9th grade chemistry questions should be fast, 9th grade chemistry questions, not difficult

It is hydrogen so that participates in the reflection of zn --h2

x=cu% = （ =35%

Let zn x g h2 y g

Zn + H2SO4 ***** ZnSO4 + H2 change.

x y we can get x+50=

65/x = 2/ y

It's good to solve it, and the next one will be sent to you a bit more.

3.It can be seen that the CO2 produced in the fourth and fifth minutes does not continue to increase.

According to the equation of reaction, mg(oh)2 reacts with hydrochloric acid and does not produce CO2, so we can tell that by the fourth minute, CaCO3 has fully reflected.

The amount of calcium carbonate can be calculated based on the amount of carbon dioxide produced.

caco3---co2

x=150g 66g

So caco3% = 150 200 x 100% = 75%The answer is not to know if it is right or not, you can hi me again.

1.A mixture of 10 g of zinc and copper is placed in a large beaker filled with sufficient dilute sulfuric acid, and after a complete reaction, the mass of the material in the beaker is reduced compared to before the reaction (zinc, copper, and dilute sulfuric acid). Find the mass fraction of copper in the original mixture.

Solution: According to the conservation of system mass, the gram of material reduced in the beaker is the mass of hydrogen that runs away, and the equation is used to calculate the mass of zinc (copper does not react with dilute sulfuric acid), and then divide the mass of zinc by the total mass of the original mixture of 10g, and multiply it by 100% to obtain the mass fraction of copper.

2.A certain mass of zinc and 50g of dilute sulfuric acid just completely react to generate zinc sulfate and hydrogen, and the mass of the solution after the reaction is weighed as, and the quality of the zinc reacted and the mass of hydrogen produced are obtained.

Let the mass of zinc be x grams, and the equation calculates that the mass of hydrogen running away is 2x 65 grams, and x+50= is obtained according to the conservation of mass, and x is solved

1.Copper and dilute sulfuric acid do not react, so only zinc and dilute sulfuric acid react, and the reduced mass is the mass of hydrogen produced, and the equation is calculated that zinc has, that is, copper has, 35%.

2.According to the equation 1molZN reacts with dilute sulfuric acid to generate 1mol of hydrogen, that is, 65g of zinc produces 2g of hydrogen, let the added zinc be xmol, and the zinc reacted by the equation can be obtained to generate hydrogen.

3.Excess hydrochloric acid means that hydrochloric acid will not be insufficient, so when the mass of the generated gas does not increase, it means that CaCO3 is completely reacted, and the gas mass is the same as when it is 5 minutes and 4 minutes, so it is completely reacted.

The mass fraction of CaCO3 is calculated by the column equation as 75%Zn+H2SO4=ZnSO4+H2

caco3+2hcl=cacl2+h2o+co2

1.What is reduced is the mass of hydrogen, and only hydrogen escapes, because hydrogen is more reactive than copper, so copper is not involved in the reaction.

zn+2hso4=h2+znso4

So zinc copper a%=35%.

2.If the reaction is complete, then the only amount that is reduced is hydrogen.

That is, the mass of zinc + the mass of dilute sulfuric acid - the mass of the hydrogen that escapes = the mass of the remaining solution after the reaction.

Set zinc 65xg

zn+2hso4=h2+znso4

65xg 2xg

So 65xg-2xg+50g=

x = So the reaction uses up zinc to produce hydrogen.

3.(1) Because the mass no longer changes and the hydrochloric acid is excessive, the reaction is complete.

2) Although magnesium hydroxide also reacts with hydrochloric acid, the product has no effect on the measurement results, so it is not considered.

Set calcium carbonate xg

caco3+2hcl=cacl2+h2o+co2xg 66g

100/xg=44/66g

x=150g

a%=150g/200g=75%

1.What is reduced is the mass of hydrogen, and only hydrogen escapes, because hydrogen is more reactive than copper, so copper is not involved in the reaction.

zn+2hso4=h2+znso4

So zinc copper a%=35%.

2.If the reaction is complete, then the only amount that is reduced is hydrogen.

