"Integration of this Chapter" Understanding Chemical Science PPT

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"Integration of this Chapter" Understanding Chemical Science PPT

Part One: Breakthrough 1

Comparative method to learn the two oxides of sodium

Typical Example 1 The correct statement about sodium oxide and sodium peroxide is ()

A. Sodium oxide is more stable than sodium peroxide

B. Can react with carbon dioxide and water to produce oxygen

C. The chemical valence of all oxygen elements contained is -2

D. The number ratio of anions and cations is 1:2

Answer:D

Analysis: Sodium oxide can be oxidized by oxygen to sodium peroxide, and sodium peroxide is more stable than sodium oxide. Sodium oxide can react with carbon dioxide and water, but no oxygen is produced. The valence of the oxygen element in sodium oxide is -2, and the valence of the oxygen element in sodium peroxide is -1. The anions and cations in sodium oxide are O2- and Na+ respectively, and the anions and cations in sodium peroxide are and Na+ respectively, and their number ratio is 1:2.

Knowledge induction

1. Comparison of sodium oxide and sodium peroxide

Special Tips (1) Compounds that react with acids to produce only salt and water are called basic oxides. The reaction between Na2O2 and acid not only generates salt and water, but also generates O2, so Na2O2 is not an alkaline oxide.

(2) Na2O2 is often used as a bleaching agent, which can bleach colored substances. Therefore, when Na2O2 is added to water containing phenolphthalein, the solution will first turn red and then fade.

2. Several important relationships regarding the reaction of Na2O2 with CO2 and H2O

(1) Relationship between the quantity of matter

Whether it is a single substance of CO2 or H2O or a mixture of the two, when a sufficient amount of Na2O2 is passed through, the ratio of the amount of CO2 or H2O participating in the reaction to the amount of O2 released is 2:1.

(2) Gas volume relationship

If a mixed gas (or a single gas) of CO2 and water vapor passes through a sufficient amount of Na2O2, the reduction in gas volume or the original mixed gas volume is the volume of oxygen generated.

(3) Solid mass relationship

When solid Na2O2 reacts with CO2 and H2O, it is equivalent to Na2O2 only absorbing the "CO" in CO2 and the "H2" in H2O; it can be seen that the corresponding reaction occurs:

This chapter integrates PPT, the second part of the content: Breakthrough 2

Avogadro's law and its corollary

Typical Example 2 Under the same temperature and pressure, container A contains H2 and container B contains NH3. If the total number of atoms they contain is equal, the ratio of the volumes of the two containers is ()

A.2:1 B.1:2

C.2∶3 D.1∶3

Answer:A

Analysis: If the total number of atoms in container A and container B is equal, then the ratio of the number of molecules is 2:1. According to Avogadro's law, the volume ratio of the two containers is 2:1.

Summary of methods 1. Avogadro’s law

Avogadro's Law: At the same temperature and pressure, the same volume of any gas contains the same number of molecules (i.e. "four identical"). It can be abbreviated as "three are the same and one is the same", that is, for gases, among the four quantities of temperature, pressure, volume, and number of molecules (amount of matter), if three quantities are the same, the fourth quantity must be the same.

2. Corollary to Avogadro’s law

The main corollary can be summarized as "three direct proportions and one inverse proportion"

Point training 2-1 Comparing methane and ammonia of equal mass, which of the following conclusions is wrong ()

A. The ratio of the number of molecules of the two gases is 17:16

B. Under the same temperature and pressure, the volume ratio of the two gases is 17:16

C. The ratio of the number of hydrogen atoms in the two gases is 17:12

D. Under the same temperature and pressure, the density ratio of the two gases is 17:16

Answer:D

Analysis: For equal masses of methane and ammonia, the ratio of the amounts of substances is inversely proportional to the molar mass, that is, the ratio of the amounts of substances is 17:16; then under the same temperature and pressure, the ratios of volumes and number of molecules of equal masses of gases are the same is 17:16, the ratio of the number of hydrogen atoms = 17×4:16×3=17:12; the density ratio is equal to the molar mass ratio, that is, the density ratio of the two gases is 16 g·mol-1:17 g ·mol-1=16∶17.

This chapter integrates PPT, the third part: Breakthrough Three

Chemical calculations centered on the amount of matter

Typical example 3 (1) The molar mass of a gas composed of a certain diatomic molecule is M g·mol-1, the mass of the gas is m g, and Avogadro's constant is NA, then:

①The volume of this gas under standard conditions is _______________L.

②The density of this gas under standard conditions is ____________g·L-1.

