High School Chemistry Inference Questions, Chemistry Inference Questions, High School

Eldest brother. I'm also in my third year of high school. And the grades are very poor.

b is s. c is for oxygen . d is hydrogen sulfide.

e is sulphur dioxide look. f is for water. The process is:

e can make the rainwater pH drop . Then E is CO2 or SO2. Whereas E is generated by BC.

b, c electron shell less than 3Then it won't be CO2It's SO2...

At this time b is a yellowish solid. So yes s. So c is oxygen.

Again f is liquid. Composed of oxygen element. So f is water.

Then a is hydrogen. Hydrogen +s to produce hydrogen sulfide. Hydrogen sulfide + SO2 is S + water Don't divide it.

Hehe. Hope it helps!

First of all, to determine the three kinds of elemental matter, 1, b atom and c atom have the same number of electrons in the outermost shell, c is a gas at room temperature, b is solid at room temperature, light yellow, easy to push, b is sulfur, c is oxygen.

2, a is a gas at room temperature, we know that the gas elements in the first three cycles are hydrogen, helium, nitrogen, fluorine, chlorine, argon, and hydrogen that can be ignited with oxygen. So A is hydrogen, and at the same time, F is water.

3. Hydrogen and sulfur are heated to produce hydrogen sulfide, and d can be known to be hydrogen sulfide.

4. Oxygen and sulfur ignite to form SO2

In summary, A hydrogen, B sulfur, C oxygen, D hydrogen sulfide, E sulfur dioxide, F water.

b is a yellow element, which is sulfur.

B and C atoms have the same number of electrons in the outermost shell, C is a non-toxic element, or a gas, then C is oxygen.

b+c ignites to generate e, then e is sulfur dioxide, and the conditions for making the rainwater acidic are met, then ok, the previous inference should be reliable.

A is a elemental gas, non-toxic, and can be ignited, guess it is hydrogen, and then look at the conditions, the number of ABC electron layers is different and not greater than 3, then B is 3 layers, C is 2 layers, and A is one layer. Determine that a is hydrogen.

A + B generates D, and D is hydrogen sulfide.

A+C is F, and F is water.

1) A contains Fe and C elements, and G is Na2O2

2) NH3+H2O = +OH - (Equal sign is replaced with an invertible sign, it cannot be played) 3) 2Fe3+ +3CO32- +3H2O = 2Fe(OH)3 +3CO2

4)8 nh3 + 6 no2 = 7 n2 + 12 h2o5)fe3c

fe3c+ 22hno3== 3fe(no3)3 + co2 ↑+13no2 ↑+11h2o

Hope it helps!

Do you have two D's on your ** and an L on the right?

①a:fe,b:hcl,c:fecl2,d:h2,e:fecl3,f:cl2

The breakthrough in solving the problem is a pale flame.

a:mg，b:co2，c:mgo，d:c，e:mg(no3)2，f:hno3

The breakthrough is reddish-brown gas.

A:C,B:SiO2,C:Co,D:Si,E:CO2,F:O2 The breakthrough is a toxic gas that can bind to hemoglobin, because generally only CO and nitrogen oxides are available.

:fe b:hcl c:fecl2 d:h2 e:fecl3 f:cl2 fill in the blanks.

mg b:co2 c:mgo d:c e:mg(no3)2 f:hno3

It is not clear what crystal is for element C and E for CO2

I'll give you examples of what you want to do inside.

There's analytics inside.

1), A is (NH4)2S, Y is O2, Z is H2O2), X is a strong acid, E is to generate C: 2H2S+3O2=2SO2+2H2O;

3) X is a strong base, E is the generation of C: 4NH3 + 5O2 = 4NO+6H2O

1. If A is elemental, the element that makes up a elemental substance is the element that forms the most compounds in nature. Then b belongs to a (non-polar) molecule (fill in "polar" and "non-polar"), and whether all atoms in b molecule satisfy the 8-electron structure (yes) (fill in: yes, no).

B (standard condition) is introduced into a 50mol L NaOH solution and the solute in the solution after the reaction is (Na2CO3 or NaHCO3 or Na2CO3 and NaHCO3) (chemical formula). If the heat emitted by the complete combustion of 4g a element, write the thermochemical equation (c+o2==(ignition)==CO2+) that represents the combustion reaction of a elemental element.

Analysis] If A is elemental, the element that makes up A is the element that forms the most compounds in nature, it can be known that A is carbon C, and the reaction of excess E and a small amount of E is different, E can be judged to be oxygen, and the volume of NaOH solution in B is not known to 50mol L of NaOH solution, so at this time, the solute is uncertain, Na2CO3 or NaHCO3 or Na2CO3 and NaHCO3, a small amount of CO2 is Na2CO3, and excess CO2 is NaHCO3.

The electronic formula of CO2 is.

o ::c::o

2. If a is the chloride formed by the elements in the third cycle. Then: write the ionic equation for solution A b (Al3+ +4OH-==alo2- +2H2O); Write the ionic equation for the reaction of A and B in solution (Al3+ +3alo2-+ 6H2O===4Al(OH)3 precipitate).

Analysis] The chloride here can be known to be AlCl3Different reactions can be carried out in small quantities and in excess by a strong base such as NaOH.

3. If A is a common metal, there is passivation in the cold and concentrated solution of E. It is known that X is an ionic compound containing non-polar bonds, and 1molX contains 38mol of electrons, and the amount of solid X and D is added to the D solution, and the ionic equation of the Kar reaction is (3F- +Fe3+ +3H2O===3HF gas + Fe(OH)3 precipitation).

Analysis] A is a common metal, and there is passivation in the cold and concentrated solution of E, and it can be seen that A is Fe or Al (here it should be Fe) and E is nitric acid.

D is generated by a small amount of nitric acid, that is, Fe(NO3)2, X is an ionic compound containing non-polar bonds, and 1molX contains 38mol of electrons, then X may be BeCl2 or CaF2, BeCl2 does not react, so it is a weak base or achloroic acid with Fe3+, double hydrolysis reaction.

1.Non-polar (A is O, B is CO2).

4oh- =alo2- +2h2o

Al3+ +3alo2- +6H2O = 4Al(OH)3 FE2+ +2OH- =FE(OH)2 (X should be CA(OH)2) I have a high school degree and a third degree in high school, please believe in the level of professional operating machines).

The solute after the reaction is Na2CO3

The thermal reaction formula is C+O2=(ignition)==CO2; ▲h=