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When using the drainage method, the density of the gas has nothing to do with the density of the air, it is necessary to enter and exit the gas directly, the gas directly enters the bottom of the gas collection cylinder and squeezes the water out of the short mouth, or the drainage method can only use a conduit, as long as the bottle mouth is immersed in water.

Air exhaust method, if the gas density is higher than the air density, the upward air discharge method is used, the bottle mouth is facing up, the long conduit is in, and the short is out. In this way, the gas is at the bottom of the bottle, and due to its high density, it will not run upward, and the air will be squeezed out.

When the density of the gas is higher than that of the air, there is no need to exhaust the air downward, because the gas will sink to the mouth of the bottle, so it will be discharged, and it will be difficult to collect.

I think it seems to be gas mass, right?

When using drainage, the quality of the gas has nothing to do with the long in and short out, as long as it does not react with the water.

With the air exhaust method, the mass of the gas is larger than the air, and the bottle mouth is upward, and it grows in: the mass is large, and it will sink, so the bottle mouth is selected upward, and in order to improve the purity of the gas, the lighter gas is lifted up from below, and if it is on the top, it will be pressed below, and the purity is reduced (this is my personal opinion).

When the mass of the gas is small, the bottle mouth is downward and grows in, and the reasons are probably as above

In the case of the drainage method, the air pipe is placed just close to the mouth of the bottle, and with the air exhaust method, it is necessary to distinguish between the upward and downward exhaust gases, and when the gas density is higher than the air density, it will be longer in and shorter, and when the gas density is higher than the air density, it will be shorter in and out.

The drainage method is all long-in, short-out. In the air exhaust method, the dense one will sink, so the one with a small density will be discharged upwards (up the bottle mouth); The same goes for the downward one.

It is very good to say that the relative molecular mass of NaHCO3 is 23 + 1 + 12 + 16 * 3 = 84, while the relative atomic mass of Na is 23, so the mass fraction of sodium in sodium bicarbonate is 23 84 * 100% = the equation in the question shows that N(NaHCO3): N(CaCO3)=1:1.

And N(CaCO3)=10g 100G mol=i.e., N(NaHCO3)=, so the mass of NaHCO3 is.

According to the second question and the question, there is sodium bicarbonate in the 9g sample, so its mass fraction is. It is obvious that less than 95%, and the mass fraction of sodium bicarbonate in the bag does not match the data on the bag.

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1.Silver 2Aluminum and copper.

3.Aluminum 4Iron, aluminum, zinc, titanium, silver.

5.Aluminum 6Iron.

7.Aluminum 8Silver.

9.Titanium 10Silver.

It is known that x, y, and z are three atomic number connected elements.

The relative acidity of the most ** oxide corresponding to the hydrate is: HxO4>H2YO4>H3ZO4

These two conditions can be known:

Non-metallic x y z

Then the stronger the non-metallic properties, the higher the hydride stability.

A is true B is false.

This is the order on the same cycle.

So the atomic radius gradually increases.

c False Correct.

c Using the limit method If all N02 is N02, 46g is 1mol, and N02 contains three atoms (one nitrogen atom and two oxygen atoms) which is 3Na

If all of them are N2O4, 46g is 3Na, and N2O4 contains six atoms (two nitrogen atoms and four oxygen atoms).

Therefore, no matter what ratio the two are mixed, 46g of the gas mixture will always contain an atomic number of 3 Na

There is also the chemical equation 2NO2===N2O4 which can be converted into each other.

D Fe powder and sufficient amount of water vapor reaction at high temperature should be 3Fe + 4H2O(G) === High temperature ===Fe3O4 + 4H2 According to the chemical equation calculation, it can be seen that the number of molecules generated should be D, so D is wrong.

Write out the electrode reaction:

The positive electrode Cu2++2E-==Cu

Negative zn-2e-==zn2+

Draw a diagram of the installation.

zn graphite |

--cuso4 solution |

The positive electrode Cu2++2E-==Cu

Negative zn-2e-==zn2+

The device diagram requires copper and zinc to be connected by wires and in the same electrolyte.

The positive electrode Cu2++2E-==Cu

Negative zn-2e-==zn2+

Draw a diagram of the installation.