A chemistry problem is a process

This black solid may be Cu(OH)2!!98/

Please, give me the score! I'm old and young!! Please!! My answer is definitely right!! 1Please!! The others are wrong! 1. I'm a professor of chemistry.

From the inscription, it can be seen that the blue precipitate becomes a black solid after heating, then the blue precipitate should be Cu(OH)2, and the black solid should be Cuo

The two reactions before and after are respectively.

2naoh + cuso4 = cu(oh)2↓ +na2so4cu(oh)2 △= cuo + h2o

where the atomic weight of Cu is, the atomic weight of oxygen is 16, the atomic weight of S is 32, and in grams Cuo, the mass of the copper-containing element is grams, so the mass fraction of copper in the raw waste liquid is.

This black solid may be copper oxide, and the blue precipitate may be copper hydroxide.

cu(oh)2=cuo+h2o

The mass fraction of copper in copper oxide is = copper copper oxide*100% because the mass of the black solid (copper oxide) is grams.

So the mass of copper in a black solid (copper oxide) = the mass of a black solid (copper oxide) * the mass fraction of copper in copper oxide.

Gram. That is, the mass of the copper element in the original waste liquid is grams.

The mass fraction of copper in the raw waste liquid = the mass of copper in the raw waste liquid and the total mass of the wastewater * 100%.

The presence of blue precipitate means that there is a copper sulfate solution in the original waste solution, CuSO4 + 2NaOH = Cu(OH)2 + Na2SO4

Heating an insoluble base produces basic oxides and water Cu(OH)2 =CuO+H2OCuo Cu

x The mass of Cu is the mass fraction of copper in the raw waste liquid = the mass of Cu The mass of the waste liquid * 100% =

am2o7^x-+3s^2-+ch^+===2m^3++es↓+fh2o

2a=2 a=1

3=e, so the number of electrons transferred by the s ion is 3*2=6, and the valency of the original m = (6+2*3) 2=6

Question 1: The mass ratio of the carbon element is 2 : 1, so the ratio of the molar mass of the carbon atom is 2:

1 Because carbon monoxide and carbon dioxide both contain a carbon atom, the molar ratio of carbon monoxide and carbon dioxide is 2:1, and then multiply by the relative molecular masses of carbon monoxide and carbon dioxide, which are 28 and 44, respectively, and the result is 28 times 2 : 44 times 1 The answer is 14:

11 Question 2: First of all, you understand what x with %75 in c1 means, it means that the mass fraction is %75

The rest of the equation also refers to the mass fraction, and you assume that the relative atomic masses of x and y are x and y, respectively

Since the chemical formula C1 is xy4, then you can calculate x and y from the mass fraction above

The column is as follows x 4y

-75% and another is --25%.

x+4y x+4y

The solution gives x=12 y =1 so x is carbon and y is hydrogen.

And then you take c2 with n carbon atoms and z hydrogen atoms in c2.

12 * n z*1

-80% -20% also solves the system of equations

12n+z*1 12n+ z*1

The solution of C2 is C2H6 and is called ethane!

Question 3: Use the limit method and hypothetical thinking to add all CO2 then the ratio of carbon to oxygen is.

12:32 or 3:8 and suppose it's all co, then the ratio is 3:4, so you say, what's the answer?

Heh c The other self can do the same thing.

Question 4 is the same as Question 3

You first calculate the mass fraction of n in ammonium nitrate (NH4NO3) and then.

Look at it with 38% which is larger if it is greater than 38 then it is definitely a substance with n less than 38 that is mixed with it

It's finally over, give me a share of breakfast, I haven't eaten breakfast yet

1.The mass ratio of carbon is 2:1, i.e. n(co):n(CO2)=2:1, m(co):m(CO2)=2 28:1 44=14:11 (i.e. the mass ratio of two bottles).

2, X, Y are C, H respectively, C1 is CH4, C2 is C2H63, N(C): N(O)=3 12:5 16=4:5, may be a combination of Co and Co2 according to the number of molecules 3:1, so choose B.

4. The N content of NH4NO3, NH4HCO3, NH4Cl, CO(NH2)2 and (NH4)2SO4 is 35% respectively, and the N content of impurities contained in NH4NO3 should be 35% high, and C is selected.

