Chemistry is great, come in and help me, the topic of the first year of high school, please 3Q

There are three common pale yellow solids in middle school: sodium peroxide, sulfur elemental and silver bromide, and this question should be sulfur elemental in line with the topic! According to the title, we can know that A is sodium sulfide, B is sodium sulfite, C is sodium sulfate, D is sulfur dioxide, and E is hydrogen sulfide The chemical reactions involved are as follows: 1. B—(oxidation) C,——Na2SO3 is easy to be oxidized to Na2SO4, and different reactions of oxidants are different 2. A + dilute sulfuric acid—C + D (gas), —Na2S + H2SO4 = Na2SO4 + H2S (gas) 3. B + dilute sulfuric acid—C + E (gas), —Na2SO3 + H2SO4 = Na2SO4 + SO2 (gas) + H2O 4, D + E—X (precipitation) + water – 2 H2S + SO2 = 3S (precipitation) + 2H2O 5, X + NaOH – (heating) A + B + water – 3S + 6NaOH = (heating) 2Na2S + Na2SO3 + 3H2O

It is known that X is a pale yellow solid, A and B are sodium salts, C is sulfate, and D and E are colorless gases, and the relationship between them is as follows: 1, B - (oxidized to) C, 2, A + dilute sulfuric acid - C + D (gas), 3, B + dilute sulfuric acid - C + E (gas), 4, D + E - X (precipitation) + water 5, X + NaOH - (heating) A + B + water From 4, it can be known that X is a solid that is insoluble in water and does not react with water, is S, by 5 and 3S + 6NaOH = (heating) 2Na2S+Na2SO3+3H2O can know that AB is one of Na2S and Na2SO3, from which it can be seen that the gas must not contain sodium, the sodium element in A and B is transferred to sulfate C, C is Na2SO4, and B can be known from 1 to Na2SO3, because Na2S is not easy to be directly oxidized to Na2SO4, then A is Na2S, and there is a reaction Na2S + H2SO4 = Na2SO4 + H2S (gas), Na2SO3 + H2SO4 = Na2SO4+SO2 (gas) + H2O, we can know that D is H2S, E is SO2, and 2H2S + SO2 = 3S (precipitation) + 2H2O is obtained by substituting 4

1) Because both NO and NO2 are derived from the reduction of nitric acid to obtain electrons, and in the redox reaction, the gain and loss of electrons are conserved, in this problem, only copper is increased by the valence of electrons, because copper becomes copper ions, which increases the 2 valence, and because copper is, it can be assumed that it has gained electrons, so nitric acid also loses electrons, and what does nitric acid become when it loses electrons? There are only no and no2, so let no be and no2 be ymol, and the equation x+y = sum can be listed.

x+3y= solution to get x=, y=, and finally no2= no=.

2) List the equations to solve, according to the equations.

Cu + 4 Hno3 (concentrated) = Cu (No3) 2 + 2 No2 + 2H2O3Cu + 8Hno3 (dilute) = 3 Cu (No3) 2 + 2 No + 4H2o According to the first equation, because No2 is then nitric acid is.

According to the second equation, nitric acid is because no is.

Since the volume is 140ml, C = per liter.

Landlord, please look at my hard typing, my answer!!

1) First of all, Cu must be oxidized to 2-valent Cu ions, and the valency increases by 2 and loses 2 electrons. CU totally. So cu loses electrons in total, and the electrons are n elements, n in Hno3 is +5 valence, no is +2 valence, no2 is +4 valence, total no2 + no=, let no2 = xmol, then no=( According to the previous electrons, we get:

x+( so x=, so no2= no=

2) If the first question is solved, the chemical equation column is solved. It is advisable to do it yourself.

The equation gives you:

The equation for the reaction of metallic copper with concentrated nitric acid and dilute nitric acid is as follows:

Cu + 4 Hno3 (Thick) = Cu (No3) 2 + 2 No2 + 2H2O3Cu + 8Hno3 (Diluted) = 3 Cu (No3) 2 + 2No + 4H2O Pure Handmade, I hope you can understand, and I hope so.

