Chemistry problems in the third year of junior high school, urgent !! About junior high school chemi

The metathesis reaction, generally speaking, is a reaction with constant valence, where the copper reaction is (Cu 2-)-2 valence before and after, and the CO2 reaction is +2 valence before and after the reaction.

In this example, it is actually the neutralization of acid and base, and there are many metathesis reactions, which are commonly used in junior high school: (conditions that do not occur in parentheses).

Basically, there are 3 categories:

HCl + NaOH == NaCl + H2O (acid-base neutralization reaction).

agno3 + nacl == nano3 + agcl produce precipitate).

CaCO3 + 2HCl == CaCl2 + H2O + CO2 produces gas).

It is important to note that there are times when metathesis cannot occur:

CaCl2 + H2O + CO2 == CaCO3 2HCl (Error).

Weak acid cannot replace strong acid, although precipitation is produced, but it cannot react!

CO2 + Cu(OH)2 = CuCO3 + H2O is a metathesis reaction.

The four basic types of reactions are classified as follows:

Chemical reaction ab

Decomposition reaction AB a b

Displacement reaction A BC AC B

Metathesis reaction: AB CD AD CB

Remember the above formula and just put the reaction formula in.

In other words, this should be the reaction of calcium hydroxide with carbon dioxide.

This is the reaction of the base with an acidic oxide, which is the same principle as the absorption of carbon dioxide by sodium hydroxide. The product must be the acid oxide corresponding to the acid oxide and water. The same is true for alkaline oxides that react with acids.

sio2+2naoh=na2sio3+h2o

It should be a metathesis reaction rather than a neutralization reaction.

(1) Relative atomic mass = 1 12 of the mass of an atom of an element and the mass of an atomic carbon element

2) Let the chemical formula of a compound be AMBN

Its relative molecular mass is the relative atomic mass of a m b n relative atomic mass n

The mass ratio of element A to element B is the relative atomic mass of a m:The relative atomic mass of b is n

The mass fraction of element a = the relative atomic mass of a m and the relative molecular mass of ambn.

3) The mass fraction (purity) of a substance in the mixture = the mass of the pure substance 100% of the total mass of the mixture

4) Gas density under standard conditions (g l) = gas mass (g) gas volume (l).

5) Purity = mass of pure substance Total mass of the mixture 100% =

The mass of the pure substance (the mass of the pure substance + the mass of the impurities) 100%=

1- Mass fraction of impurities.

6) Mass fraction of solute = solute mass Solution mass 100% = solute mass (solute mass + solvent mass) 100%.

7) Dilution and concentration of the solution.

m concentrated a% concentrated = m dilute b% dilute = (m concentrated + increased solvent mass) b% dilute.

8) Mixing of two solutions with different mass fractions relative to the solute.

m concentrated a% concentrated + m dilute b% dilute = (m concentrated + m dilute) c%.

9) The quality of the solute in the solution.

The mass fraction of the solute in the solution.

The volume of the solution The density of the solution.

m(fe3o4)=2000*(

m(fe)=1450/(56*3+16*4)*56*3=1050t

m (pig iron) = 1050 (iron content of pig iron, take your own value) Iron content of pig iron, take your own value).

It is about equal to 414 and multiplied by 2000, which is equal to 550t. That is, Fe3O4 has a total of 550T, and then the relative atomic mass is brought in, and finally multiplied by 4%. Absolutely, yes!