A junior high school chemistry problem. Chemistry Problems, Junior High School?

You can according to the law of the periodic table, each period, each main group (subgroup) from left to right oxidation increases, reducibility decreases, from top to bottom, oxidation weakens, reducibility increases, so you can judge that the relationship between those ions is good!

Common in junior high school.

OH- and: H+, NH4+, Mg2+, Al3+, Fe2+, Zn2+, Fe3+, Cu2+, AG+

cl- and ag+

SO42- and: Ba2+, AG+, Ca2+

CO32- and: NH4+, MG2+, Al3+, Fe2+, Zn2+, Fe3+, Cu2+, AG+, Ca2+, Ba2+

For details, please refer to the solubility table, where those marked with "no, micro" will precipitate, and those marked with "weak" will be difficult to ionize.

ag+ and cl-

BA2+ and SO42-

H+ and CO32-

Mg2+ and OH-

al3+ and oh-

Mn2+ and OH-

ZN2+ and OH-

Fe2+ and OH-

Fe3+ and OH-

Cu2+ and OH-

Ba2+ and CO32-

Ca2+ and CO32-

Mn2+ and CO32-

ZN2+ and CO32-

Fe2+ and CO32-

AG+ and CO32-

In fact, all it takes is to produce precipitation, gas or water......

The solubility of oxygen in the aqueous solution is very small, and after normal washing, it enters the pressure overflow from port A and is pumped out at port B. When the bottle is filled with water, the air pressure of the liquid column of pipe A in the bottle is level with the water surface, and the air intake from port B needs to maintain a constant positive pressure to keep the gas saturated so that the pure water can flow out of port A after washing (the bottle is emptied and collected, and the water may contain a small amount of oxygen loss). It can be understood as a constant pressure phenomenon similar to that at both ends of the U-shaped pipe.

This is the drainage method to collect oxygen, if the oxygen enters from the B port, with the input of the oxygen is not transported, the water in the bottle will be pressed out by the oxygen, and the water will be strung out from the thin tube flow at the A port, and finally the oxygen left in the bottle is oxygen. If oxygen is fed in from port A, it is equivalent to the oxygen needs to bubble from the bottom of the water to the void above the bottle, one is slow, and the other is that a small part of the oxygen will be dissolved in the water when it passes through the imitation of water, resulting in oxygen loss. Therefore, it is more reasonable to enter from exit B.

Because the gas collector cylinder is filled with water, if the oxygen mold enters from the A end, it will surface and then be discharged from B, so the hail will not be able to collect oxygen. Therefore, it can only enter from the B end, and the water is discharged through the drainage gas collection method to collect oxygen.

Hello, if the bottle is filled with water, then the drainage method is used to collect the oxygen of the Shen hood, the oxygen density is smaller than the water, the water is below, so the gas enters from the B port, and the water in the bottle is discharged from the A mouth by pressure, and the oxygen will be obtained from the bottle Min, I hope it will help you.

When there is water in the bottle, it is the drainage gas collection method from the B mouth, and the water can be discharged from the A mouth by the B inlet, so that the spring shed can collect the purer gas, if the gas enters from A, the water can not be discharged from the spring forest liquid, the gas will escape, and it will not be collected.

When the multi-functional bottle is filled with water, it is to use the drainage method to collect oxygen, because the density of oxygen is smaller than the water absolute lead front, so the oxygen will float above the water, from the short tube can make the oxygen float directly on the water surface, with the increase of oxygen The pressure in the bottle continues to increase, and the water will be pressed into the long tube to exclude the bottle. In this way, the purpose of collecting oxygen by the drainage method can be realized.

When there is water in the bottle, the gas is introduced from the B port, the density of the gas is smaller than that of the water, and it will stay above the water, and then the water is squeezed into the bottom, and then the water is discharged from the A pipe after the nucleus is opened, if the gas enters from A, it will float up, and then go out directly from B, which is equivalent to no collection.

Oxygen is insoluble in water, and entering from port B will discharge the water in the bottle from port A, so as to achieve the purpose of collecting oxygen. From the A mouth into the Qi Qi to the spring banquet mausoleum can drain the water in the bottle.

Oxygen enters from the B mouth, and floats on the water after entering, when the oxygen enters more and more, the pressure in the bottle will become greater and larger, through the action of pressure, the water will be discharged from the A mouth, and when the water is discharged, the remaining gas in the bottle is oxygen.

Because the water is heavier, it is always below, if you take in the air from port B, the water can be discharged from the relatively long A pipe, if you take in the air from A, the gas goes up, and the straight field sparrow regrets going out from the B mouth again, and the water can't be discharged.

Because the density of water is higher than that of oxygen, it is easy to fill the bottle from top to bottom when collecting oxygen, and drain the water.

Because oxygen is denser than water, it is necessary to drain it downward, so oxygen should enter through the pipe.

There is no brother to fight with water, it falls at the bottom of the bottle, and slowly reaches the B mouth.

There is water, and it is still from A, and it can only be envied by the auspicious code of B. Mouth B, float through the bottle, and slowly squeeze the water out from A until the lowest end of the A tube is level.

The liquid is filtered from port A to filter oxygen, and the liquid is entered from port B, which is filled with water, to collect oxygen.

1.Let the valency of x be n, and y be m because the volume ratio of H2 produced under the same banquet conditions is 4:3, so the mass ratio of H2 is also 4:

The chemical equations of X,Y and dilute hydrochloric acid are x+nhcl==xcln+n 2 h2 9 n x 4 x 4=9 n x=36 n y+mhcl==yclm=m 2 h2 8 m y 3 y 3=8 m y=24 m because of the equal mass x and y respectively.