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25*（1-20％）=20g

Let the mass of the carbon dioxide gas produced be x

The mass of calcium chloride produced is y

CaCO3+2HCl==CaCl2+CO2 up arrow +H2020G y x

20g y x

The solution is x= y=

Answer: The mass of carbon dioxide gas is:

The mass of calcium chloride produced is .

The marble participating in the reaction is 25g multiplied by its percentage (1-20) as 20gCaCO3 The relative molecular mass is: 40 + 12 + 16 * 3 = 100 CaCl2 The relative molecular weight of CaCl2 is: 40+

The relative molecular mass of CO2 is: 12+16*2=44, then the relative molecular weight of CaCO3 is relative to the actual mass = the relative molecular mass of CO2 is compared to the actual mass =

The relative molecular mass of CaCl2 is relative to the actual mass.

The key is to understand what is the mass of the marble participating in the reaction.

is the total mass - impurities).

It is OK to find other unknown quantities according to the relative molecular mass and known quantities Chemistry in junior high school is still a simple foundation, and this kind of problem is just a matter of using mathematical knowledge to remember to subtract impurities or the mass of catalysts.

Let it generate x grams of carbon dioxide and Y grams of calcium chloride.

caco3+2hcl=cacl2+h2o+co2↑25*20% y x

100:44=5:x

x=100:111=5:y

Y = then carbon dioxide is generated, and the volume under the standard condition is calcium chloride.

It also doesn't tell if the status is a standard or not. If it's not a standard condition, let's use the standard gas equation of state to do the math yourself. )

CaCO3+2HCl=CaCl2+ H2O+CO2 CO2= (Volume is calculated by yourself).

Calcium chloride = marble involved in the reaction is 25g * (100%-20%) = 20g

Chiral carbon atom refers to a carbon atom that is connected to four distinct atoms or groups, and four different ones must be connected, as shown in the figure below.

It's just an organic.

1. The maximum mass of the solution is required to reflect the smallest mass of the gas or precipitate, and if the mass of the original solid is 1, the mass of hydrogen generated by the reaction A is 2 56;Reaction b produces hydrogen with a mass of 6 54;The mass of barium hydroxide precipitate generated by reaction c is 230 171;Reaction d produces carbon dioxide gas with a mass of 44 106. It can be seen that the mass of the residual solution of reaction C is the smallest, and the mass of gas generated by reaction A is the smallest, so the mass of the remaining solution is the largest.

2. From the meaning of the title, it can be seen that there is enough iron filings, so there are three directions for M grams of iron filings, one is to replace copper (CuCl2 + Fe = FeCl2 + Cu), and the ratio of iron filings to copper is 1:1;the second is to oxidize ferric chloride to ferrous chloride (2FeCl3+Fe=3FeCl2), and the ratio of iron filings to the amount of ferric chloride is 1:2; The third is the remaining iron filings.

Therefore, let the amount of iron filings reflected with copper chloride be x, the amount of iron filings reflected with ferric chloride is y, and the amount of remaining iron filings is z, then 56(x+y+z)=m, 64x+56z=m, so 56(x+y+z)=64x+56z, so 56y=8x, so x=7y, that is, x y=7:1, because the amount of copper chloride in the original mixture = x, the amount of ferric chloride = 2y, so the ratio of the amount of copper chloride to ferric chloride in the original mixture is x 2y=7:2.

3. From the inscription, it can be seen that the mass of water in the original crystal hydrate is p-q, so the amount of the hydrate is (p-q) 18, so the molecular weight of the crystal hydrate is 18p (p-q).

It's called 525594......Yes, the first question is is barium hydroxide good?!

Calcium hydroxide white.

Barium hydroxide white.

Copper hydroxide blue.

Iron hydroxide reddish-brown.

Ferrous hydroxide.

Pure ferrous hydroxide is white.

However, it is generally difficult to see, and it is often seen in light green.

Because there is oxygen dissolved in the solution and the reaction system (open system) and the outside world there is a material exchange and energy exchange, and the oxygen in the air is constantly dissolved into the solution, and the newly formed ferrous hydroxide is quickly oxidized to iron hydroxide, so it is difficult to observe the formation of white ferrous hydroxide precipitation during the experiment, and only the mixture of oxidized ferrous hydroxide and newly formed ferrous hydroxide can be seen: some gray-green flocculent precipitates.

White, white, light blue, brown or reddish-brown, white.

Calcium hydroxide is a white precipitate.

Barium hydroxide also has a white precipitate.

Copper hydroxide is black precipitate.

Iron hydroxide ferrous hydroxide is reddish-brown.