1) Let the volumes of NO and NO2 in the reaction be x, y (according to the conservation of electrons of gain and loss) x+y=, respectively

x/ ×1=(

The solution is x= l y= l

2) According to the fact that there is only one solute nano3 in the solution after the reaction, the amount and concentration of the substance of the original nitric acid is c

Conservation of nitrogen) then 10-3 V a + the concentration of orthonitric acid is: c= (10-3 va +3) From the initial and final state of the reaction, copper loses electrons in the reaction, hydrogen peroxide gains electrons in the reaction, and 30% of the mass of hydrogen peroxide is required to be m (according to the conservation of electrons gained).

Solution: m = g

Cu stands for conservation of electrons: Cu loses electrons.

No gets 3mol electrons and No2 gets 1mol electrons.

So a total of electrons are obtained, and there are 3n(no)+n(no2)= and n(no)+n(no2)= according to the conservation of elements.

The solution is n(no)=,n(no2)=

v(no)=,v(no2)=

Parse: N2O5(G) -N2O3 (G) O2(G)5 x x x

n2o3---n2o ＋ o2x－ x－ x－

x＋x－x＝3 （mol/l）

The equilibrium concentration of N2O5 is 2 and the conversion rate of N2O5 is = x 5 = 60%.

The equilibrium concentration of N2O is =x-1,6=

Therefore, the equilibrium concentration of DN2O is .

n(n2o3)=

The amount of N2O5 required to produce this part of N2O3 is, and the total amount of N2 produced at the same time N(O2)=

The amount of substances that produce N2O and O2 at the same time is .

Equation (1) + (2).

N2O5 (g) = N2O(G) + 2O2 (G) The amount of the substance that consumes N2O5 is, and the generation of N2O is.

The equilibrium concentration of N2O5 (

N2O5 conversion rate (

N2O equilibrium concentration.

Because the data given is the ,.. when the balance is balancedThe amount of N2O3 generated, the amount of oxygen generated (the first equation), and the amount of O2 (the second equation) is decomposed by N2O3, and the amount of oxygen matter is directly calculated), so the reaction generates oxygen according to the equation. The same goes for n2o3=.

So in the end, the remaining first step reaction generates xmoln2o3, so, o2=x+. Solution. x=, so the first step is to generate and the amount of N2O3 and the last substance is , so the reaction is made, so it is generated.

So d is correct. A is wrong because the original N2O5 = the last remainder. Not 3

b Answer conversion rate = actual reactant Initially it total number = ... c answer, because there is at the end, so the concentration should be, I don't know how to continue to ask. Thank you.

Let no nitrous oxide decompose x, nitrous oxide decomposes y, and the container volume is v then x+y=

x-y=so x=3v

y = so nitrous oxide is generated, ie.

D is correct, ABC can't be sure.

According to known conditions: the original concentration of N2O5 is: According to the formula, it can be known:

For each N2O generated, 2 O2 are generated, and the O2 concentration in the container is N2O*2+N2O3, that is, the N2O equilibrium concentration is (, the consumed N2O5 is N2O+N2O3=, the remaining N2O5=, and the conversion rate is =

1.There are three atoms, H, D, and T, each of which can form the diatomic molecules H2, D2, and T2, and the relationship between them is filled in on the following horizontal line.

Under standard conditions, the ratio of their elemental density is 1:2:3 and the ratio of density is the relative molecular mass ratio).

When each of the three elemental substances is 1mol, the ratio of their proton numbers is 1:1:1 and the ratio of neutron numbers is 0:1:2

When each of the three elemental elements is 1 gram, the ratio of their protons is 6:3:2 and the ratio of neutrons is 0:3:4

At the same temperature and pressure, the ratio of the number of electrons contained in 1L of various elemental gases is 1:1:1

2.If an element is composed of two isotopes, the ratio of its atomic numbers is 9:1, the first isotope has 10 neutrons and 10 protons in the nucleus, and the number of neutrons in the nucleus of the second isotope is 2 more than the number of neutrons in the nucleus of the first isotope, then the approximate relative atomic mass of the element is (10+10)*9 10 + 10+12)*1 10=

3.Write the symbol for a particle with the same number of protons and the same number of electrons as Na+: NH4+ (the same number of protons and electrons is the same if the number of charges is the same).