③The total number of atoms contained in this gas is ____________.

④The mass of one molecule of this gas is ____________g.

(2) Under standard conditions, the volume of 1.6 g of a certain gaseous oxide RO2 is 0.56 L. The molar mass of the gas is ____________.

Method summary The essence of chemical calculation is to use chemical knowledge to find the quantitative relationship between known quantities and unknown quantities, and then calculate and solve it. In the process of solving problems, complex problems can be simplified by finding out the quantitative relationship between known quantities and unknown quantities based on conditions. The problem-solving method is based on the fact that the amount of a certain particle material remains unchanged before and after the change, and the conservation relationship is listed to solve it.

1. The amount of matter is the bridge and link between macroscopic quantities (such as mass, volume) and microscopic quantities (such as the number of particles), and is the core of calculations. When answering, it is necessary to clarify the relationship between them.

2. We must firmly grasp the core of calculation "amount of matter". Firmly establish the thinking mode of "converting other physical quantities into quantities of matter" and make full use of the following "universal identities" and their deformations to calculate each quantity.

In the formula, n is the amount of the substance, the unit is mol; m is the mass of the substance, the unit is g; M is the molar mass, the unit is g·mol-1; V (g) is the gas volume, the unit is L; Vm is the gas Molar volume, the unit is L·mol-1; N is the number of particles; NA is Avogadro's constant, the unit is mol-1; x is the mass of the saturated solution, the unit is g; S is the solubility, the unit is g ;c is the quantity concentration of the substance, in mol·L-1; V(aq) is the volume of the solution, in L.

This chapter integrates PPT, the fourth part: Breakthrough Four

Commonly used methods in chemical calculations

Typical Example 4: There is 100 mL of mixed gas of CO2 and O2. After sufficient Na2O2 is introduced, the volume becomes 60 mL. What is the volume of O2 in the original mixed gas?

Solve using difference method:

2CO2+2Na2O2 2Na2CO3+O2　ΔV

twenty one

V(CO2) 100 mL-60 mL

2/(V"(" CO_2 ")" )=1/(40" " mL), then V(CO2)=80 mL.

Then V(O2) in the original mixed gas=100 mL-80 mL=20 mL.

Answer: The volume of O2 in the original mixed gas is 20 mL.

Summary of methods 1. Calculating the amount of relevant substances based on chemical equations

(1) Write the chemical equation related to the reaction.

(2) Find the ratio of stoichiometric numbers of related substances.

(3) Correspond to the stoichiometric number and find the amount of the relevant substance.

(4) List the proportional formula for calculation.

2. Relational method

The basis of chemical calculations is the proportional relationship between the quantities of substances, which can usually be obtained from chemical equations or chemical formulas. However, for complex problems, such as the relationship between the known object and the object to be sought through many reactions, it is necessary to directly determine the proportional relationship between the known quantity and the unknown quantity, that is, the "relational expression". For the calculation of multi-step reactions, it is easier to use "relational expressions".

There are many kinds of "relational expressions", the common ones are: mass or mass fraction relational expression, substance quantity or particle number relational expression, gas volume relational expression, etc.

A general method for determining the relationship between a known quantity and an unknown quantity.

(1) Determine the relationship formula based on the chemical equation: first write the chemical equation, and then extract certain relationships from the equation as needed. For example, MnO2+4HCl (concentrated) MnCl2+Cl2↑+2H2O, the following relationship can be obtained: 4HCl~Cl2.

(2) In the reaction between Na and H2O or acid, determine the relationship formula according to the conservation principle: 2Na~H2.

3.Difference method

The difference method is to find the so-called "theoretical difference" based on the changes in the relevant physical quantities of the substance before and after the chemical reaction. Such as the mass before and after the reaction, the volume of the gas, the amount of the substance, the pressure of the gas, the heat change during the reaction, etc. The size of this difference is proportional to the relative amounts of reactants. The difference method is a calculation method that uses this proportional relationship to solve certain quantitative changes. The key to solving this type of question is to determine the "theoretical difference" based on the meaning of the question, and then list the proportional formula based on the "actual difference" provided by the question to find the answer.

4. Calculation of overdose problem

Chemical reactions of substances are carried out according to a certain quantitative relationship, but if the quantities of two or more reactants are given in the question, to solve such problems, you should first determine which substance is in excess, and then solve the problem based on the insufficient amount of substances.

(1) Write the chemical equation of the reaction.

(2) Determine which reactant is in excess.

(3) Solve based on the insufficient amount of material.

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