1. Let the CO carbon element 2a grams, then the CO2 carbon element A grams, we can know that the CO mass is 28*2A 12, the CO2 mass is 44A 12, and the mass ratio of the two is 56:44 (14:11), and the CO medium carbon is 2mol, and the CO2 medium carbon is 1mol, and the CO is 56 grams, and the CO2 is 44 grams.

2. x 4y=75 25, x ny=80 20, these two formulas are a system of simultaneous equations, we can know that n is 3, and its chemical formula is xy3

3. If the mass ratio is 3:5, the quantity ratio or number ratio of the substance is 4:5, and the number ratio in CO is 1:

1, CO2 in 1:2, 4:5 between the two values, in line with the requirements, so B reasonable, (without the quantity of the substance can be calculated by the mass ratio, the key is that the mixing ratio should be between the minimum value and the maximum value.)

4. The N content in ammonium nitrate is 35%, indicating that the N content in this impurity is greater than that of ammonium nitrate, and only C (46%) is consistent.

I solved it as a high school student, it seems that this should be a junior high school question, but you can ignore the quantity calculation method of matter, calculate the value range of the mixture, and directly use limit thinking to calculate the maximum and minimum values, I wish you success in your studies.

The mass reduction is extremely oxygen production, and the potassium chlorate content in the mixture is x2kclo3 = 2kcl + 3O2

xx = the amount of manganese dioxide is.

Oxygen is produced after a few moments of heating.

2kclo3= 2kcl + 3o2

x ' yx'= y=

Potassium chlorate is consumed to produce potassium chloride, and the amount of manganese dioxide is.

The amount of potassium chloride is:'=

1. Let m(n2) = x grams, m(nh3) = y grams.

Conservation of mass: x + y = 42 + 9 + 17 - 3 = 65

Conservation of Nitrogen Mass:

x + y * 14/17 = 42 + 17 * 14 /17 = 56 ②

Lianli Solution:

x = 14；y = 51

2. Conservation of mass

a + b + c = 42 + 9 + 17 = 68 ①

Ratio of the mass of the elements n and h:

a + 14c / 17) :b + 3c / 17) = (42 + 17 * 14 / 17) :9 + 3 * 17 / 17) ②

The above two relationships can be satisfied.

The formula is simplified as: 3a = 14b

Formulating this as a = 14b 3, substituting in gives b,c relation:

17b + 3c = 204

n2+3h2=2nh3

Reaction molar ratio: 1 3 2 Reaction mass ratio: 28 6 34 Equilibrium concentration is: 42 9 17 Equilibrium molar number:

1) (9-3)=6, 34g of ammonia is required to generate 6g of hydrogen, and it can be seen that the initial ammonia is: 17+34=41g; Nitrogen is: 42-28 = 14g.

2）a:b=28:6=14:3； （b+6）：（c-34）=9:17

1.At the end, there are 6 grams of hydrogen more, so 34 grams of ammonia are needed to decompose, so 14g 51g

a+14c/17=42

Therefore, 17a-17b+11c=561

Let NH3 with AMOL at the beginning react xmol

2nh3 *****===n2+ 3h2a 0 0

x x/2 3x/2

a-x x/2 3x/2

The front-to-rear pressure ratio is the ratio of the total amount of gaseous substances.

a/(a+x)=

Derives x a=

The degree of dissociation is. You can do the rest yourself

Hope it helps

From a macroscopic point of view, the precipitated crystal contains the addition of anhydrous RSO4 and a small part of the precipitated solution (since the crystal has been precipitated, the remaining solution is saturated like the original solution). That is to say, subtracting 8 g from the middle, the mass ratio of RSO4 to water in the remaining is obtained from this, and in addition to the anhydrous RSO4 added, it is also precipitated to be about 4 g. With the addition of 8G, the crystal contains a total of RSO4 12G, H2O.

Therefore, the mass ratio of RSO4 to water is 12, and the relative mass of 7 water molecules is 18*7=126, then the relative molecular mass of RSO4 is 126 and the relative atomic mass is 24

So choose D. Hope you get it!

There is also a simple way: substituting the relative atomic masses of the four options, only the relative molecular mass of MgSO4·7H2O is 246

100 ml mol l co-sulfuric acid, 300 ml mol l co-sulfuric acid, total sulfuric acid, h+ altogether, the amount of substances in the solution h+ concentration =