba(OH)2 solution, blue color; Dropwise added to K2CO3 solution, red; Add dropwise to CaCl2, purple color. ba(oh)2 = ba2+ +2oh- ；co3(2-) h2o ó hco3- +oh-

2.(1) Part of the powder is added to water, shaken, and white precipitate is generated. (There must be k2co3).

2) Excessive dilute nitric acid is added to the suspension of (1), and the white precipitate disappears and bubbles are generated.

3) Take a small amount of (2) solution and drop it into the Agno3 solution, and a white precipitate is generated. (There must be cacl2).

According to the above experimental phenomena, determine what substances must be contained in the original white powder, what substances may be contained, and write the relevant ion equations.

Answer: There must be K2CO3 and CaCl2 and possibly Ba(Oh)2

00, with hard work for you to do it in word, awesome!

Choose D,The non-metallic nature of Br in IBR is stronger than that of I, so the electronic ability is stronger, so the electron pair will be biased towards BR, so that BR is -1 valence, so I is +1 valence, and the whole reaction is whole, Bromine-1 valence in HBR and I +1 valence in HIO are not valence, not redox, not oxidant and reducing agent.

Option A, according to the oxidation of the oxidant (the reduced valency of the reactant) is stronger than the oxidation of the oxidation product (the substance with the increased valency of the product), 2H2S + O2 = 2S + 2H2O: O2 (oxidant) >S (oxidation product); In the same way, in 4NaI+O2+2H2SO4=2I2+2Na2SO4+2H2O, O2 (oxidant) > I2 (oxidation product);

Na2S+I2=2Nai+S in I2 (oxidant) > S (oxidation product), in summary: O2>I2>S

Select B (1) Cl2 + 2Ki = 2Ki + I2 Medium oxidation: Cl2> I2 (2) 2FeCl2 + Cl2 = 2FeCl3, Cl2 >Fe3+

3) 2FeCl3+2Ki=2FeCl2+2Kcl+I2 Fe3+>I2 (4)I2+SO2+2H2O=H2SO4+2HI I2>H2SO4, but the comparison in the title is SO2, there is a rule, in the same reaction, the oxidation of the oxidant is stronger than that of the reducing agent, so I2> SO2, rightfully so.

In summary: Cl2 Fe3+ I2 SO2

In summary: choose B to choose CD, the valency of H2O in BR2+SO2+2H2O=H2SO4+2HBR neither increases nor decreases, neither oxidant nor reducing agent, reclaimed water is neither an oxidant nor a reducing agent, and the valency changes only the oxygen in sodium peroxide, which is both increased and decreased; Middle.

Water is neither an oxidant nor a reducing agent, and the valence is 4Fe(OH)2 and O2; The valency of oxygen in the reclaimed water is increased, and the water is used as a reducing agent, and the original reaction of the reaction is.

2al+2naoh+6h2o=2naalo2+3h2 +4h2o, in order to facilitate writing this as you wrote, so the hydrogenation valency in the water is reduced, and it is used as an oxidant. Select CD

In IBR, look at the electronegativity (this will be learned later, the same family, the electronegativity is decreasing from top to bottom, the electronegativity of BR is greater than that of I, so BR is -1 valence), I is +1 valence, it can be seen that this is not a redox reaction, and the valency does not change. Choose D,

The oxidation of the oxidant is greater than the oxidation of the oxidation product, and it is a reducing agent because the electron loss of the reactant with increased valency is oxidized to reflect its reducibility. The electrons obtained from the reactants with reduced valency are reduced to reflect their oxidation and are oxidants. Judging from this, oxidizing O2 S, O2 I2, I2 S, so the answer is A

The method is the same as above, oxidizing Cl2 I2, Cl2 Fe3+, Fe3+ I2, I2 H2SO4, is this topic wrong? The SO2 in the option should be H2SO4, so change it to B

Br decreases by 1 2 valencies and S increases by 2 valences, and the water valency in the subject stem remains unchanged, which is neither an oxidant nor a reducing agent. The O-1 of a Na2O2 is reduced to the NaOH-2 valence, and the -2 of H2O is increased to the O2 0 valence, and H2O is a reducing agent.

The valency of B H2O remains unchanged, and the same as the stem of the question C H2O to O2, the reducing agent D H2O to H2, the oxidizing agent. Only B is the same as the question stem, so choose ACD

（1）d 。On both sides of the chemical equation, the valency of I and BR does not change, so IBR is neither an oxidizing agent nor a reducing agent.

2）a。The first equation can be derived from the oxidation of O2 and the oxidation of s; The second equation can be obtained from the O2 oxidation >> I2 oxidation; The third equation can be obtained from the oxidation of I2 and the oxidation of s; In summary, a is obtained.

3）c。The first equation gives Cl2 oxidation >> i2 oxidation; The second equation yields Cl2 oxidation >> Fe3+ oxidation; The third equation yields Fe3+ oxidation >> i2 oxidation; In the fourth equation, i2 is SO2 is the oxidizing agent and I2 is the reducing agent, so SO2 oxidation >> I2 oxidation; In summary, c.

4）c。In the equation BR2+SO2+2H2O=H2SO4+2HBR, the valency of H and O in water does not change, so water is neither an oxidizing agent nor a reducing agent. In ABCD, only the valency of water in C has changed, and the valency of O has changed from -2 valence to 0 valence.

So choose C.

a Electrolytes do not necessarily conduct electricity, but only when dissolved in water or in a molten state are ionized out of free-moving ions.

b is incorrect, in a certain concentration range we think that it is proportional to the concentration, but the migration rate of the ion is related to the charge, radius, and hydration radius of the ion itself.

c Incorrect For example, acid salts, strong acids and weak alkaline salts can ionize hydrogen ions. The cations ionized are all hydrogen anions.

d Incorrect The presence of anions and cations means that they are ionic compounds Translated as saying that electrolytes are ionic compounds.

For example, sulfurous acid is a molecular compound that exists in molecular form, but ionizes in water, and is a weak electrolyte.

Incorrect can only say that sodium hydroxide is an electrolyte.

b is incorrect, like sodium hydroxide ionized when dissolved in water, not because it is ionized after electricity is applied.

c True, for example, ammonia conducts electricity, but the compound ammonia is not an electrolyte.

dIncorrect, insoluble substances can also be electrolytes, but they are dissolved in water and are less dissolved, but the dissolved parts are also fully ionized, such as barium sulfate.

Incorrect kclo3=k + clo3 - 3