That is, the mass of zinc + the mass of dilute sulfuric acid - the mass of the hydrogen that escapes = the mass of the remaining solution after the reaction.

Set zinc 65xg

zn+2hso4=h2+znso4

65xg 2xg

So 65xg-2xg+50g=

x = So the reaction uses up zinc to produce hydrogen.

3.(1) Because the mass no longer changes and the hydrochloric acid is excessive, the reaction is complete.

2) Although magnesium hydroxide also reacts with hydrochloric acid, the product has no effect on the measurement results, so it is not considered.

Set calcium carbonate xg

caco3+2hcl=cacl2+h2o+co2xg 66g

100/xg=44/66g

x=150g

a%=150g/200g=75%

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What do you mean? What about the topic? The problem?

1.The percentage content of copper sulfate is (the atomic weight of copper is calculated as 64) (16*4) (16*4+64+32)=64 160=40%, assuming that the mass of xg of copper sulfate (CuSO4) contains 32 grams of oxygen, there is x*40%=32

Get x = 80g

That is, copper sulfate (CuSO4) with a mass of 80g contains 32 grams of oxygen element 2The mass fraction of nitrogen in urea CO(NH2)2 (14*2) [12+16+(14+1*2)*2] = approximately equal) ammonium sulfate (NH4)2SO4 in the mass fraction of nitrogen (14*2) [(14+1*4)*2+32+16*4]=approximately equal) Assuming that the nitrogen content of y tons of ammonium sulfate is equal to the nitrogen content of 1 ton of urea, there is y*

We get y=approximately equal to)

That is, the nitrogen content of one ton of ammonium sulfate is equal to the nitrogen content of one ton of urea.

This question is very simple, and the identification is complete. Do it yourself, it's easy.

1 32g/(64/160)=80g

The mass fraction of nitrogen in 2 CO(NH2)2 28 60*100%=(NH4)2SO4 28 132*100%=1t*

The nitrogen content of one ton of ammonium sulfate is equal to the nitrogen content of 1 ton of urea.

This question is about the order of displacement reaction in chemical reactions, Ag ions preferentially replace with elemental iron (i.e. iron powder), when Ag ions are completely replaced, Fe will have a replacement reaction with Cu ions, and when Cu ions are replaced, excess iron powder will be left. As long as you understand the order of reactions, this problem will be solved.

1) The addition of dilute hydrochloric acid has gas generation, and it is obvious that H2 is generated, while Ag and Cu will not react with dilute hydrochloric acid, so there is residual iron powder in the filter residue (that is, there is a surplus of iron powder, or excessive iron powder).

Answer: The content in the filter residue is (Fe, Ag, Cu)), and the solute in the filtrate is (Fe(NO3)2).

2) There is a white precipitate added to dilute hydrochloric acid, which is obviously a precipitate formed by the reaction of chloride ions and silver ions, so it is concluded that the iron powder added is too small to completely replace the Ag ions, and of course it is impossible to react with Cu ions.

Answer: Then the solute must be contained in the filtrate is (Fe(NO3)2, Cu(NO3)2, AGNo3) and (AG) in the filter residue

It's very clear, I hope it helps you!!

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If the filter residue obtained by A is added to dilute hydrochloric acid, gas is generated, indicating that there is an excess of Fe powder, so the Ag+ and Cu2+ in the solution are replaced, so the filter residue contains Ag, Cu and Fe, and the solute in the filtrate is only Fe(NO3)2

The second question is to add the filtrate obtained by B to dilute hydrochloric acid, there is a white precipitate, indicating that the precipitated AgCl is generated, so there must be Ag+ in the filtrate, we know that when Fe reacts with Ag+ and Cu2+ at the same time, Fe preferentially reacts with Ag+, and only when there is no Ag+ in the solution can it continue to react with Cu2+, since there is Ag+ left in the solution, then there may only be AG elements replaced by Fe in the filter residue, and there are AgNO3 and Cu(NO3)2 in the filtrate. and Fe(NO3)2, which was later generated

Iron first replaces silver, then copper (silver is less mobile than copper, and the weaker one is replaced first).

