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The priming gas mixture happens to be completely reactive--- indicating that H2:O2=2:1 should have a total mass of .

So h2= o2= can be obtained

2na202+4h2o=4naoh+o2xx=

2na+2h2o=2naoh+h2yy=

So mna2o2= mna=

So mna2o=

The mass ratio of sodium, sodium oxide, and sodium peroxide is: 31:39

First of all, the reaction is the combustion of hydrogen and oxygen, then n(h2):n(o2)= 2 :1 then it can be calculated that h2 = is then o2 =

Then sodium is.

Sodium peroxide is.

Then sodium oxide is.

na is na2o2 is.

Na2O is.

Because from the reaction with a sufficient amount of water to form a gas mixture, it can be seen that the primer gas mixture happens to be completely reacted, H2 and O2 are respectively.

Then according to the reaction ratio, it can be obtained.

3Cl2+8NH3=N2+6NH4Cl NH4Cl is a solid pressure ignored.

According to pv=nrt v r t unchanged so.

p-post-p=npre-npost=11:1

The pressure ratio in the vessel before and after the reaction is closest to 11:1

Al2O3+N2+3C===2AlN+3Co N2 is an oxidizing agent and C is a reducing agent.

For every 1 mol of ain, 3 mol of electrons need to be transferred.

The valency of nitrogen in AIN is -3

The molar mass of AIN is 41g mol

Detailed process of b:

From the original equation, we can see that n2 2aln

In this process, 6 mol of electrons are transferred, and 2mol of ALN is generated, so 3 mol of electrons need to be transferred for every 1 mol of AIN generated.

Al2O3+N2+3C===2AlN+3Co, in this reaction, the number of electrons transferred, taking n as an example, changes from N2 to Aln, the valency changes from 0 valence to -3 valence, and every n becomes Aln, and the number of transferred electrons is 3E.

Therefore, for every 1 mol of ain, 3 mol of electrons need to be transferred.

The transfer of electrons is the valence of valence.

al2o3+n2+3c===2aln+3co1mol 2*3e-

Caption: 1molAl2O3 reaction Then N2 either 0 valence becomes -3 valence or 3 valence transfers 3mol electrons, and there are 2 n decreasing to -3 valence n, which is 2 times 3 e-

So it is to transfer 6mol electrons.

Item A, the drop is reduced, N2 is used as an oxidant, and the valence state of each element in Al2O3 remains unchanged, which is not an oxidant. Mistake.

B term, 1molAln is generated, the valence state of N element decreases from 0 to minus three, 3mol electrons are obtained, C is increased from 0 to 2, due to the generation of 1molAln consumption, a total of electrons are lost, and one loss is a transfer, a total of 3mol electrons are transferred, correct.

c term, n valency should be -3, false.

D, the molar mass should be 41g mol, false. Pick B

Correct answer. Option A, the valency of N changes from 0 to -3, decreasing, N2 is the oxidizing agent, and the valency of aluminum oxide does not change.

Option b is correct.

The valency of Al has not changed.

1molaln is generated, n is generated to obtain three mol electrons, and 3mol electrons are transferred.

option c, which should be -3, shows the positive valence of the metal in the compound.

Option d The molar mass should be 41g mol, the unit is not right.

Choosing B, if 2molALN is generated in the equation, the valence of N is reduced by 6 valence, then 1mol is generated and 3mol electrons are transferred.

Item B: The valency of n in AIN is 3 valent. So for every 1 molain generated, 3 mol of electrons need to be transferred, and b is correct.

In the first stage, Na2CO3 + HCl = NAHC3 + NaCl In the second stage, NaHCO3 + HCl = NaCl + H2O + CO2

Let Na2CO3 be XMOL and NaHCO3 be YMOL106x+84Y=

x=1*x+y=1*(

The solution is x=y=, i.e., a 1:1 mixture.

Mass fraction, na2co3 = 106 190, nahco3 = 84 190 You can notice that the condition gives one more, and the original mass can not be given.

n (HCL stage 1) =

n(HCL stage 2) = (

na2co3+hcl=nahco3+naclx y

x= y=nahco3+hcl=nacl+h2o+co2z

z=n(nahco3)=

m(nahco3)=

m(na2co3)=

nahco3%=

na2co3%=

The answer "at least 12 atoms in the same plane" means that the single bond of the vinyl group attached to the benzene is turned. But have you ever thought about it, there are a total of 16 atoms in the system, 16-12 = 4, 12 coplanar, what about the other 4, if the other four are coplanar, isn't that at least 4? The answer is then clearly wrong, and the answer only considers the atoms including the benzene ring, and not the vinyl group alone.

But in reality there are more than 4 atoms coplanar. First of all, vinyl 5 atoms are coplanar. The same goes for the C on the benzene ring attached to the vinyl.

But considering that the single key in the middle is used as the axis of rotation. So this is all coplanar with the points on the axis, right? Then the c on the benzene ring that is in opposition to the vinyl group and the 1 h connected to it are also coplanar, a total of 8 atoms.

That is, a minimum of 8 and a maximum of 16.

Only the benzene ring and the 12 atoms directly attached to the benzene ring must be in the same plane.

The reason is that all single bonds can be rotated, and will be in the most stable spatial position, taking styrene as an example, the 6 carbon atoms and 5 hydrogen atoms on the benzene ring are in a plane, and the carbon atom in the c=c linked to the benzene ring is also in that plane, but the c-c single bond formed by the c link in the benzene ring and a c link in c=c can be rotated, so that at least 12 can be in the plane, and, in general, in the spatial structure of organic matter, The distance between the non-bonded atoms is as far away as possible to reduce the repulsion between the atoms and to achieve the most stable state.

And then it's very simple at most, all the atoms are coplanar, because the six carbons and five hydrogen on the benzene and the carbon attached to the benzene on the vinyl group are necessarily coplanar, but the atoms of ethylene are in the same plane, so after rotation, all the atoms can be coplanar.

The styrene molecule has a maximum of 16 atoms in the same plane and a minimum of 13.

At most, the atoms on benzene and ethylene are in one plane, a total of 15, and at least the carbon and ethylene atoms on benzene are in one plane, a total of 11.

A maximum of 16 atoms are in the same plane and a minimum of 8 atoms are in the same plane.

There are at least 12 of them, and all atoms where the benzene ring is located must be in one plane.

At most, there must be all 16, and the benzene ring and ethylene double bonds are in the same plane.

The crux of the matter is that both the benzene ring and the double bond are sp2 hybridized, both are planar conformations, the double bond cannot be rotated, and the single bond can rotate freely.