Write the chemical formula CH4 for a particle consisting of 5 atoms with 10 protons and 10 electrons

(1) CaCO3 = High Temperature = CAO+CO2

Cao + SO2 = High temperature = CaSO3

2caso3 + o2 = 2caso4

2) Choose D as the answer

A false. n is the most ** is +5 valence, and the hydrate corresponding to the oxide is nitric acid, which is a strong acid.

B false. Oxides have salt-forming oxides, not salt-forming oxides, and salt-forming oxides can react with strong alkalis to form oxygenates, which is also a characteristic reaction to judge whether an oxide belongs to salt-forming oxides.

For example, if CO2 reacts with NaOH to form the oxynate Na2CO3, then CO2 is a salt-forming oxide.

For example, CO and NaOH cannot form oxygenates under general conditions, and sodium formate can only be generated under very harsh conditions such as high temperature and high pressure, which is also an organic salt. Therefore, CO is considered to be a non-salt oxide.

The oxides formed by N, C and Si are: NO2, NO, N2O4, NO2, etc.; co2，co ； sio2

Among them, CO and NO are non-salt oxides, which are considered not to react with NaOH; SiO2 reacts with NaOH at room temperature to form NaSiO3, but the reaction rate is very slow.

C false. The stronger the non-metallic nature of the element, the greater the stability of the gaseous hydride formed.

Such as: halogen stability hf hcl hbr hi

n, c, si, non-metallic: c n si

From this, it can be concluded that the stability of the gaseous hydride formed is: NH3 CH4 SIH4

However, the spatial structure of CH4 is very special, and the spatial structure of CH4 is a regular tetrahedral type with 4 H atoms as the 4 vertices and C atoms as the center, which has a high degree of spatial symmetry, large C-H bond energy, and good CH4 stability.

The spatial structure of NH3 is a tetrahedral structure with N atom as the center, 3 H atoms and a lone pair of electrons of N as the 4 vertices, in which 4 atoms in the NH3 molecule form a trigonal pyramid. Due to the large repulsion of lone pairs, the Nh3 molecule and a lone pair of N's lone pairs do not form a tetrahedron, and the repulsion of n-H bonds by lone pairs of electrons decreases the N-H bond energy and NH3 stability.

The stability of CH4 is stronger than that of NH3.

Therefore, the stability of the gaseous hydrides formed is from large to small: CH4 NH3 SIH4

d correct. C in CH4, NH3, SIH4 is -4 valence; n manifest -3 valence; Si explicit -4 valence, all of which are reducible, and O2 can undergo redox reaction, that is, CH4, NH3, and SiH4 can all be burned in O2 and release a large amount of heat, so the heat of reaction is greater than 0

1.Because of the high temperature generated when coal is burned. Causes limestone to decompose into calcium oxide.

The reaction equation of this step is: CaCO3 = high temperature = Cao + CO2 and under high temperature conditions, calcium oxide will react with sulfur dioxide to form calcium sulfite. The reaction equation for this step is:

Cao + SO2 = High temperature = CaSO3Therefore, the amount of sulfur dioxide in the exhaust gas can be reduced.

2.The answer is C A counter-example of the answer is: The hydrate of the most ** oxide of n is HNO3 and nitric acid is a strong acid.

The counter-example of the answer b is: carbon monoxide, the oxide of c, is not a salt oxide and cannot react with a strong base d The answer is obviously wrong and does not need to be prompted again.

Hope the answer will help you.

Chemistry enthusiasts sincerely answer for you.

1.Limestone caCO3=high temperature=cao+CO2 cao+so2=caso3

2.Option A, false, Hno3 is a strong acid.

Option B, wrong, acidic oxides can react with bases, and such as CO, NO are not acidic oxides C options, right.

d option, false, the reaction is exothermic, the heat of reaction is less than 0