Find a chemistry problem solution, find a chemistry problem solution

The sulfur content of 1t coal is 1%.

So there is sulfur 1000*1%=10kg

According to the relational formula.

s---caco3---caso4

10---x---y

32 10 = 100 x, 32 10 = 136 y solution x , y =

At least kilograms of limestone should be added to 1t of coal, and kilograms of calcium sulfate can be added after the reaction.

Write in terms of conservation.

The sulfur content of 1t coal is 1%.

So there is sulfur 1000x1%=10kg

Write out the reaction equations.

2cao+2so2+o2=2caso4

2x60 2x32 2x136

x 10 y

120/x=64/10=272/y

x=y=limestone:

Calcium sulfate:

The sulfur content of 1t coal is 1%.

So there is sulfur 1000*1%=10kg

s---so2---caso4 caco3---caso410 x y x

32 10=136 x 100 y=136 x solution x solution =

Therefore, at least limestone should be added to 1t of coal, and calcium sulfate can be obtained after the reaction.

Why do you say "some kind of carbon", and say yes?"A ton of coal"This?

Is there a mistake in writing?

CaCO3 + 2 S + 5 2 O2 ==CaSO4 + CO2 + SO2 This is the chemical equation of the reaction.

It's wrong and it's a strange book.

Redox reactions.

The number of electrons gained and lost is the same.

The total number of valence rises and falls is equal.

Adipic acid is reducible, then adipic acid loses electrons, and VO2+ electrons are completely reacted to a 1:2 relationship according to the title.

The coefficient of VO2+ is 2, and the coefficient of VON+ is 2, and according to the conservation of charge, n=2

When an organic substance undergoes oxidation and reduction reactions, it is calculated by the number of carbon electrons transferred. hon

The valency of X(FCLBR) is set at +1, respectively

1, and then calculate the valency of c according to the sum of the valency of 0 (in organic matter, the valency of c ranges from -4 to +4, sometimes there is a fraction, sometimes there is a 0 valence).

According to the title, 20ml

H2C2O4 (C is +3 valence) is used as a reducing agent, and the C valency can only rise to +4 valence in the reaction (the most ** of C is +4) - C loses 1*2* electrons.

4 10-3mol VO2

v is +5 valence), and the reduced product is vo

n+ (v is 2+n valence).

v gives (5-2-n)*4 10-3mol has 1*2*

5-2-n）*4×10-3mol

The solution is n=2

Hello, under the standard condition, the volume of a mole of gas is liters, and that liter of gas is 1/mole, so the mass of a mole of gas is grams, that is, the molar mass of the gas, so the relative molecular mass is also so much, so the relative atomic mass is divided by n

PV=RRT=(M M)RT, deformed to PM=DRT

p is the pressure, m is the relative atomic mass sought, r is the amount of matter, v is the volume, r is the molar gas constant, t is the temperature, d is the density, m is the mass), and finally the pressure, temperature, and density are all known conditions, r=; The rest is algebra.

The first equation is the equation of state of an ideal gas.

Since the title clearly tells us to "classify from the types of elements involved in the transfer of electrons", then it is obvious that we can know:

The chemical equation for oxygen production in option A is: 2kclo3=mno3 =2kCl+3O2, and the elements participating in the electron transfer are: Cl, O, and in B, the chemical equation is:

2 H2O2 = 2H2O+ O2, the elements participating in the electron transfer are only O, while in the C option, the chemical equation is: 2kmNO4 K2Mno4+Mno2+O2, the elements participating in the electron transfer are Mn, O, and the chemical equation in the D option is: 2 Hgo = 2 Hg + O2, and the elements participating in the electron transfer are Hg and O, so B is selected for this question

Just compare the changes in valence.

A:O from -2 to 0 (between O and Cl).

B: O from -1 liter to 0 (and is a transfer between oxygen and oxygen itself) C: O from -2 liters to 0 (between O and Mn).

d:o from -2 to 0 (between O and Hg).

Whereas, the change in valency is directly proportional to the electron transfer.

So choose B

Electron transfer in a occurs between the elements chlorine and oxygen.

Electron transfer occurs only between oxygen (-1 to 0, -1 to -2) in b, and electron transfer occurs between manganese and oxygen in c.

Electron transfer occurs between the element mercury and the element oxygen.