1. When the filter residue is added to dilute hydrochloric acid, gas is generated, indicating that there is iron in the filter residue, and the copper and silver in the filtrate have all been replaced, so there is only ferrous nitrate in the filtrate.

2. The filtrate is added to dilute hydrochloric acid, and there is a white precipitate generated, indicating that there is still silver nitrate left, so the filtrate must contain ferrous nitrate and copper nitrate, and there is only silver in the filter residue.

1) If the filter residue obtained by A is added to dilute hydrochloric acid, gas is generated, and the content of the filter residue is (Ag, Cu, Fe), and the solute in the filtrate is (Fe(NO3)2).

The presence of gas means that there is an excess of iron powder, and all AgNO3 and Cu(NO3)2 react light, leaving only ferrous nitrate in the solution.

2) If the filtrate obtained by B is added to dilute hydrochloric acid, there is a white precipitate and generated, then the solute contained in the filtrate must be (silver nitrate, copper nitrate, ferrous nitrate) and (AG) in the filter residue

There is a precipitation, that is, the precipitation of silver chloride, which means that Agno3 has not completed the reaction, and the iron powder here is a small amount of iron powder, and the iron powder here is first reacted with silver nitrate, and the silver nitrate has not been rereacted, so copper nitrate will not participate in the reaction, so there are three solutes, and the filter residue only has a small amount of silver to participate in the reaction.

(1) Excess Fe The filter residue contains Fe and Cu, and the solute is Fe(NO3)3

2) Fe is not excessive, the solute is AgNO3, Fe(NO3)3 filter residue has AG

According to the order of activity of metals: iron, copper, silver.

So the order of the reaction is that iron reacts with silver nitrate first, and the reaction is completed. Iron is reacting with copper nitrate.

1.If there are bubbles, there must be iron, which means that there is too much iron, and the filtrate is only ferrous nitrate.

2。The presence of a white precipitate indicates that there are silver ions in it, indicating that the iron is a trace amount, so there must be silver nitrate in the filtrate, copper nitrate, ferrous nitrate, and only silver in the filter residue.

Analysis: This question involves 2 reactions, but the product precipitates Cu(OH)2 and the solute in the solution is only Na2SO4, both of which are produced by NaOH.

It is completely possible to solve the problem with the relational method.

Solution: Let the mass of Cu(OH)2 produced by the precipitate be m1:

2naoh---cu(oh)2

50g-25g)*16% m1

Find m1 = and let the mass of Na2SO4 be m2:

2NaOH---Na2SO480 14250g *16% m2 Find m2=

Mass of solution after reaction: law of conservation of mass).

Na2SO4% = (

To make the solution 10%, water can be added, and the added water quality is x: (the solute mass remains unchanged before and after the solution is diluted with water).

Find x = A: (1) The mass of the precipitate produced during the reaction.

2) The mass fraction of the solute in the solution obtained after the reaction is 16%.

3) The addition of water can make the mass fraction of the solute of the solution obtained after the reaction to 10%.

25g*16%=4g (50g-25g)*16%=4g2naoh+h2so4=na2so4+2h2o80 1424g x80/4g=142/x

x=cuso4+2naoh==cu(oh)2↓+na2so480 98 142

4g y z

80/4g=98/y=142/z

y=z=1) The mass of the precipitate produced during the reaction.

2) The mass fraction of the solute in the solution obtained after the reaction (

3) That is, the mass fraction of the solute of the solution obtained after the reaction can be 10% by adding water

1.Iron wire burns in oxygen.

Iron + oxygen ignited

Ferric oxide.

2.Sulfur burns in oxygen.

Sulfur + oxygen ignited

Sulfur dioxide. 3.Decompose.

Hydrogen peroxide. Oxygen is produced in solution.

Hydrogen peroxide Manganese dioxide.

Water + oxygen. 4.Respiration of living beings.

Glucose + oxygen enzyme carbon dioxide + water.

Between the arrows are the catalyst or reaction conditions.

Literal expressions are not used.

Trim, no gas or precipitation symbols, no equal signs.