Calcium hydroxide: white solid.

Barium hydroxide: white solid.

Copper hydroxide: blue or blue-green gel or light blue crystal, blackening after a long time in dry air.

Iron hydroxide: brown or reddish-brown powder (dry) or dark brown flocculent precipitate (in solution).

Ferrous hydroxide: The pure substance is white, but it is very unstable and easily oxidized to iron hydroxide rapidly.

Calcium hydroxide white.

Barium hydroxide White powder.

Copper hydroxide blue.

Iron hydroxide reddish-brown.

Ferrous hydroxide white.

Is it relative molecular mass and right? is 2*(31*2+16*5)=284

Is your 2 in mols? 2mol is 142 grams, and 2 is 142 na (Afojandro's constant).

The relative molecular mass is 284, and the actual mass is multiplied by 10 to the minus twenty-fifth power.

In this question, I feel that the quality of the alloy is a bit problematic, but the second question does not use this condition, let no2 xmol no ymol x+y= ( cu mmol ag nmol

Conservation of electron gain and loss 2m+n=3x+y= why= Because the last ones produced are copper nitrate and silver nitrate 2m+n= so x= y=

1) The amount of the reduced Hno3 substance is mol, the total amount of Hno3 is mol, and the combination with copper and silver is mol, let the alloy contain copper xmol, silver ymol: 64x+108y= 2x+y= solution x= y= they lose electrons in total, let the gas contain no2 amol, no bmol: a+b= a+3b= solution a= b= so:

2) Gas in the product: NO2; no 。

3) The mass of ag in the alloy is grams.

fe2i3·h3o?

I count it according to fe2o3·h2o.

The solid mass difference is.

fe2o3·h2o=fe2o3+h2o

If the quality difference is caused by crystal water, the required Fe2O3·H2O is FeCO3=FeO+CO2

If the poor quality is caused by carbon dioxide, FeCO3 is required.

And there are only iron nails in the first place, so there must be ferrous carbonate.

The solid that disperses after heating is Fe2O3 (set to X

mol), feo (set to y

mol)18x+44y=poor quality, caused by the combination of crystalline water and carbon dioxide) The residue is completely dissolved and gasless, then the elemental iron does not react with sulfuric acid at all. fe2o3h2so4

fe2(so4)3

3h20feoh2so4

feso4h20

fe2(so4)3fe

3feso4

From the above equation, it is known that elemental iron does not consume sulfuric acid. Sulfuric acid is consumed entirely by two oxides.

then 3x+y=5*

Solve two equations to know: x=

gThe mass of the iron nail before rusting is 56*(2x+y)+

Namely. Ferrous carbonate is present in rust.

The mass of Fe2O3·H2O punch slag in the corroded iron nail is g, and the quality of the iron nail before rusting is at least.

First of all, the four chemical reagents are soluble, the light green solution is FeCl, and then the three drops are served with No. 3 like its chain, which changes from white to light green and then to reddish-brown Fe(OH).

precipitation, and then you can know which bottle is BA (OH).

ba(oh)

Add dropwise to the two reagents, and there is a white deep talk about the na

SO4, the other is NACL. So your answer is fine.

Dissolve first, the green solution is FeCl2, and the spine has a precipitate, and the precipitate quickly turns red

Ba(OH)2, which only generates precipitate, does not change the color of Na2CO4, and the rest is NaCl

Remember the law: the drop is to return oxygen, and the increase is to lose oxygen (the valency decreases to gain electrons, and the reduction is an oxidant: the increase in valency and the loss of electrons is oxidized is a reduction hunger agent).

a) HCl is neither an oxidizing agent nor a reducing agent (the valency of the oxidant is reduced and the chemical valence of the reducing agent is increased).

b) HCL (Cl- is the lowest valency, can only be available and can only be used as a reducing agent) (H+) to H2, the valency is reduced, and it is an oxidizing agent.

c) HCl, Cl- is oxidized to Cl2 with increased valence and is a reducing agent.

d) The early rock is neither an oxidant nor a reducing agent.

I guess your main purpose is to know how to judge this kind of topic, right? Then trust me because I'm a chemistry genius!!

Remember: the oxidant is reduced by the reducing valence, and the reducing agent is oxidized and the chemical valence is raised. You can judge all the topics from this!

For the analysis of this question, you can refer to the answer to the "Blue Melancholy" in the banquet book.

I'll give you a way for me to memorize the song"Loss of high oxygen and low origin"

The loss of electrons and the increase in valency are the oxidation reactions.

Electrons are obtained, and the valency decrease is a reduction reaction.