High One Chemistry requires a detailed explanation, High One Chemistry explains in detail

1. Contains iron ions (+3 valence), oxidized to ferric oxide, i.e., iron oxide precipitation. P.S. In fact, iron ions are yellow solutions, but the amount is small, so they are transparent and clear.

2. The iron scale is ferric oxide further oxidized into ferric oxide, that is, rust. The iron oxide that falls below is red, because it is difficult to make contact with the air one step further.

3 Alum is kal(SO4)2·12H2O, which is adsorption.

4. The hardness of water in the north is high, that is, there are many ions, such as calcium and potassium, which are more active than iron, and it will take a long time to wait until it is oxidized.

The brownish-red flocculent precipitate is Fe(OH)3, then the rust is Fe2O3. i.e., the identification of iron ions, Fe3+ +6(scn)-=[Fe(scn)6]3-. Interpretation.

There are iron ions in the water, and the iron ions will be hydrolyzed, Fe3+ +3H2O = (reversible) Fe(OH)3+3H+. This is the sophomore content. Hydrolysis is very small, but after a long time, the water will partially evaporate, and the equilibrium will shift to the right, (as will be learned in the principle of chemical reaction), so a precipitate is formed.

The rust scale is formed by the decomposition of the precipitated part due to 2Fe(OH)3=Fe2O3+3H2O. Due to the adsorption of Mingfan. Because if it is the acidity of Mingfan, the iron ions can be precipitated when the pH is about 3 in the solution, and the acidity of Mingfan due to its hydrolysis is not enough.

The reason for the removal with alum is that alum is potassium aluminum sulfate dodecahydrate, in which Al will be hydrolyzed to form Al(OH)3

Whereas, aluminum hydroxide has an adsorption effect, not because of acidity. I think this phenomenon is related to the temperature in the north and south, the higher temperature in the south is conducive to Fe3+ hydrolysis, and it is also conducive to the decomposition of Fe(OH)3 into rust.

Do it with the stool of the conservation of gains and losses electrons.

Amount of O2 substance = =

1mol O2 reaction gives 4mol electrons.

So O2 reaction co-transfer = electrons.

1mol Ag reacts and loses 1mol electrons.

The 1mol Cu reaction loses 2mol electrons.

So the disturbance is: n(ag) +2*n(cu) = and because 108*n(ag) +64*n(cu) = 7g, n(cu) = is solved

So Qing Li Brigade m(cu) =64* = choose a

A xuanff's solution steps are exactly right!

Set: Sodium bicarbonate reacts with the solid carnage fiber mixture of sodium hydroxide vertical hall, and the mass is reduced by xg

nahco3+naoh==na2co3+h2o↑ △mx x=

Hence NAOH overdose.

nahco3+naoh==na2co3+h2o↑y y=

The mass of sodium hydroxide in the mixture.

There is one Mg2+ and two Cl- in MgCl2

So theoretically it should be 3*(2+1) = 9

But by two is repeated, so there are 7 kinds.

They were: 24mg + 35cl + 35cl, the relative molecular weight was 9424mg + 35cl + 37cl, the relative molecular weight was 9624mg + 37cl + 37cl, the relative molecular weight was 9825mg + 35cl + 35cl, the relative molecular weight was 9525mg + 35cl + 37cl, the relative molecular weight was 9725mg + 37cl + 37cl, the relative molecular weight was 9926mg + 37cl + 37cl, The relative molecular mass is 100

Solution: The total volume is 2L, and the total amount of gas mol is:. Since it is a gas, the volume can be similar, then the volume is: va=

vb=vc=

Let the A consumed when generating C be Xmol and B be Ymol;

4a（g）+b（g）==2c（g）

x yx=4*

y=1*then the amount before the reaction of a is.

b The amount before the reaction is.

4a（g）+b（g）==2c（g）

x y 0 before reaction

In equilibrium, it can be calculated that b reacts and a reacts , then a originally has and b has va=(

vb=（ /

vc= /

Choose CNaOH+CO2=NA2CO3+H2O

2. Na2CO3 + CO2 + H2O = 2NaHCO3 Assuming that NaOH is all converted to NaCO3 (i.e., the reaction is complete according to formula 1), then there is.

2naoh ~ na2co3 ~ co2

m1 m2 gives m1=

m2=m1 means that CO2 is sufficient, continue to react according to the dicatus.

na2co3 ~ 2nahco3~ co2 ~ mm3

The solution is m3=

mco2=m2+m3=

Put the iron powder into the solution containing Cu2+, Ag+, Fe3+, when the reaction is completed, if there is iron powder left at the bottom of the beaker, the metal ions that must not be contained in the solution are Cu2+, Ag+, Fe3+ If there is no iron powder left in the beaker, the metal ions that must be contained in the solution are Fe2+ If the solution does not contain Cu2+, the metal ions that must not be contained are Ag+, Fe3+

Fe reacts with all three ions, Fe remains, and the ions react completely naturally.

No iron powder remaining. Then FE reacts. After the reaction, no matter how much it reacts with that ion, Fe2+ is generated.

The oxidation of Ag+ and Fe3+ is stronger than that of Cu2+. The one with strong oxidation reacts with Fe first. Cu2+ finally reacts, Cu2+ is gone, and naturally Ag+ and Fe3+ are gone.

Fe reacts with Cu2+, Ag+ and Fe3+.

So when there is Fe leftover, the solution must only contain Fe2+ in the end, and the precipitate has Ag, Cu and Fe

Then the metal ions that are not contained in the solution are Cu2+, Ag+ and Fe3+

If there is no Fe left, it means that there is a small amount or an appropriate amount of Fe.

When Fe is just completely reacted with Fe3+, Cu2+, and Ag+, there is only Fe2+ in the solution, so it must contain Fe2+

The absence of Cu2+ means that Fe has completely reacted Cu2+, and the reoxidation of AG+ Fe3+ Cu2+ means that there must be no A+ and Fe3+ in the solution, because Cu2+ is also gone.

Therefore, it must be free of AG+, Fe3+, Cu2+

At first, when I did the question, I felt that there was something wrong with the answer, because all three answers could be solved in one of the three situations, but after checking it, I decided that it was correct.

Avogadro's constant na = number of oxygen atoms The amount of oxygen atom of the substance, the number of oxygen atoms = m

The amount of matter of oxygen atom = n 16

Avogadro's constant na=16m n

The oxygen element has equal mass, and both contain 6 moles of oxygen atoms.

Then SO2 is 3 moles 192 grams, and SO3 is 2 moles 160 grams 192:160=6:5