Therefore, B is different and B is chosen

A: 2kClO3 = 2kCl+3O2 O Valency: +2-0B:

2H2O2=2H2O+O2 O Valency: 1-0C: 2kmNO4 = K2Mno4+Mno2+O2 O Valency:

D: 2Hgo = 2Hg + O2 O Valence: +2-0 In summary: a, c, and d are all pyrolysis and the valency changes from +2 valence to 0 valence. And b does not change from +1 to 0, so the answer is b

Reject the experience party to copy the answer!

C and A are both decomposed by heating, and although mercury oxide is rare, it is almost known that it is heated to separate oxygen, right?

The types of elements involved in the transfer of electrons are non-metals in b and all other metals are involved.

a, c, and d are all two elements involved in electron transfer, that is, electrons are transferred from one element to another; And b has only one element, oxygen, and the disproportionation reaction occurs.

b There is only a change in the valency of oxygen elements, some of which are increased and some of which are decreased, which is their own redox reaction; The valency of the other three oxygen elements increased after valency reaction.

,: The amount of matter is.

,: The amount of matter is.

The amount of substance is: Boling.

,: The amount of matter is.

Therefore, the quantity ratio of matter is: containment.

Leaky orange mol: :

1 50 : 1 80 : 1 320 : 1 560

Simplify it again.

The mass of c is 5*36%=. All of this carbon is converted into caco3The quality of all caco3 c is.

The mass of CaCO3 is 15g.

The content of oxygen is 64%, and the content of C is 36% of the white precipitate, that is, CaCO3, 5 grams * 36% to obtain the content of C.

And then you can count what caco3 is, I forgot some data that can't be calculated.

In the mixed gas of CO and CO2, the mass fraction of oxygen is 64%, the carbon content is 1-64% = 36%, and its mass is 5g * 36% =, carbon monoxide reacts with iron oxide to form carbon dioxide, and the mass of carbon elements does not increase.

c---caco3

x12/x=15g

Select A, the number of hydrogen atoms contained in ethanol in aqueous ethanol solution is equal to the number of hydrogen atoms contained in water, because there are 6 hydrogen atoms in ethanol and 2 hydrogen atoms in water, the ratio of the number of molecules in ethanol and water is 1:3, so the mass fraction of ethanol in the solution is 1*46 1*46+3*18=46%.

The number of hydrogen atoms contained in ethanol in aqueous ethanol solution is equal to the number of hydrogen atoms contained in water, assuming that there is one ethanol molecule, then there are three water molecules, so the molecular mass of the ethanol molecule is divided by the mass of the ethanol and water molecules, multiplied by 100%.

The process of dissolution filtration evaporation is used to separate one soluble and one insoluble, and A, can B, can.

c No, because it is not soluble.

dNo, because they are all dissolved.

So choose AB

The answer is a, manganese dioxide and sodium chloride and b, copper oxide and potassium carbonate.

Manganese dioxide in A is insoluble in water, and copper oxide in B is insoluble in water. Dissolve separately - filter out manganese dioxide and copper oxide, and evaporate the filtrate to obtain sodium chloride and potassium carbonate respectively.

After dissolving them together, the manganese dioxide is filtered out, and then the sodium chloride is evaporated.

a. Manganese dioxide can be filtered out and sodium chloride can be evaporated.

b No, both are soluble and react.

c No, neither is soluble.

dNo, both are soluble.

The volume of the mixed solution is equal to the sum of the volumes of the two solutions before mixing.

That is to say, 1 L of 2mol L of copper sulfate solution and 1L of 1mol L of sulfuric acid are still 2L after mixing

Then the amount concentration of copper sulfate = 1L * 2mol 2L = 1mol L

The amount concentration of the substance of sulfuric acid = 1L*1mol L 2L=

The amount of hydrogen ions in the mixture concentration (two hydrogen ions in each sulfuric acid) = 1mol*2 2=1mol l

The amount of copper ions in the mixture (one copper ion in each copper sulfate) = 2mol l*1 2=1mol l

The concentration of sulfate in the mixture = (2mol l*1+1mol l*1) 2=

fe2(so4)3

The concentration of the species of iron ions (every three sulfuric acid roots can bind to 2 iron ions) = 1mol l

1）1mol/l;

2) (The following answers do not consider hydrolysis).

c(h+)=1mol/l;c(cu2+)=1mol/l;c(so4 2-)=

3) (Negative charge cancellation of ferrous sulfate and sulfate before covering).